3.477 \(\int \frac{\sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=305 \[ -\frac{b^3 \left (-29 a^2 b^2+20 a^4+12 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{3 b \left (-7 a^2 b^2+2 a^4+4 b^4\right ) \tan (c+d x)}{2 a^4 d \left (a^2-b^2\right )^2}+\frac{\left (a^2+12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^5 d}+\frac{\left (-10 a^2 b^2+a^4+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}+\frac{b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]

[Out]

-((b^3*(20*a^4 - 29*a^2*b^2 + 12*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*(a - b)^(5/2)*(
a + b)^(5/2)*d)) + ((a^2 + 12*b^2)*ArcTanh[Sin[c + d*x]])/(2*a^5*d) - (3*b*(2*a^4 - 7*a^2*b^2 + 4*b^4)*Tan[c +
d*x])/(2*a^4*(a^2 - b^2)^2*d) + ((a^4 - 10*a^2*b^2 + 6*b^4)*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*(a^2 - b^2)^2*d
) + (b^2*Sec[c + d*x]*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + (b^2*(7*a^2 - 4*b^2)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.0777, antiderivative size = 305, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2802, 3055, 3001, 3770, 2659, 205} \[ -\frac{b^3 \left (-29 a^2 b^2+20 a^4+12 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{3 b \left (-7 a^2 b^2+2 a^4+4 b^4\right ) \tan (c+d x)}{2 a^4 d \left (a^2-b^2\right )^2}+\frac{\left (a^2+12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^5 d}+\frac{\left (-10 a^2 b^2+a^4+6 b^4\right ) \tan (c+d x) \sec (c+d x)}{2 a^3 d \left (a^2-b^2\right )^2}+\frac{b^2 \left (7 a^2-4 b^2\right ) \tan (c+d x) \sec (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}+\frac{b^2 \tan (c+d x) \sec (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^3*(20*a^4 - 29*a^2*b^2 + 12*b^4)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^5*(a - b)^(5/2)*(
a + b)^(5/2)*d)) + ((a^2 + 12*b^2)*ArcTanh[Sin[c + d*x]])/(2*a^5*d) - (3*b*(2*a^4 - 7*a^2*b^2 + 4*b^4)*Tan[c +
d*x])/(2*a^4*(a^2 - b^2)^2*d) + ((a^4 - 10*a^2*b^2 + 6*b^4)*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*(a^2 - b^2)^2*d
) + (b^2*Sec[c + d*x]*Tan[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + (b^2*(7*a^2 - 4*b^2)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Cos[c + d*x]))

