3.1.15 \(\int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))^3} \, dx\) [15]

Optimal. Leaf size=458 \[ \frac {2 a^3 \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} (a c-b d)^3 f}-\frac {2 d^3 (3 a c-2 b d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 (c-d)^{3/2} (c+d)^{3/2} (a c-b d)^2 f}-\frac {d^3 \left (c^2+2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 (c-d)^{5/2} (c+d)^{5/2} (a c-b d) f}-\frac {2 d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 \sqrt {c-d} \sqrt {c+d} (a c-b d)^3 f}-\frac {d^3 \sin (e+f x)}{2 c (a c-b d) \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}+\frac {3 d^4 \sin (e+f x)}{2 c (a c-b d) \left (c^2-d^2\right )^2 f (d+c \cos (e+f x))}+\frac {d^2 (3 a c-2 b d) \sin (e+f x)}{c (a c-b d)^2 \left (c^2-d^2\right ) f (d+c \cos (e+f x))} \]

[Out]

-2*d^3*(3*a*c-2*b*d)*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/c^2/(c-d)^(3/2)/(c+d)^(3/2)/(a*c-b*d)
^2/f-d^3*(c^2+2*d^2)*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/c^2/(c-d)^(5/2)/(c+d)^(5/2)/(a*c-b*d)
/f-1/2*d^3*sin(f*x+e)/c/(a*c-b*d)/(c^2-d^2)/f/(d+c*cos(f*x+e))^2+3/2*d^4*sin(f*x+e)/c/(a*c-b*d)/(c^2-d^2)^2/f/
(d+c*cos(f*x+e))+d^2*(3*a*c-2*b*d)*sin(f*x+e)/c/(a*c-b*d)^2/(c^2-d^2)/f/(d+c*cos(f*x+e))+2*a^3*arctan((a-b)^(1
/2)*tan(1/2*f*x+1/2*e)/(a+b)^(1/2))/(a*c-b*d)^3/f/(a-b)^(1/2)/(a+b)^(1/2)-2*d*(3*a^2*c^2-3*a*b*c*d+b^2*d^2)*ar
ctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/c^2/(a*c-b*d)^3/f/(c-d)^(1/2)/(c+d)^(1/2)

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Rubi [A]
time = 0.70, antiderivative size = 458, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2907, 3031, 2738, 211, 2743, 2833, 12, 214} \begin {gather*} \frac {2 a^3 \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{f \sqrt {a-b} \sqrt {a+b} (a c-b d)^3}-\frac {2 d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 f \sqrt {c-d} \sqrt {c+d} (a c-b d)^3}-\frac {2 d^3 (3 a c-2 b d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 f (c-d)^{3/2} (c+d)^{3/2} (a c-b d)^2}+\frac {d^2 (3 a c-2 b d) \sin (e+f x)}{c f \left (c^2-d^2\right ) (a c-b d)^2 (c \cos (e+f x)+d)}+\frac {3 d^4 \sin (e+f x)}{2 c f \left (c^2-d^2\right )^2 (a c-b d) (c \cos (e+f x)+d)}-\frac {d^3 \sin (e+f x)}{2 c f \left (c^2-d^2\right ) (a c-b d) (c \cos (e+f x)+d)^2}-\frac {d^3 \left (c^2+2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 f (c-d)^{5/2} (c+d)^{5/2} (a c-b d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Cos[e + f*x])*(c + d*Sec[e + f*x])^3),x]

[Out]

(2*a^3*ArcTan[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a*c - b*d)^3*f) - (2*d^3*
(3*a*c - 2*b*d)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(c^2*(c - d)^(3/2)*(c + d)^(3/2)*(a*c - b
*d)^2*f) - (d^3*(c^2 + 2*d^2)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(c^2*(c - d)^(5/2)*(c + d)^
(5/2)*(a*c - b*d)*f) - (2*d*(3*a^2*c^2 - 3*a*b*c*d + b^2*d^2)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c +
d]])/(c^2*Sqrt[c - d]*Sqrt[c + d]*(a*c - b*d)^3*f) - (d^3*Sin[e + f*x])/(2*c*(a*c - b*d)*(c^2 - d^2)*f*(d + c*
Cos[e + f*x])^2) + (3*d^4*Sin[e + f*x])/(2*c*(a*c - b*d)*(c^2 - d^2)^2*f*(d + c*Cos[e + f*x])) + (d^2*(3*a*c -
2*b*d)*Sin[e + f*x])/(c*(a*c - b*d)^2*(c^2 - d^2)*f*(d + c*Cos[e + f*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

