### 3.896 $$\int \frac{F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx$$

Optimal. Leaf size=196 $\frac{2 e^{\frac{1}{2} (2 d+2 e x+i \pi )} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (2,\frac{b c \log (F)}{e}+1;\frac{b c \log (F)}{e}+2;-e^{\frac{1}{2} (2 d+2 e x+i \pi )}\right )}{3 e^2 f^2}+\frac{b c \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) F^{c (a+b x)}}{6 e^2 f^2}+\frac{\tanh \left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) F^{c (a+b x)}}{6 e f^2}$

[Out]

(2*E^((2*d + I*Pi + 2*e*x)/2)*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/e, 2 + (b*c*Log[F])/e, -E^
((2*d + I*Pi + 2*e*x)/2)]*(e - b*c*Log[F]))/(3*e^2*f^2) + (b*c*F^(c*(a + b*x))*Log[F]*Sech[d/2 + (I/4)*Pi + (e
*x)/2]^2)/(6*e^2*f^2) + (F^(c*(a + b*x))*Sech[d/2 + (I/4)*Pi + (e*x)/2]^2*Tanh[d/2 + (I/4)*Pi + (e*x)/2])/(6*e
*f^2)

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Rubi [A]  time = 0.114658, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.12, Rules used = {5496, 5490, 5492} $\frac{2 e^{\frac{1}{2} (2 d+2 e x+i \pi )} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (2,\frac{b c \log (F)}{e}+1;\frac{b c \log (F)}{e}+2;-e^{\frac{1}{2} (2 d+2 e x+i \pi )}\right )}{3 e^2 f^2}+\frac{b c \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) F^{c (a+b x)}}{6 e^2 f^2}+\frac{\tanh \left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}+\frac{i \pi }{4}\right ) F^{c (a+b x)}}{6 e f^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[F^(c*(a + b*x))/(f + I*f*Sinh[d + e*x])^2,x]

[Out]

(2*E^((2*d + I*Pi + 2*e*x)/2)*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/e, 2 + (b*c*Log[F])/e, -E^
((2*d + I*Pi + 2*e*x)/2)]*(e - b*c*Log[F]))/(3*e^2*f^2) + (b*c*F^(c*(a + b*x))*Log[F]*Sech[d/2 + (I/4)*Pi + (e
*x)/2]^2)/(6*e^2*f^2) + (F^(c*(a + b*x))*Sech[d/2 + (I/4)*Pi + (e*x)/2]^2*Tanh[d/2 + (I/4)*Pi + (e*x)/2])/(6*e
*f^2)

Rule 5496

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sinh[(d_.) + (e_.)*(x_)])^(n_.), x_Symbol] :> Dist[2^n*f^n
, Int[F^(c*(a + b*x))*Cosh[d/2 + (e*x)/2 - (f*Pi)/(4*g)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] &
& EqQ[f^2 + g^2, 0] && ILtQ[n, 0]

Rule 5490

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b
*x))*Sech[d + e*x]^(n - 2))/(e^2*(n - 1)*(n - 2)), x] + (Dist[(e^2*(n - 2)^2 - b^2*c^2*Log[F]^2)/(e^2*(n - 1)*
(n - 2)), Int[F^(c*(a + b*x))*Sech[d + e*x]^(n - 2), x], x] + Simp[(F^(c*(a + b*x))*Sech[d + e*x]^(n - 1)*Sinh
[d + e*x])/(e*(n - 1)), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*(n - 2)^2 - b^2*c^2*Log[F]^2, 0] && GtQ
[n, 1] && NeQ[n, 2]

Rule 5492

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[(2^n*E^(n*(d + e*x))*F
^(c*(a + b*x))*Hypergeometric2F1[n, n/2 + (b*c*Log[F])/(2*e), 1 + n/2 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))])
/(e*n + b*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{F^{c (a+b x)}}{(f+i f \sinh (d+e x))^2} \, dx &=\frac{\int F^{c (a+b x)} \text{sech}^4\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right ) \, dx}{4 f^2}\\ &=\frac{b c F^{c (a+b x)} \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right )}{6 e^2 f^2}+\frac{F^{c (a+b x)} \text{sech}^2\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right ) \tanh \left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right )}{6 e f^2}+\frac{\left (1-\frac{b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \text{sech}^2\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right ) \, dx}{6 f^2}\\ &=\frac{2 e^{\frac{1}{2} (2 d+i \pi +2 e x)} F^{c (a+b x)} \, _2F_1\left (2,1+\frac{b c \log (F)}{e};2+\frac{b c \log (F)}{e};-e^{\frac{1}{2} (2 d+i \pi +2 e x)}\right ) (e-b c \log (F))}{3 e^2 f^2}+\frac{b c F^{c (a+b x)} \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right )}{6 e^2 f^2}+\frac{F^{c (a+b x)} \text{sech}^2\left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right ) \tanh \left (\frac{d}{2}+\frac{i \pi }{4}+\frac{e x}{2}\right )}{6 e f^2}\\ \end{align*}