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2659

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 205

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

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x)}{(a+b \cos (c+d x))^3} \, dx &=\frac{b^2 \sec (c+d x) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{\int \frac{\left (2 \left (a^2-2 b^2\right )-2 a b \cos (c+d x)+3 b^2 \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{b^2 \sec (c+d x) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b^2 \left (7 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{\left (2 \left (a^4-10 a^2 b^2+6 b^4\right )-a b \left (4 a^2-b^2\right ) \cos (c+d x)+2 b^2 \left (7 a^2-4 b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{\left (a^4-10 a^2 b^2+6 b^4\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b^2 \sec (c+d x) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b^2 \left (7 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-6 b \left (2 a^4-7 a^2 b^2+4 b^4\right )+2 a \left (a^4+4 a^2 b^2-2 b^4\right ) \cos (c+d x)+2 b \left (a^4-10 a^2 b^2+6 b^4\right ) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{4 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac{3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{2 a^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^4-10 a^2 b^2+6 b^4\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b^2 \sec (c+d x) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b^2 \left (7 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\int \frac{\left (2 \left (a^2-b^2\right )^2 \left (a^2+12 b^2\right )+2 a b \left (a^4-10 a^2 b^2+6 b^4\right ) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{4 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{2 a^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^4-10 a^2 b^2+6 b^4\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b^2 \sec (c+d x) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b^2 \left (7 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac{\left (a^2+12 b^2\right ) \int \sec (c+d x) \, dx}{2 a^5}-\frac{\left (b^3 \left (20 a^4-29 a^2 b^2+12 b^4\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{2 a^5 \left (a^2-b^2\right )^2}\\ &=\frac{\left (a^2+12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^5 d}-\frac{3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{2 a^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^4-10 a^2 b^2+6 b^4\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b^2 \sec (c+d x) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b^2 \left (7 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}-\frac{\left (b^3 \left (20 a^4-29 a^2 b^2+12 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^5 \left (a^2-b^2\right )^2 d}\\ &=-\frac{b^3 \left (20 a^4-29 a^2 b^2+12 b^4\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^5 (a-b)^{5/2} (a+b)^{5/2} d}+\frac{\left (a^2+12 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 a^5 d}-\frac{3 b \left (2 a^4-7 a^2 b^2+4 b^4\right ) \tan (c+d x)}{2 a^4 \left (a^2-b^2\right )^2 d}+\frac{\left (a^4-10 a^2 b^2+6 b^4\right ) \sec (c+d x) \tan (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{b^2 \sec (c+d x) \tan (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}+\frac{b^2 \left (7 a^2-4 b^2\right ) \sec (c+d x) \tan (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.19163, size = 427, normalized size = 1.4 \[ \frac{b^4 \sin (c+d x)}{2 a^3 d (a-b) (a+b) (a+b \cos (c+d x))^2}+\frac{3 \left (3 a^2 b^4 \sin (c+d x)-2 b^6 \sin (c+d x)\right )}{2 a^4 d (a-b)^2 (a+b)^2 (a+b \cos (c+d x))}+\frac{b^3 \left (-29 a^2 b^2+20 a^4+12 b^4\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{a^5 d \left (a^2-b^2\right )^2 \sqrt{b^2-a^2}}+\frac{\left (-a^2-12 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^5 d}+\frac{\left (a^2+12 b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^5 d}-\frac{3 b \sin \left (\frac{1}{2} (c+d x)\right )}{a^4 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{3 b \sin \left (\frac{1}{2} (c+d x)\right )}{a^4 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{1}{4 a^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{1}{4 a^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^3*(20*a^4 - 29*a^2*b^2 + 12*b^4)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(a^5*(a^2 - b^2)^2*S
qrt[-a^2 + b^2]*d) + ((-a^2 - 12*b^2)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(2*a^5*d) + ((a^2 + 12*b^2)*Lo
g[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(2*a^5*d) + 1/(4*a^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - (3*b
*Sin[(c + d*x)/2])/(a^4*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - 1/(4*a^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2])^2) - (3*b*Sin[(c + d*x)/2])/(a^4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (b^4*Sin[c + d*x])/(2*a^3*(a
- b)*(a + b)*d*(a + b*Cos[c + d*x])^2) + (3*(3*a^2*b^4*Sin[c + d*x] - 2*b^6*Sin[c + d*x]))/(2*a^4*(a - b)^2*(
a + b)^2*d*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Maple [B]  time = 0.133, size = 845, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+b*cos(d*x+c))^3,x)

[Out]