Rule 2833

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b
^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 2907

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> In
t[(a + b*Sin[e + f*x])^m*((d + c*Sin[e + f*x])^n/Sin[e + f*x]^n), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Int
egerQ[n]

Rule 3031

Int[((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(g*sin[e + f*x])^p*(a + b*sin[e + f*x])^m*(c + d*sin[e + f*x])
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[b*c - a*d, 0] && (IntegersQ[m, n] || IntegersQ[m, p
] || IntegersQ[n, p]) && NeQ[p, 2]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \cos (e+f x)) (c+d \sec (e+f x))^3} \, dx &=\int \frac {\cos ^3(e+f x)}{(a+b \cos (e+f x)) (d+c \cos (e+f x))^3} \, dx\\ &=\int \left (\frac {a^3}{(a c-b d)^3 (a+b \cos (e+f x))}-\frac {d^3}{c^2 (a c-b d) (d+c \cos (e+f x))^3}+\frac {d^2 (3 a c-2 b d)}{c^2 (a c-b d)^2 (d+c \cos (e+f x))^2}-\frac {d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right )}{c^2 (a c-b d)^3 (d+c \cos (e+f x))}\right ) \, dx\\ &=\frac {a^3 \int \frac {1}{a+b \cos (e+f x)} \, dx}{(a c-b d)^3}+\frac {\left (d^2 (3 a c-2 b d)\right ) \int \frac {1}{(d+c \cos (e+f x))^2} \, dx}{c^2 (a c-b d)^2}-\frac {d^3 \int \frac {1}{(d+c \cos (e+f x))^3} \, dx}{c^2 (a c-b d)}-\frac {\left (d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right )\right ) \int \frac {1}{d+c \cos (e+f x)} \, dx}{c^2 (a c-b d)^3}\\ &=-\frac {d^3 \sin (e+f x)}{2 c (a c-b d) \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}+\frac {d^2 (3 a c-2 b d) \sin (e+f x)}{c (a c-b d)^2 \left (c^2-d^2\right ) f (d+c \cos (e+f x))}-\frac {\left (d^2 (3 a c-2 b d)\right ) \int \frac {d}{d+c \cos (e+f x)} \, dx}{c^2 (a c-b d)^2 \left (c^2-d^2\right )}-\frac {d^3 \int \frac {-2 d+c \cos (e+f x)}{(d+c \cos (e+f x))^2} \, dx}{2 c^2 (a c-b d) \left (c^2-d^2\right )}+\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{(a c-b d)^3 f}-\frac {\left (2 d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{c+d+(-c+d) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{c^2 (a c-b d)^3 f}\\ &=\frac {2 a^3 \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} (a c-b d)^3 f}-\frac {2 d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 \sqrt {c-d} \sqrt {c+d} (a c-b d)^3 f}-\frac {d^3 \sin (e+f x)}{2 c (a c-b d) \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}+\frac {3 d^4 \sin (e+f x)}{2 c (a c-b d) \left (c^2-d^2\right )^2 f (d+c \cos (e+f x))}+\frac {d^2 (3 a c-2 b d) \sin (e+f x)}{c (a c-b d)^2 \left (c^2-d^2\right ) f (d+c \cos (e+f x))}-\frac {d^3 \int \frac {c^2+2 d^2}{d+c \cos (e+f x)} \, dx}{2 c^2 (a c-b d) \left (c^2-d^2\right )^2}-\frac {\left (d^3 (3 a c-2 b d)\right ) \int \frac {1}{d+c \cos (e+f x)} \, dx}{c^2 (a c-b d)^2 \left (c^2-d^2\right )}\\ &=\frac {2 a^3 \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} (a c-b d)^3 f}-\frac {2 d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 \sqrt {c-d} \sqrt {c+d} (a c-b d)^3 f}-\frac {d^3 \sin (e+f x)}{2 c (a c-b d) \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}+\frac {3 d^4 \sin (e+f x)}{2 c (a c-b d) \left (c^2-d^2\right )^2 f (d+c \cos (e+f x))}+\frac {d^2 (3 a c-2 b d) \sin (e+f x)}{c (a c-b d)^2 \left (c^2-d^2\right ) f (d+c \cos (e+f x))}-\frac {\left (d^3 \left (c^2+2 d^2\right )\right ) \int \frac {1}{d+c \cos (e+f x)} \, dx}{2 c^2 (a c-b d) \left (c^2-d^2\right )^2}-\frac {\left (2 d^3 (3 a c-2 b d)\right ) \text {Subst}\left (\int \frac {1}{c+d+(-c+d) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{c^2 (a c-b d)^2 \left (c^2-d^2\right ) f}\\ &=\frac {2 a^3 \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} (a c-b d)^3 f}-\frac {2 d^3 (3 a c-2 b d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 (c-d)^{3/2} (c+d)^{3/2} (a c-b d)^2 f}-\frac {2 d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 \sqrt {c-d} \sqrt {c+d} (a c-b d)^3 f}-\frac {d^3 \sin (e+f x)}{2 c (a c-b d) \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}+\frac {3 d^4 \sin (e+f x)}{2 c (a c-b d) \left (c^2-d^2\right )^2 f (d+c \cos (e+f x))}+\frac {d^2 (3 a c-2 b d) \sin (e+f x)}{c (a c-b d)^2 \left (c^2-d^2\right ) f (d+c \cos (e+f x))}-\frac {\left (d^3 \left (c^2+2 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{c+d+(-c+d) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{c^2 (a c-b d) \left (c^2-d^2\right )^2 f}\\ &=\frac {2 a^3 \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} (a c-b d)^3 f}-\frac {2 d^3 (3 a c-2 b d) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 (c-d)^{3/2} (c+d)^{3/2} (a c-b d)^2 f}-\frac {d^3 \left (c^2+2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 (c-d)^{5/2} (c+d)^{5/2} (a c-b d) f}-\frac {2 d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^2 \sqrt {c-d} \sqrt {c+d} (a c-b d)^3 f}-\frac {d^3 \sin (e+f x)}{2 c (a c-b d) \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}+\frac {3 d^4 \sin (e+f x)}{2 c (a c-b d) \left (c^2-d^2\right )^2 f (d+c \cos (e+f x))}+\frac {d^2 (3 a c-2 b d) \sin (e+f x)}{c (a c-b d)^2 \left (c^2-d^2\right ) f (d+c \cos (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 2.57, size = 319, normalized size = 0.70 \begin {gather*} \frac {(d+c \cos (e+f x)) \sec ^3(e+f x) \left (-\frac {4 a^3 \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-a^2+b^2}}\right ) (d+c \cos (e+f x))^2}{\sqrt {-a^2+b^2}}+\frac {2 d \left (-6 a b c^3 d+b^2 d^2 \left (2 c^2+d^2\right )+a^2 \left (6 c^4-5 c^2 d^2+2 d^4\right )\right ) \tanh ^{-1}\left (\frac {(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) (d+c \cos (e+f x))^2}{\left (c^2-d^2\right )^{5/2}}-\frac {d^3 (a c-b d)^2 \sin (e+f x)}{c (c-d) (c+d)}+\frac {d^2 (a c-b d) \left (6 a c^3-4 b c^2 d-3 a c d^2+b d^3\right ) (d+c \cos (e+f x)) \sin (e+f x)}{c (c-d)^2 (c+d)^2}\right )}{2 (a c-b d)^3 f (c+d \sec (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*Cos[e + f*x])*(c + d*Sec[e + f*x])^3),x]

[Out]

