3.2.92 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^4} \, dx\) [192]

Optimal. Leaf size=189 \[ \frac {a \left (2 c^2-2 c d+d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{(c-d)^{5/2} (c+d)^{7/2} f}+\frac {a \tan (e+f x)}{3 (c+d) f (c+d \sec (e+f x))^3}+\frac {a (2 c-3 d) \tan (e+f x)}{6 (c-d) (c+d)^2 f (c+d \sec (e+f x))^2}+\frac {a (c-4 d) (2 c-d) \tan (e+f x)}{6 (c-d)^2 (c+d)^3 f (c+d \sec (e+f x))} \]

[Out]

a*(2*c^2-2*c*d+d^2)*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/(c-d)^(5/2)/(c+d)^(7/2)/f+1/3*a*tan(f*
x+e)/(c+d)/f/(c+d*sec(f*x+e))^3+1/6*a*(2*c-3*d)*tan(f*x+e)/(c-d)/(c+d)^2/f/(c+d*sec(f*x+e))^2+1/6*a*(c-4*d)*(2
*c-d)*tan(f*x+e)/(c-d)^2/(c+d)^3/f/(c+d*sec(f*x+e))

________________________________________________________________________________________

Rubi [A]
time = 0.32, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4088, 12, 3916, 2738, 214} \begin {gather*} \frac {a \left (2 c^2-2 c d+d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f (c-d)^{5/2} (c+d)^{7/2}}+\frac {a (c-4 d) (2 c-d) \tan (e+f x)}{6 f (c-d)^2 (c+d)^3 (c+d \sec (e+f x))}+\frac {a (2 c-3 d) \tan (e+f x)}{6 f (c-d) (c+d)^2 (c+d \sec (e+f x))^2}+\frac {a \tan (e+f x)}{3 f (c+d) (c+d \sec (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c + d*Sec[e + f*x])^4,x]

[Out]

(a*(2*c^2 - 2*c*d + d^2)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/((c - d)^(5/2)*(c + d)^(7/2)*f)
+ (a*Tan[e + f*x])/(3*(c + d)*f*(c + d*Sec[e + f*x])^3) + (a*(2*c - 3*d)*Tan[e + f*x])/(6*(c - d)*(c + d)^2*f*
(c + d*Sec[e + f*x])^2) + (a*(c - 4*d)*(2*c - d)*Tan[e + f*x])/(6*(c - d)^2*(c + d)^3*f*(c + d*Sec[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 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 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4088