Mathematica [A]  time = 3.01077, size = 255, normalized size = 1.3 $\frac{F^{c (a+b x)} \left (\cosh \left (\frac{1}{2} (d+e x)\right )+i \sinh \left (\frac{1}{2} (d+e x)\right )\right ) \left ((1-i) \left (e^2-b^2 c^2 \log ^2(F)\right ) \left (\cosh \left (\frac{1}{2} (d+e x)\right )+i \sinh \left (\frac{1}{2} (d+e x)\right )\right )^3 \left (-1+(1+i) \, _2F_1\left (1,\frac{b c \log (F)}{e};\frac{b c \log (F)}{e}+1;-i (\cosh (d+e x)+\sinh (d+e x))\right )\right )+2 \sinh \left (\frac{1}{2} (d+e x)\right ) \left (e^2-b^2 c^2 \log ^2(F)\right ) \left (\cosh \left (\frac{1}{2} (d+e x)\right )+i \sinh \left (\frac{1}{2} (d+e x)\right )\right )^2+e (b c \log (F)+i e) \left (\cosh \left (\frac{1}{2} (d+e x)\right )+i \sinh \left (\frac{1}{2} (d+e x)\right )\right )+2 e^2 \sinh \left (\frac{1}{2} (d+e x)\right )\right )}{3 e^3 (f+i f \sinh (d+e x))^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[F^(c*(a + b*x))/(f + I*f*Sinh[d + e*x])^2,x]

[Out]

(F^(c*(a + b*x))*(Cosh[(d + e*x)/2] + I*Sinh[(d + e*x)/2])*(e*(I*e + b*c*Log[F])*(Cosh[(d + e*x)/2] + I*Sinh[(
d + e*x)/2]) + (1 - I)*(-1 + (1 + I)*Hypergeometric2F1[1, (b*c*Log[F])/e, 1 + (b*c*Log[F])/e, (-I)*(Cosh[d + e
*x] + Sinh[d + e*x])])*(e^2 - b^2*c^2*Log[F]^2)*(Cosh[(d + e*x)/2] + I*Sinh[(d + e*x)/2])^3 + 2*e^2*Sinh[(d +
e*x)/2] + 2*(e^2 - b^2*c^2*Log[F]^2)*(Cosh[(d + e*x)/2] + I*Sinh[(d + e*x)/2])^2*Sinh[(d + e*x)/2]))/(3*e^3*(f
+ I*f*Sinh[d + e*x])^2)

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Maple [F]  time = 0.06, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{c \left ( bx+a \right ) }}{ \left ( f+if\sinh \left ( ex+d \right ) \right ) ^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d))^2,x)

[Out]

int(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d))^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d))^2,x, algorithm="maxima")

[Out]