((d + c*Cos[e + f*x])*Sec[e + f*x]^3*((-4*a^3*ArcTanh[((a - b)*Tan[(e + f*x)/2])/Sqrt[-a^2 + b^2]]*(d + c*Cos[
e + f*x])^2)/Sqrt[-a^2 + b^2] + (2*d*(-6*a*b*c^3*d + b^2*d^2*(2*c^2 + d^2) + a^2*(6*c^4 - 5*c^2*d^2 + 2*d^4))*
ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(d + c*Cos[e + f*x])^2)/(c^2 - d^2)^(5/2) - (d^3*(a*c - b
*d)^2*Sin[e + f*x])/(c*(c - d)*(c + d)) + (d^2*(a*c - b*d)*(6*a*c^3 - 4*b*c^2*d - 3*a*c*d^2 + b*d^3)*(d + c*Co
s[e + f*x])*Sin[e + f*x])/(c*(c - d)^2*(c + d)^2)))/(2*(a*c - b*d)^3*f*(c + d*Sec[e + f*x])^3)

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Maple [A]
time = 3.35, size = 412, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {\frac {2 a^{3} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a c -b d \right )^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 d \left (\frac {-\frac {\left (6 a^{2} c^{3}+a^{2} c^{2} d -2 a^{2} c \,d^{2}-10 a b \,c^{2} d -2 a b c \,d^{2}+2 a b \,d^{3}+4 b^{2} c \,d^{2}+b^{2} d^{3}\right ) d \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 \left (c -d \right ) \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (6 a^{2} c^{3}-a^{2} c^{2} d -2 a^{2} c \,d^{2}-10 a b \,c^{2} d +2 a b c \,d^{2}+2 a b \,d^{3}+4 b^{2} c \,d^{2}-b^{2} d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 \left (c +d \right ) \left (c -d \right )^{2}}}{\left (c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-d \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-c -d \right )^{2}}-\frac {\left (6 a^{2} c^{4}-5 a^{2} c^{2} d^{2}+2 a^{2} d^{4}-6 a b \,c^{3} d +2 b^{2} c^{2} d^{2}+b^{2} d^{4}\right ) \arctanh \left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{2 \left (c^{4}-2 d^{2} c^{2}+d^{4}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (a c -b d \right )^{3}}}{f}\)	\(412\)
default	\(\frac {\frac {2 a^{3} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a c -b d \right )^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 d \left (\frac {-\frac {\left (6 a^{2} c^{3}+a^{2} c^{2} d -2 a^{2} c \,d^{2}-10 a b \,c^{2} d -2 a b c \,d^{2}+2 a b \,d^{3}+4 b^{2} c \,d^{2}+b^{2} d^{3}\right ) d \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 \left (c -d \right ) \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (6 a^{2} c^{3}-a^{2} c^{2} d -2 a^{2} c \,d^{2}-10 a b \,c^{2} d +2 a b c \,d^{2}+2 a b \,d^{3}+4 b^{2} c \,d^{2}-b^{2} d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 \left (c +d \right ) \left (c -d \right )^{2}}}{\left (c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-d \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-c -d \right )^{2}}-\frac {\left (6 a^{2} c^{4}-5 a^{2} c^{2} d^{2}+2 a^{2} d^{4}-6 a b \,c^{3} d +2 b^{2} c^{2} d^{2}+b^{2} d^{4}\right ) \arctanh \left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{2 \left (c^{4}-2 d^{2} c^{2}+d^{4}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (a c -b d \right )^{3}}}{f}\)	\(412\)
risch	\(\text {Expression too large to display}\)	\(1658\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

1/f*(2*a^3/(a*c-b*d)^3/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*f*x+1/2*e)/((a-b)*(a+b))^(1/2))+2*d/(a*c-b*d)^
3*((-1/2*(6*a^2*c^3+a^2*c^2*d-2*a^2*c*d^2-10*a*b*c^2*d-2*a*b*c*d^2+2*a*b*d^3+4*b^2*c*d^2+b^2*d^3)*d/(c-d)/(c^2
+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3+1/2*d*(6*a^2*c^3-a^2*c^2*d-2*a^2*c*d^2-10*a*b*c^2*d+2*a*b*c*d^2+2*a*b*d^3+4*b
^2*c*d^2-b^2*d^3)/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e))/(c*tan(1/2*f*x+1/2*e)^2-d*tan(1/2*f*x+1/2*e)^2-c-d)^2-1/2*
(6*a^2*c^4-5*a^2*c^2*d^2+2*a^2*d^4-6*a*b*c^3*d+2*b^2*c^2*d^2+b^2*d^4)/(c^4-2*c^2*d^2+d^4)/((c+d)*(c-d))^(1/2)*
arctanh((c-d)*tan(1/2*f*x+1/2*e)/((c+d)*(c-d))^(1/2))))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \cos {\left (e + f x \right )}\right ) \left (c + d \sec {\left (e + f x \right )}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e))**3,x)