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a
*B)*(m + 2)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^4} \, dx &=\frac {a \tan (e+f x)}{3 (c+d) f (c+d \sec (e+f x))^3}-\frac {\int \frac {\sec (e+f x) (-3 a (c-d)-2 a (c-d) \sec (e+f x))}{(c+d \sec (e+f x))^3} \, dx}{3 \left (c^2-d^2\right )}\\ &=\frac {a \tan (e+f x)}{3 (c+d) f (c+d \sec (e+f x))^3}+\frac {a (2 c-3 d) \tan (e+f x)}{6 (c-d) (c+d)^2 f (c+d \sec (e+f x))^2}+\frac {\int \frac {\sec (e+f x) (2 a (3 c-2 d) (c-d)+a (2 c-3 d) (c-d) \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx}{6 \left (c^2-d^2\right )^2}\\ &=\frac {a \tan (e+f x)}{3 (c+d) f (c+d \sec (e+f x))^3}+\frac {a (2 c-3 d) \tan (e+f x)}{6 (c-d) (c+d)^2 f (c+d \sec (e+f x))^2}+\frac {a (c-4 d) (2 c-d) \tan (e+f x)}{6 (c-d)^2 (c+d)^3 f (c+d \sec (e+f x))}-\frac {\int -\frac {3 a (c-d) \left (2 c^2-2 c d+d^2\right ) \sec (e+f x)}{c+d \sec (e+f x)} \, dx}{6 \left (c^2-d^2\right )^3}\\ &=\frac {a \tan (e+f x)}{3 (c+d) f (c+d \sec (e+f x))^3}+\frac {a (2 c-3 d) \tan (e+f x)}{6 (c-d) (c+d)^2 f (c+d \sec (e+f x))^2}+\frac {a (c-4 d) (2 c-d) \tan (e+f x)}{6 (c-d)^2 (c+d)^3 f (c+d \sec (e+f x))}+\frac {\left (a \left (2 c^2-2 c d+d^2\right )\right ) \int \frac {\sec (e+f x)}{c+d \sec (e+f x)} \, dx}{2 (c-d)^2 (c+d)^3}\\ &=\frac {a \tan (e+f x)}{3 (c+d) f (c+d \sec (e+f x))^3}+\frac {a (2 c-3 d) \tan (e+f x)}{6 (c-d) (c+d)^2 f (c+d \sec (e+f x))^2}+\frac {a (c-4 d) (2 c-d) \tan (e+f x)}{6 (c-d)^2 (c+d)^3 f (c+d \sec (e+f x))}+\frac {\left (a \left (2 c^2-2 c d+d^2\right )\right ) \int \frac {1}{1+\frac {c \cos (e+f x)}{d}} \, dx}{2 (c-d)^2 d (c+d)^3}\\ &=\frac {a \tan (e+f x)}{3 (c+d) f (c+d \sec (e+f x))^3}+\frac {a (2 c-3 d) \tan (e+f x)}{6 (c-d) (c+d)^2 f (c+d \sec (e+f x))^2}+\frac {a (c-4 d) (2 c-d) \tan (e+f x)}{6 (c-d)^2 (c+d)^3 f (c+d \sec (e+f x))}+\frac {\left (a \left (2 c^2-2 c d+d^2\right )\right ) \text {Subst}\left (\int \frac {1}{1+\frac {c}{d}+\left (1-\frac {c}{d}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{(c-d)^2 d (c+d)^3 f}\\ &=\frac {a \left (2 c^2-2 c d+d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{(c-d)^{5/2} (c+d)^{7/2} f}+\frac {a \tan (e+f x)}{3 (c+d) f (c+d \sec (e+f x))^3}+\frac {a (2 c-3 d) \tan (e+f x)}{6 (c-d) (c+d)^2 f (c+d \sec (e+f x))^2}+\frac {a (c-4 d) (2 c-d) \tan (e+f x)}{6 (c-d)^2 (c+d)^3 f (c+d \sec (e+f x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 3.34, size = 247, normalized size = 1.31 \begin {gather*} -\frac {a (1+\cos (e+f x)) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \left (6 \left (2 c^2-2 c d+d^2\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))^3-\frac {1}{2} \sqrt {c^2-d^2} \left (6 c^4-12 c^3 d+2 c^2 d^2-15 c d^3+10 d^4+6 d \left (2 c^3-7 c^2 d+2 c d^2+d^3\right ) \cos (e+f x)+\left (6 c^4-12 c^3 d-2 c^2 d^2+3 c d^3+2 d^4\right ) \cos (2 (e+f x))\right ) \sin (e+f x)\right )}{12 (c-d)^2 (c+d)^3 \sqrt {c^2-d^2} f (d+c \cos (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x]))/(c + d*Sec[e + f*x])^4,x]

[Out]

-1/12*(a*(1 + Cos[e + f*x])*Sec[(e + f*x)/2]^2*(6*(2*c^2 - 2*c*d + d^2)*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sq
rt[c^2 - d^2]]*(d + c*Cos[e + f*x])^3 - (Sqrt[c^2 - d^2]*(6*c^4 - 12*c^3*d + 2*c^2*d^2 - 15*c*d^3 + 10*d^4 + 6
*d*(2*c^3 - 7*c^2*d + 2*c*d^2 + d^3)*Cos[e + f*x] + (6*c^4 - 12*c^3*d - 2*c^2*d^2 + 3*c*d^3 + 2*d^4)*Cos[2*(e
+ f*x)])*Sin[e + f*x])/2))/((c - d)^2*(c + d)^3*Sqrt[c^2 - d^2]*f*(d + c*Cos[e + f*x])^3)