1/2/d/a^3/(tan(1/2*d*x+1/2*c)-1)^2+1/2/d/a^3/(tan(1/2*d*x+1/2*c)-1)+3/d/a^4/(tan(1/2*d*x+1/2*c)-1)*b-1/2/d/a^3
*ln(tan(1/2*d*x+1/2*c)-1)-6/d/a^5*ln(tan(1/2*d*x+1/2*c)-1)*b^2-1/2/d/a^3/(tan(1/2*d*x+1/2*c)+1)^2+1/2/d/a^3/(t
an(1/2*d*x+1/2*c)+1)+3/d/a^4/(tan(1/2*d*x+1/2*c)+1)*b+1/2/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)+6/d/a^5*ln(tan(1/2*d*
x+1/2*c)+1)*b^2+10/d*b^4/a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1
/2*d*x+1/2*c)^3+1/d*b^5/a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/
2*d*x+1/2*c)^3-6/d*b^6/a^4/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2
*d*x+1/2*c)^3+10/d*b^4/a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2
*d*x+1/2*c)-1/d*b^5/a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*
x+1/2*c)-6/d*b^6/a^4/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b+a+b)^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1
/2*c)-20/d*b^3/a/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))+
29/d*b^5/a^3/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-12/d
*b^7/a^5/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 16.4117, size = 3380, normalized size = 11.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(((20*a^4*b^5 - 29*a^2*b^7 + 12*b^9)*cos(d*x + c)^4 + 2*(20*a^5*b^4 - 29*a^3*b^6 + 12*a*b^8)*cos(d*x + c
)^3 + (20*a^6*b^3 - 29*a^4*b^5 + 12*a^2*b^7)*cos(d*x + c)^2)*sqrt(-a^2 + b^2)*log((2*a*b*cos(d*x + c) + (2*a^2
- b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)
^2 + 2*a*b*cos(d*x + c) + a^2)) - ((a^8*b^2 + 9*a^6*b^4 - 33*a^4*b^6 + 35*a^2*b^8 - 12*b^10)*cos(d*x + c)^4 +
2*(a^9*b + 9*a^7*b^3 - 33*a^5*b^5 + 35*a^3*b^7 - 12*a*b^9)*cos(d*x + c)^3 + (a^10 + 9*a^8*b^2 - 33*a^6*b^4 + 3
5*a^4*b^6 - 12*a^2*b^8)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((a^8*b^2 + 9*a^6*b^4 - 33*a^4*b^6 + 35*a^2*b^
8 - 12*b^10)*cos(d*x + c)^4 + 2*(a^9*b + 9*a^7*b^3 - 33*a^5*b^5 + 35*a^3*b^7 - 12*a*b^9)*cos(d*x + c)^3 + (a^1
0 + 9*a^8*b^2 - 33*a^6*b^4 + 35*a^4*b^6 - 12*a^2*b^8)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(a^10 - 3*a^8
*b^2 + 3*a^6*b^4 - a^4*b^6 - 3*(2*a^7*b^3 - 9*a^5*b^5 + 11*a^3*b^7 - 4*a*b^9)*cos(d*x + c)^3 - (11*a^8*b^2 - 4
3*a^6*b^4 + 50*a^4*b^6 - 18*a^2*b^8)*cos(d*x + c)^2 - 4*(a^9*b - 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*cos(d*x + c)
)*sin(d*x + c))/((a^11*b^2 - 3*a^9*b^4 + 3*a^7*b^6 - a^5*b^8)*d*cos(d*x + c)^4 + 2*(a^12*b - 3*a^10*b^3 + 3*a^
8*b^5 - a^6*b^7)*d*cos(d*x + c)^3 + (a^13 - 3*a^11*b^2 + 3*a^9*b^4 - a^7*b^6)*d*cos(d*x + c)^2), -1/4*(2*((20*
a^4*b^5 - 29*a^2*b^7 + 12*b^9)*cos(d*x + c)^4 + 2*(20*a^5*b^4 - 29*a^3*b^6 + 12*a*b^8)*cos(d*x + c)^3 + (20*a^
6*b^3 - 29*a^4*b^5 + 12*a^2*b^7)*cos(d*x + c)^2)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)