[Out]

Integral(1/((a + b*cos(e + f*x))*(c + d*sec(e + f*x))**3), x)

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Giac [A]
time = 0.74, size = 770, normalized size = 1.68 \begin {gather*} \frac {\frac {2 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{3}}{{\left (a^{3} c^{3} - 3 \, a^{2} b c^{2} d + 3 \, a b^{2} c d^{2} - b^{3} d^{3}\right )} \sqrt {a^{2} - b^{2}}} + \frac {{\left (6 \, a^{2} c^{4} d - 6 \, a b c^{3} d^{2} - 5 \, a^{2} c^{2} d^{3} + 2 \, b^{2} c^{2} d^{3} + 2 \, a^{2} d^{5} + b^{2} d^{5}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )}}{{\left (a^{3} c^{7} - 3 \, a^{2} b c^{6} d - 2 \, a^{3} c^{5} d^{2} + 3 \, a b^{2} c^{5} d^{2} + 6 \, a^{2} b c^{4} d^{3} - b^{3} c^{4} d^{3} + a^{3} c^{3} d^{4} - 6 \, a b^{2} c^{3} d^{4} - 3 \, a^{2} b c^{2} d^{5} + 2 \, b^{3} c^{2} d^{5} + 3 \, a b^{2} c d^{6} - b^{3} d^{7}\right )} \sqrt {-c^{2} + d^{2}}} - \frac {6 \, a c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, a c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, b c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, b c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + b d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 6 \, a c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 5 \, a c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, b c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, a c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, b c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, a d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (a^{2} c^{6} - 2 \, a b c^{5} d - 2 \, a^{2} c^{4} d^{2} + b^{2} c^{4} d^{2} + 4 \, a b c^{3} d^{3} + a^{2} c^{2} d^{4} - 2 \, b^{2} c^{2} d^{4} - 2 \, a b c d^{5} + b^{2} d^{6}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}^{2}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(f*x+e))/(c+d*sec(f*x+e))^3,x, algorithm="giac")

[Out]

(2*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*f*x + 1/2*e) - b*tan(1/2*f*x + 1/2*e))
/sqrt(a^2 - b^2)))*a^3/((a^3*c^3 - 3*a^2*b*c^2*d + 3*a*b^2*c*d^2 - b^3*d^3)*sqrt(a^2 - b^2)) + (6*a^2*c^4*d -
6*a*b*c^3*d^2 - 5*a^2*c^2*d^3 + 2*b^2*c^2*d^3 + 2*a^2*d^5 + b^2*d^5)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*c
- 2*d) + arctan((c*tan(1/2*f*x + 1/2*e) - d*tan(1/2*f*x + 1/2*e))/sqrt(-c^2 + d^2)))/((a^3*c^7 - 3*a^2*b*c^6*
d - 2*a^3*c^5*d^2 + 3*a*b^2*c^5*d^2 + 6*a^2*b*c^4*d^3 - b^3*c^4*d^3 + a^3*c^3*d^4 - 6*a*b^2*c^3*d^4 - 3*a^2*b*
c^2*d^5 + 2*b^3*c^2*d^5 + 3*a*b^2*c*d^6 - b^3*d^7)*sqrt(-c^2 + d^2)) - (6*a*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 - 5
*a*c^2*d^3*tan(1/2*f*x + 1/2*e)^3 - 4*b*c^2*d^3*tan(1/2*f*x + 1/2*e)^3 - 3*a*c*d^4*tan(1/2*f*x + 1/2*e)^3 + 3*
b*c*d^4*tan(1/2*f*x + 1/2*e)^3 + 2*a*d^5*tan(1/2*f*x + 1/2*e)^3 + b*d^5*tan(1/2*f*x + 1/2*e)^3 - 6*a*c^3*d^2*t
an(1/2*f*x + 1/2*e) - 5*a*c^2*d^3*tan(1/2*f*x + 1/2*e) + 4*b*c^2*d^3*tan(1/2*f*x + 1/2*e) + 3*a*c*d^4*tan(1/2*
f*x + 1/2*e) + 3*b*c*d^4*tan(1/2*f*x + 1/2*e) + 2*a*d^5*tan(1/2*f*x + 1/2*e) - b*d^5*tan(1/2*f*x + 1/2*e))/((a
^2*c^6 - 2*a*b*c^5*d - 2*a^2*c^4*d^2 + b^2*c^4*d^2 + 4*a*b*c^3*d^3 + a^2*c^2*d^4 - 2*b^2*c^2*d^4 - 2*a*b*c*d^5
+ b^2*d^6)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/2*e)^2 - c - d)^2))/f