________________________________________________________________________________________

Maple [A]
time = 0.49, size = 271, normalized size = 1.43 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))/(c+d*sec(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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(sec(f*x+e)*(a+a*sec(f*x+e))/(c+d*sec(f*x+e))^4,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*c^2-4*d^2>0)', see `assume?`
for more de

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 621 vs. \(2 (181) = 362\).
time = 2.51, size = 1304, normalized size = 6.90 \begin {gather*} \left [\frac {3 \, {\left (2 \, a c^{2} d^{3} - 2 \, a c d^{4} + a d^{5} + {\left (2 \, a c^{5} - 2 \, a c^{4} d + a c^{3} d^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (2 \, a c^{4} d - 2 \, a c^{3} d^{2} + a c^{2} d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (2 \, a c^{3} d^{2} - 2 \, a c^{2} d^{3} + a c d^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) + 2 \, {\left (2 \, a c^{4} d^{2} - 9 \, a c^{3} d^{3} + 2 \, a c^{2} d^{4} + 9 \, a c d^{5} - 4 \, a d^{6} + {\left (6 \, a c^{6} - 12 \, a c^{5} d - 8 \, a c^{4} d^{2} + 15 \, a c^{3} d^{3} + 4 \, a c^{2} d^{4} - 3 \, a c d^{5} - 2 \, a d^{6}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (2 \, a c^{5} d - 7 \, a c^{4} d^{2} + 8 \, a c^{2} d^{4} - 2 \, a c d^{5} - a d^{6}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{12 \, {\left ({\left (c^{10} + c^{9} d - 3 \, c^{8} d^{2} - 3 \, c^{7} d^{3} + 3 \, c^{6} d^{4} + 3 \, c^{5} d^{5} - c^{4} d^{6} - c^{3} d^{7}\right )} f \cos \left (f x + e\right )^{3} + 3 \, {\left (c^{9} d + c^{8} d^{2} - 3 \, c^{7} d^{3} - 3 \, c^{6} d^{4} + 3 \, c^{5} d^{5} + 3 \, c^{4} d^{6} - c^{3} d^{7} - c^{2} d^{8}\right )} f \cos \left (f x + e\right )^{2} + 3 \, {\left (c^{8} d^{2} + c^{7} d^{3} - 3 \, c^{6} d^{4} - 3 \, c^{5} d^{5} + 3 \, c^{4} d^{6} + 3 \, c^{3} d^{7} - c^{2} d^{8} - c d^{9}\right )} f \cos \left (f x + e\right ) + {\left (c^{7} d^{3} + c^{6} d^{4} - 3 \, c^{5} d^{5} - 3 \, c^{4} d^{6} + 3 \, c^{3} d^{7} + 3 \, c^{2} d^{8} - c d^{9} - d^{10}\right )} f\right )}}, \frac {3 \, {\left (2 \, a c^{2} d^{3} - 2 \, a c d^{4} + a d^{5} + {\left (2 \, a c^{5} - 2 \, a c^{4} d + a c^{3} d^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (2 \, a c^{4} d - 2 \, a c^{3} d^{2} + a c^{2} d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (2 \, a c^{3} d^{2} - 2 \, a c^{2} d^{3} + a c d^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) + {\left (2 \, a c^{4} d^{2} - 9 \, a c^{3} d^{3} + 2 \, a c^{2} d^{4} + 9 \, a c d^{5} - 4 \, a d^{6} + {\left (6 \, a c^{6} - 12 \, a c^{5} d - 8 \, a c^{4} d^{2} + 15 \, a c^{3} d^{3} + 4 \, a c^{2} d^{4} - 3 \, a c d^{5} - 2 \, a d^{6}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (2 \, a c^{5} d - 7 \, a c^{4} d^{2} + 8 \, a c^{2} d^{4} - 2 \, a c d^{5} - a d^{6}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \, {\left ({\left (c^{10} + c^{9} d - 3 \, c^{8} d^{2} - 3 \, c^{7} d^{3} + 3 \, c^{6} d^{4} + 3 \, c^{5} d^{5} - c^{4} d^{6} - c^{3} d^{7}\right )} f \cos \left (f x + e\right )^{3} + 3 \, {\left (c^{9} d + c^{8} d^{2} - 3 \, c^{7} d^{3} - 3 \, c^{6} d^{4} + 3 \, c^{5} d^{5} + 3 \, c^{4} d^{6} - c^{3} d^{7} - c^{2} d^{8}\right )} f \cos \left (f x + e\right )^{2} + 3 \, {\left (c^{8} d^{2} + c^{7} d^{3} - 3 \, c^{6} d^{4} - 3 \, c^{5} d^{5} + 3 \, c^{4} d^{6} + 3 \, c^{3} d^{7} - c^{2} d^{8} - c d^{9}\right )} f \cos \left (f x + e\right ) + {\left (c^{7} d^{3} + c^{6} d^{4} - 3 \, c^{5} d^{5} - 3 \, c^{4} d^{6} + 3 \, c^{3} d^{7} + 3 \, c^{2} d^{8} - c d^{9} - d^{10}\right )} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c+d*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