*sin(d*x + c))) - ((a^8*b^2 + 9*a^6*b^4 - 33*a^4*b^6 + 35*a^2*b^8 - 12*b^10)*cos(d*x + c)^4 + 2*(a^9*b + 9*a^7
*b^3 - 33*a^5*b^5 + 35*a^3*b^7 - 12*a*b^9)*cos(d*x + c)^3 + (a^10 + 9*a^8*b^2 - 33*a^6*b^4 + 35*a^4*b^6 - 12*a
^2*b^8)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((a^8*b^2 + 9*a^6*b^4 - 33*a^4*b^6 + 35*a^2*b^8 - 12*b^10)*cos
(d*x + c)^4 + 2*(a^9*b + 9*a^7*b^3 - 33*a^5*b^5 + 35*a^3*b^7 - 12*a*b^9)*cos(d*x + c)^3 + (a^10 + 9*a^8*b^2 -
33*a^6*b^4 + 35*a^4*b^6 - 12*a^2*b^8)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(a^10 - 3*a^8*b^2 + 3*a^6*b^4
- a^4*b^6 - 3*(2*a^7*b^3 - 9*a^5*b^5 + 11*a^3*b^7 - 4*a*b^9)*cos(d*x + c)^3 - (11*a^8*b^2 - 43*a^6*b^4 + 50*a
^4*b^6 - 18*a^2*b^8)*cos(d*x + c)^2 - 4*(a^9*b - 3*a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*cos(d*x + c))*sin(d*x + c))/
((a^11*b^2 - 3*a^9*b^4 + 3*a^7*b^6 - a^5*b^8)*d*cos(d*x + c)^4 + 2*(a^12*b - 3*a^10*b^3 + 3*a^8*b^5 - a^6*b^7)
*d*cos(d*x + c)^3 + (a^13 - 3*a^11*b^2 + 3*a^9*b^4 - a^7*b^6)*d*cos(d*x + c)^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.37061, size = 1081, normalized size = 3.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(20*a^4*b^3 - 29*a^2*b^5 + 12*b^7)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1
/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/((a^9 - 2*a^7*b^2 + a^5*b^4)*sqrt(a^2 - b^2)) + 2*
(a^7*tan(1/2*d*x + 1/2*c)^7 + 4*a^6*b*tan(1/2*d*x + 1/2*c)^7 - 13*a^5*b^2*tan(1/2*d*x + 1/2*c)^7 - 2*a^4*b^3*t
an(1/2*d*x + 1/2*c)^7 + 33*a^3*b^4*tan(1/2*d*x + 1/2*c)^7 - 17*a^2*b^5*tan(1/2*d*x + 1/2*c)^7 - 18*a*b^6*tan(1
/2*d*x + 1/2*c)^7 + 12*b^7*tan(1/2*d*x + 1/2*c)^7 + 3*a^7*tan(1/2*d*x + 1/2*c)^5 + 4*a^6*b*tan(1/2*d*x + 1/2*c
)^5 + 5*a^5*b^2*tan(1/2*d*x + 1/2*c)^5 - 26*a^4*b^3*tan(1/2*d*x + 1/2*c)^5 - 29*a^3*b^4*tan(1/2*d*x + 1/2*c)^5
+ 67*a^2*b^5*tan(1/2*d*x + 1/2*c)^5 + 18*a*b^6*tan(1/2*d*x + 1/2*c)^5 - 36*b^7*tan(1/2*d*x + 1/2*c)^5 + 3*a^7
*tan(1/2*d*x + 1/2*c)^3 - 4*a^6*b*tan(1/2*d*x + 1/2*c)^3 + 5*a^5*b^2*tan(1/2*d*x + 1/2*c)^3 + 26*a^4*b^3*tan(1
/2*d*x + 1/2*c)^3 - 29*a^3*b^4*tan(1/2*d*x + 1/2*c)^3 - 67*a^2*b^5*tan(1/2*d*x + 1/2*c)^3 + 18*a*b^6*tan(1/2*d
*x + 1/2*c)^3 + 36*b^7*tan(1/2*d*x + 1/2*c)^3 + a^7*tan(1/2*d*x + 1/2*c) - 4*a^6*b*tan(1/2*d*x + 1/2*c) - 13*a
^5*b^2*tan(1/2*d*x + 1/2*c) + 2*a^4*b^3*tan(1/2*d*x + 1/2*c) + 33*a^3*b^4*tan(1/2*d*x + 1/2*c) + 17*a^2*b^5*ta
n(1/2*d*x + 1/2*c) - 18*a*b^6*tan(1/2*d*x + 1/2*c) - 12*b^7*tan(1/2*d*x + 1/2*c))/((a^8 - 2*a^6*b^2 + a^4*b^4)
*(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^2 - a - b)^2) + (a^2 + 12*b^2
)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^5 - (a^2 + 12*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^5)/d