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Mupad [B]
time = 20.25, size = 2500, normalized size = 5.46 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c + d/cos(e + f*x))^3*(a + b*cos(e + f*x))),x)

[Out]

((tan(e/2 + (f*x)/2)^3*(2*a*d^4 + b*d^4 - 6*a*c^2*d^2 - a*c*d^3 + 4*b*c*d^3))/((c + d)^2*(a^2*c^3 - b^2*d^3 -
a^2*c^2*d + b^2*c*d^2 + 2*a*b*c*d^2 - 2*a*b*c^2*d)) - (tan(e/2 + (f*x)/2)*(2*a*d^4 - b*d^4 - 6*a*c^2*d^2 + a*c
*d^3 + 4*b*c*d^3))/((c + d)*(a^2*c^4 + b^2*d^4 - 2*a^2*c^3*d - 2*b^2*c*d^3 + a^2*c^2*d^2 + b^2*c^2*d^2 - 2*a*b
*c*d^3 - 2*a*b*c^3*d + 4*a*b*c^2*d^2)))/(f*(2*c*d - tan(e/2 + (f*x)/2)^2*(2*c^2 - 2*d^2) + tan(e/2 + (f*x)/2)^
4*(c^2 - 2*c*d + d^2) + c^2 + d^2)) + (a^3*atan(((a^3*(b^2 - a^2)^(1/2)*((8*tan(e/2 + (f*x)/2)*(b^7*d^10 - 8*a
^7*d^10 - 4*a^7*c^10 + 4*a^6*b*c^10 - 3*a*b^6*d^10 + 16*a^6*b*d^10 + 8*a^7*c*d^9 + 8*a^7*c^9*d + 7*a^2*b^5*d^1
0 - 13*a^3*b^4*d^10 + 16*a^4*b^3*d^10 - 16*a^5*b^2*d^10 + 32*a^7*c^2*d^8 - 32*a^7*c^3*d^7 - 57*a^7*c^4*d^6 + 4
8*a^7*c^5*d^5 + 52*a^7*c^6*d^4 - 32*a^7*c^7*d^3 - 24*a^7*c^8*d^2 + 4*b^7*c^2*d^8 + 4*b^7*c^4*d^6 - 12*a*b^6*c^
2*d^8 - 12*a*b^6*c^3*d^7 - 12*a*b^6*c^4*d^6 - 24*a*b^6*c^5*d^5 - 72*a^6*b*c^2*d^8 + 56*a^6*b*c^3*d^7 + 155*a^6
*b*c^4*d^6 - 108*a^6*b*c^5*d^5 - 172*a^6*b*c^6*d^4 + 104*a^6*b*c^7*d^3 + 96*a^6*b*c^8*d^2 + 10*a^2*b^5*c^2*d^8
+ 36*a^2*b^5*c^3*d^7 + 4*a^2*b^5*c^4*d^6 + 72*a^2*b^5*c^5*d^5 + 60*a^2*b^5*c^6*d^4 + 2*a^3*b^4*c^2*d^8 - 60*a
^3*b^4*c^3*d^7 + 20*a^3*b^4*c^4*d^6 - 12*a^3*b^4*c^5*d^5 - 180*a^3*b^4*c^6*d^4 - 72*a^3*b^4*c^7*d^3 - 26*a^4*b
^3*c^2*d^8 + 84*a^4*b^3*c^3*d^7 + 25*a^4*b^3*c^4*d^6 - 156*a^4*b^3*c^5*d^5 + 120*a^4*b^3*c^6*d^4 + 216*a^4*b^3
*c^7*d^3 + 36*a^4*b^3*c^8*d^2 + 62*a^5*b^2*c^2*d^8 - 72*a^5*b^2*c^3*d^7 - 139*a^5*b^2*c^4*d^6 + 180*a^5*b^2*c^