[1/12*(3*(2*a*c^2*d^3 - 2*a*c*d^4 + a*d^5 + (2*a*c^5 - 2*a*c^4*d + a*c^3*d^2)*cos(f*x + e)^3 + 3*(2*a*c^4*d -
2*a*c^3*d^2 + a*c^2*d^3)*cos(f*x + e)^2 + 3*(2*a*c^3*d^2 - 2*a*c^2*d^3 + a*c*d^4)*cos(f*x + e))*sqrt(c^2 - d^2
)*log((2*c*d*cos(f*x + e) - (c^2 - 2*d^2)*cos(f*x + e)^2 + 2*sqrt(c^2 - d^2)*(d*cos(f*x + e) + c)*sin(f*x + e)
+ 2*c^2 - d^2)/(c^2*cos(f*x + e)^2 + 2*c*d*cos(f*x + e) + d^2)) + 2*(2*a*c^4*d^2 - 9*a*c^3*d^3 + 2*a*c^2*d^4
+ 9*a*c*d^5 - 4*a*d^6 + (6*a*c^6 - 12*a*c^5*d - 8*a*c^4*d^2 + 15*a*c^3*d^3 + 4*a*c^2*d^4 - 3*a*c*d^5 - 2*a*d^6
)*cos(f*x + e)^2 + 3*(2*a*c^5*d - 7*a*c^4*d^2 + 8*a*c^2*d^4 - 2*a*c*d^5 - a*d^6)*cos(f*x + e))*sin(f*x + e))/(
(c^10 + c^9*d - 3*c^8*d^2 - 3*c^7*d^3 + 3*c^6*d^4 + 3*c^5*d^5 - c^4*d^6 - c^3*d^7)*f*cos(f*x + e)^3 + 3*(c^9*d
+ c^8*d^2 - 3*c^7*d^3 - 3*c^6*d^4 + 3*c^5*d^5 + 3*c^4*d^6 - c^3*d^7 - c^2*d^8)*f*cos(f*x + e)^2 + 3*(c^8*d^2
+ c^7*d^3 - 3*c^6*d^4 - 3*c^5*d^5 + 3*c^4*d^6 + 3*c^3*d^7 - c^2*d^8 - c*d^9)*f*cos(f*x + e) + (c^7*d^3 + c^6*d
^4 - 3*c^5*d^5 - 3*c^4*d^6 + 3*c^3*d^7 + 3*c^2*d^8 - c*d^9 - d^10)*f), 1/6*(3*(2*a*c^2*d^3 - 2*a*c*d^4 + a*d^5
+ (2*a*c^5 - 2*a*c^4*d + a*c^3*d^2)*cos(f*x + e)^3 + 3*(2*a*c^4*d - 2*a*c^3*d^2 + a*c^2*d^3)*cos(f*x + e)^2 +
3*(2*a*c^3*d^2 - 2*a*c^2*d^3 + a*c*d^4)*cos(f*x + e))*sqrt(-c^2 + d^2)*arctan(-sqrt(-c^2 + d^2)*(d*cos(f*x +
e) + c)/((c^2 - d^2)*sin(f*x + e))) + (2*a*c^4*d^2 - 9*a*c^3*d^3 + 2*a*c^2*d^4 + 9*a*c*d^5 - 4*a*d^6 + (6*a*c^
6 - 12*a*c^5*d - 8*a*c^4*d^2 + 15*a*c^3*d^3 + 4*a*c^2*d^4 - 3*a*c*d^5 - 2*a*d^6)*cos(f*x + e)^2 + 3*(2*a*c^5*d
- 7*a*c^4*d^2 + 8*a*c^2*d^4 - 2*a*c*d^5 - a*d^6)*cos(f*x + e))*sin(f*x + e))/((c^10 + c^9*d - 3*c^8*d^2 - 3*c
^7*d^3 + 3*c^6*d^4 + 3*c^5*d^5 - c^4*d^6 - c^3*d^7)*f*cos(f*x + e)^3 + 3*(c^9*d + c^8*d^2 - 3*c^7*d^3 - 3*c^6*
d^4 + 3*c^5*d^5 + 3*c^4*d^6 - c^3*d^7 - c^2*d^8)*f*cos(f*x + e)^2 + 3*(c^8*d^2 + c^7*d^3 - 3*c^6*d^4 - 3*c^5*d
^5 + 3*c^4*d^6 + 3*c^3*d^7 - c^2*d^8 - c*d^9)*f*cos(f*x + e) + (c^7*d^3 + c^6*d^4 - 3*c^5*d^5 - 3*c^4*d^6 + 3*
c^3*d^7 + 3*c^2*d^8 - c*d^9 - d^10)*f)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int \frac {\sec {\left (e + f x \right )}}{c^{4} + 4 c^{3} d \sec {\left (e + f x \right )} + 6 c^{2} d^{2} \sec ^{2}{\left (e + f x \right )} + 4 c d^{3} \sec ^{3}{\left (e + f x \right )} + d^{4} \sec ^{4}{\left (e + f x \right )}}\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{c^{4} + 4 c^{3} d \sec {\left (e + f x \right )} + 6 c^{2} d^{2} \sec ^{2}{\left (e + f x \right )} + 4 c d^{3} \sec ^{3}{\left (e + f x \right )} + d^{4} \sec ^{4}{\left (e + f x \right )}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c+d*sec(f*x+e))**4,x)

[Out]

a*(Integral(sec(e + f*x)/(c**4 + 4*c**3*d*sec(e + f*x) + 6*c**2*d**2*sec(e + f*x)**2 + 4*c*d**3*sec(e + f*x)**
3 + d**4*sec(e + f*x)**4), x) + Integral(sec(e + f*x)**2/(c**4 + 4*c**3*d*sec(e + f*x) + 6*c**2*d**2*sec(e + f
*x)**2 + 4*c*d**3*sec(e + f*x)**3 + d**4*sec(e + f*x)**4), x))

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (174) = 348\).
time = 0.53, size = 449, normalized size = 2.38 \begin {gather*} -\frac {\frac {3 \, {\left (2 \, a c^{2} - 2 \, a c d + a d^{2}\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 (c^{5} + c^{4} d - 2 \, c^{3} d^{2} - 2 \, c^{2} d^{3} + c d^{4} + d^{5}\right )} \sqrt {-c^{2} + d^{2}}} + \frac {6 \, a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 18 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 21 \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 12 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 3 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 12 \, a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 24 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 8 \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 24 \, a c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 4 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 6 \, a c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 6 \, a c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 21 \, a c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 9 \, a d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (c^{5} + c^{4} d - 2 \, c^{3} d^{2} - 2 \, c^{2} d^{3} + c d^{4} + d^{5}\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 )}^{3}}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))/(c+d*sec(f*x+e))^4,x, algorithm="giac")

[Out]