5*d^5 + 120*a^5*b^2*c^6*d^4 - 216*a^5*b^2*c^7*d^3 - 108*a^5*b^2*c^8*d^2 - 8*a^6*b*c*d^9 - 8*a^6*b*c^9*d))/(a^4
*c^11 - b^4*d^11 + a^4*c^10*d - b^4*c*d^10 - a^4*c^4*d^7 - a^4*c^5*d^6 + 3*a^4*c^6*d^5 + 3*a^4*c^7*d^4 - 3*a^4
*c^8*d^3 - 3*a^4*c^9*d^2 + 3*b^4*c^2*d^9 + 3*b^4*c^3*d^8 - 3*b^4*c^4*d^7 - 3*b^4*c^5*d^6 + b^4*c^6*d^5 + b^4*c
^7*d^4 + 4*a*b^3*c^2*d^9 - 12*a*b^3*c^3*d^8 - 12*a*b^3*c^4*d^7 + 12*a*b^3*c^5*d^6 + 12*a*b^3*c^6*d^5 - 4*a*b^3
*c^7*d^4 - 4*a*b^3*c^8*d^3 + 4*a^3*b*c^3*d^8 + 4*a^3*b*c^4*d^7 - 12*a^3*b*c^5*d^6 - 12*a^3*b*c^6*d^5 + 12*a^3*
b*c^7*d^4 + 12*a^3*b*c^8*d^3 - 4*a^3*b*c^9*d^2 - 6*a^2*b^2*c^2*d^9 - 6*a^2*b^2*c^3*d^8 + 18*a^2*b^2*c^4*d^7 +
18*a^2*b^2*c^5*d^6 - 18*a^2*b^2*c^6*d^5 - 18*a^2*b^2*c^7*d^4 + 6*a^2*b^2*c^8*d^3 + 6*a^2*b^2*c^9*d^2 + 4*a*b^3
*c*d^10 - 4*a^3*b*c^10*d) + (a^3*(b^2 - a^2)^(1/2)*((8*(2*b^10*d^15 - 4*a^10*c^15 + 8*a^9*b*c^15 - 2*a*b^9*d^1
5 + 12*a^10*c^14*d - 2*b^10*c*d^14 - 4*a^8*b^2*c^15 + 2*a^2*b^8*d^15 - 6*a^3*b^7*d^15 + 4*a^4*b^6*d^15 + 4*a^1
0*c^6*d^9 - 2*a^10*c^7*d^8 - 18*a^10*c^8*d^7 + 4*a^10*c^9*d^6 + 36*a^10*c^10*d^5 - 6*a^10*c^11*d^4 - 34*a^10*c
^12*d^3 + 8*a^10*c^13*d^2 - 6*b^10*c^4*d^11 + 6*b^10*c^5*d^10 + 4*b^10*c^6*d^9 - 4*b^10*c^7*d^8 + 10*a*b^9*c^2
*d^13 - 12*a*b^9*c^3*d^12 + 18*a*b^9*c^4*d^11 + 48*a*b^9*c^5*d^10 - 58*a*b^9*c^6*d^9 - 28*a*b^9*c^7*d^8 + 32*a
*b^9*c^8*d^7 + 8*a^2*b^8*c*d^14 - 8*a^3*b^7*c*d^14 + 34*a^4*b^6*c*d^14 - 24*a^5*b^5*c*d^14 + 32*a^7*b^3*c^14*d
- 52*a^8*b^2*c^14*d - 24*a^9*b*c^5*d^10 + 6*a^9*b*c^6*d^9 + 112*a^9*b*c^7*d^8 + 10*a^9*b*c^8*d^7 - 236*a^9*b*
c^9*d^6 - 30*a^9*b*c^10*d^5 + 240*a^9*b*c^11*d^4 + 6*a^9*b*c^12*d^3 - 100*a^9*b*c^13*d^2 - 8*a^2*b^8*c^2*d^13
+ 10*a^2*b^8*c^3*d^12 + 90*a^2*b^8*c^4*d^11 - 156*a^2*b^8*c^5*d^10 - 164*a^2*b^8*c^6*d^9 + 250*a^2*b^8*c^7*d^8
+ 80*a^2*b^8*c^8*d^7 - 112*a^2*b^8*c^9*d^6 + 28*a^3*b^7*c^2*d^13 + 84*a^3*b^7*c^3*d^12 - 224*a^3*b^7*c^4*d^11
- 252*a^3*b^7*c^5*d^10 + 612*a^3*b^7*c^6*d^9 + 284*a^3*b^7*c^7*d^8 - 634*a^3*b^7*c^8*d^7 - 108*a^3*b^7*c^9*d^
6 + 224*a^3*b^7*c^10*d^5 - 12*a^4*b^6*c^2*d^13 - 220*a^4*b^6*c^3*d^12 - 104*a^4*b^6*c^4*d^11 + 820*a^4*b^6*c^5
*d^10 + 260*a^4*b^6*c^6*d^9 - 1396*a^4*b^6*c^7*d^8 - 180*a^4*b^6*c^8*d^7 + 1042*a^4*b^6*c^9*d^6 + 32*a^4*b^6*c
^10*d^5 - 280*a^4*b^6*c^11*d^4 - 78*a^5*b^5*c^2*d^13 + 128*a^5*b^5*c^3*d^12 + 536*a^5*b^5*c^4*d^11 - 188*a^5*b
^5*c^5*d^10 - 1532*a^5*b^5*c^6*d^9 + 204*a^5*b^5*c^7*d^8 + 1992*a^5*b^5*c^8*d^7 - 236*a^5*b^5*c^9*d^6 - 1142*a
^5*b^5*c^10*d^5 + 116*a^5*b^5*c^11*d^4 + 224*a^5*b^5*c^12*d^3 + 60*a^6*b^4*c^2*d^13 + 90*a^6*b^4*c^3*d^12 - 32
0*a^6*b^4*c^4*d^11 - 668*a^6*b^4*c^5*d^10 + 708*a^6*b^4*c^6*d^9 + 1660*a^6*b^4*c^7*d^8 - 888*a^6*b^4*c^8*d^7 -
1788*a^6*b^4*c^9*d^6 + 632*a^6*b^4*c^10*d^5 + 818*a^6*b^4*c^11*d^4 - 192*a^6*b^4*c^12*d^3 - 112*a^6*b^4*c^13*
d^2 - 80*a^7*b^3*c^3*d^12 - 50*a^7*b^3*c^4*d^11 + 408*a^7*b^3*c^5*d^10 + 452*a^7*b^3*c^6*d^9 - 932*a^7*b^3*c^7
*d^8 - 1040*a^7*b^3*c^8*d^7 + 1100*a^7*b^3*c^9*d^6 + 956*a^7*b^3*c^10*d^5 - 636*a^7*b^3*c^11*d^4 - 350*a^7*b^3
*c^12*d^3 + 140*a^7*b^3*c^13*d^2 + 60*a^8*b^2*c^4*d^11 + 6*a^8*b^2*c^5*d^10 - 292*a^8*b^2*c^6*d^9 - 148*a^8*b^
2*c^7*d^8 + 646*a^8*b^2*c^8*d^7 + 334*a^8*b^2*c^9*d^6 - 708*a^8*b^2*c^10*d^5 - 252*a^8*b^2*c^11*d^4 + 346*a^8*
b^2*c^12*d^3 + 64*a^8*b^2*c^13*d^2 - 8*a*b^9*c*d^14 + 8*a^9*b*c^14*d))/(a^6*c^13 - b^6*d^13 + a^6*c^12*d - b^6
*c*d^12 - a^6*c^6*d^7 - a^6*c^7*d^6 + 3*a^6*c^8*d^5 + 3*a^6*c^9*d^4 - 3*a^6*c^10*d^3 - 3*a^6*c^11*d^2 + 3*b^6*
c^2*d^11 + 3*b^6*c^3*d^10 - 3*b^6*c^4*d^9 - 3*b...

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