-1/3*(3*(2*a*c^2 - 2*a*c*d + a*d^2)*(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)))/((c^5 + c^4*d - 2*c^3*d^2 - 2*c^2*d^3 + c*d^4 + d^5)*sqrt
(-c^2 + d^2)) + (6*a*c^4*tan(1/2*f*x + 1/2*e)^5 - 18*a*c^3*d*tan(1/2*f*x + 1/2*e)^5 + 21*a*c^2*d^2*tan(1/2*f*x
+ 1/2*e)^5 - 12*a*c*d^3*tan(1/2*f*x + 1/2*e)^5 + 3*a*d^4*tan(1/2*f*x + 1/2*e)^5 - 12*a*c^4*tan(1/2*f*x + 1/2*
e)^3 + 24*a*c^3*d*tan(1/2*f*x + 1/2*e)^3 + 8*a*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 - 24*a*c*d^3*tan(1/2*f*x + 1/2*e
)^3 + 4*a*d^4*tan(1/2*f*x + 1/2*e)^3 + 6*a*c^4*tan(1/2*f*x + 1/2*e) - 6*a*c^3*d*tan(1/2*f*x + 1/2*e) - 21*a*c^
2*d^2*tan(1/2*f*x + 1/2*e) + 9*a*d^4*tan(1/2*f*x + 1/2*e))/((c^5 + c^4*d - 2*c^3*d^2 - 2*c^2*d^3 + c*d^4 + d^5
)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/2*e)^2 - c - d)^3))/f

________________________________________________________________________________________

Mupad [B]
time = 5.28, size = 321, normalized size = 1.70 \begin {gather*} \frac {\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (2\,a\,c^2-2\,a\,c\,d+a\,d^2\right )}{{\left (c+d\right )}^3}+\frac {a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c^2-6\,c\,d+3\,d^2\right )}{\left (c+d\right )\,\left (c^2-2\,c\,d+d^2\right )}-\frac {4\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,c^2-6\,c\,d+d^2\right )}{3\,{\left (c+d\right )}^2\,\left (c-d\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (-3\,c^3-3\,c^2\,d+3\,c\,d^2+3\,d^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (-3\,c^3+3\,c^2\,d+3\,c\,d^2-3\,d^3\right )+3\,c\,d^2+3\,c^2\,d+c^3+d^3-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (c^3-3\,c^2\,d+3\,c\,d^2-d^3\right )\right )}+\frac {a\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c-2\,d\right )\,\left (c^2-2\,c\,d+d^2\right )}{2\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}\right )\,\left (2\,c^2-2\,c\,d+d^2\right )}{f\,{\left (c+d\right )}^{7/2}\,{\left (c-d\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))/(cos(e + f*x)*(c + d/cos(e + f*x))^4),x)

[Out]

((tan(e/2 + (f*x)/2)^5*(2*a*c^2 + a*d^2 - 2*a*c*d))/(c + d)^3 + (a*tan(e/2 + (f*x)/2)*(2*c^2 - 6*c*d + 3*d^2))
/((c + d)*(c^2 - 2*c*d + d^2)) - (4*a*tan(e/2 + (f*x)/2)^3*(3*c^2 - 6*c*d + d^2))/(3*(c + d)^2*(c - d)))/(f*(t
an(e/2 + (f*x)/2)^2*(3*c*d^2 - 3*c^2*d - 3*c^3 + 3*d^3) - tan(e/2 + (f*x)/2)^4*(3*c*d^2 + 3*c^2*d - 3*c^3 - 3*
d^3) + 3*c*d^2 + 3*c^2*d + c^3 + d^3 - tan(e/2 + (f*x)/2)^6*(3*c*d^2 - 3*c^2*d + c^3 - d^3))) + (a*atanh((tan(
e/2 + (f*x)/2)*(2*c - 2*d)*(c^2 - 2*c*d + d^2))/(2*(c + d)^(1/2)*(c - d)^(5/2)))*(2*c^2 - 2*c*d + d^2))/(f*(c
+ d)^(7/2)*(c - d)^(5/2))

________________________________________________________________________________________