(16*I*F^(a*c)*b^2*c^2*e*log(F)^2 + 16*I*F^(a*c)*b*c*e^2*log(F))*integrate(F^(b*c*x)/(-I*b^3*c^3*f^2*log(F)^3 +
9*I*b^2*c^2*e*f^2*log(F)^2 - 26*I*b*c*e^2*f^2*log(F) + 24*I*e^3*f^2 + (b^3*c^3*f^2*e^(5*d)*log(F)^3 - 9*b^2*c
^2*e*f^2*e^(5*d)*log(F)^2 + 26*b*c*e^2*f^2*e^(5*d)*log(F) - 24*e^3*f^2*e^(5*d))*e^(5*e*x) + (-5*I*b^3*c^3*f^2*
e^(4*d)*log(F)^3 + 45*I*b^2*c^2*e*f^2*e^(4*d)*log(F)^2 - 130*I*b*c*e^2*f^2*e^(4*d)*log(F) + 120*I*e^3*f^2*e^(4
*d))*e^(4*e*x) - 10*(b^3*c^3*f^2*e^(3*d)*log(F)^3 - 9*b^2*c^2*e*f^2*e^(3*d)*log(F)^2 + 26*b*c*e^2*f^2*e^(3*d)*
log(F) - 24*e^3*f^2*e^(3*d))*e^(3*e*x) + (10*I*b^3*c^3*f^2*e^(2*d)*log(F)^3 - 90*I*b^2*c^2*e*f^2*e^(2*d)*log(F
)^2 + 260*I*b*c*e^2*f^2*e^(2*d)*log(F) - 240*I*e^3*f^2*e^(2*d))*e^(2*e*x) + 5*(b^3*c^3*f^2*e^d*log(F)^3 - 9*b^
2*c^2*e*f^2*e^d*log(F)^2 + 26*b*c*e^2*f^2*e^d*log(F) - 24*e^3*f^2*e^d)*e^(e*x)), x) + (16*F^(a*c)*b*c*e*log(F)
+ 16*F^(a*c)*e^2 - 4*(F^(a*c)*b^2*c^2*e^(2*d)*log(F)^2 - 7*F^(a*c)*b*c*e*e^(2*d)*log(F) + 12*F^(a*c)*e^2*e^(2
*d))*e^(2*e*x) - (16*I*F^(a*c)*b*c*e*e^d*log(F) - 64*I*F^(a*c)*e^2*e^d)*e^(e*x))*F^(b*c*x)/(b^3*c^3*f^2*log(F)
^3 - 9*b^2*c^2*e*f^2*log(F)^2 + 26*b*c*e^2*f^2*log(F) - 24*e^3*f^2 + (b^3*c^3*f^2*e^(4*d)*log(F)^3 - 9*b^2*c^2
*e*f^2*e^(4*d)*log(F)^2 + 26*b*c*e^2*f^2*e^(4*d)*log(F) - 24*e^3*f^2*e^(4*d))*e^(4*e*x) + (-4*I*b^3*c^3*f^2*e^
(3*d)*log(F)^3 + 36*I*b^2*c^2*e*f^2*e^(3*d)*log(F)^2 - 104*I*b*c*e^2*f^2*e^(3*d)*log(F) + 96*I*e^3*f^2*e^(3*d)
)*e^(3*e*x) - 6*(b^3*c^3*f^2*e^(2*d)*log(F)^3 - 9*b^2*c^2*e*f^2*e^(2*d)*log(F)^2 + 26*b*c*e^2*f^2*e^(2*d)*log(
F) - 24*e^3*f^2*e^(2*d))*e^(2*e*x) + (4*I*b^3*c^3*f^2*e^d*log(F)^3 - 36*I*b^2*c^2*e*f^2*e^d*log(F)^2 + 104*I*b
*c*e^2*f^2*e^d*log(F) - 96*I*e^3*f^2*e^d)*e^(e*x))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (6 \, e^{2} e^{\left (e x + d\right )} +{\left (-2 i \, b^{2} c^{2} e^{\left (2 \, e x + 2 \, d\right )} - 4 \, b^{2} c^{2} e^{\left (e x + d\right )} + 2 i \, b^{2} c^{2}\right )} \log \left (F\right )^{2} - 2 i \, e^{2} +{\left (-2 i \, b c e e^{\left (2 \, e x + 2 \, d\right )} - 2 \, b c e e^{\left (e x + d\right )}\right )} \log \left (F\right )\right )} F^{b c x + a c} +{\left (3 \, e^{3} f^{2} e^{\left (3 \, e x + 3 \, d\right )} - 9 i \, e^{3} f^{2} e^{\left (2 \, e x + 2 \, d\right )} - 9 \, e^{3} f^{2} e^{\left (e x + d\right )} + 3 i \, e^{3} f^{2}\right )}{\rm integral}\left (\frac{{\left (2 i \, b^{3} c^{3} \log \left (F\right )^{3} - 2 i \, b c e^{2} \log \left (F\right )\right )} F^{b c x + a c}}{3 \,{\left (e^{3} f^{2} e^{\left (e x + d\right )} - i \, e^{3} f^{2}\right )}}, x\right )}{3 \, e^{3} f^{2} e^{\left (3 \, e x + 3 \, d\right )} - 9 i \, e^{3} f^{2} e^{\left (2 \, e x + 2 \, d\right )} - 9 \, e^{3} f^{2} e^{\left (e x + d\right )} + 3 i \, e^{3} f^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d))^2,x, algorithm="fricas")

[Out]

((6*e^2*e^(e*x + d) + (-2*I*b^2*c^2*e^(2*e*x + 2*d) - 4*b^2*c^2*e^(e*x + d) + 2*I*b^2*c^2)*log(F)^2 - 2*I*e^2
+ (-2*I*b*c*e*e^(2*e*x + 2*d) - 2*b*c*e*e^(e*x + d))*log(F))*F^(b*c*x + a*c) + (3*e^3*f^2*e^(3*e*x + 3*d) - 9*
I*e^3*f^2*e^(2*e*x + 2*d) - 9*e^3*f^2*e^(e*x + d) + 3*I*e^3*f^2)*integral(1/3*(2*I*b^3*c^3*log(F)^3 - 2*I*b*c*
e^2*log(F))*F^(b*c*x + a*c)/(e^3*f^2*e^(e*x + d) - I*e^3*f^2), x))/(3*e^3*f^2*e^(3*e*x + 3*d) - 9*I*e^3*f^2*e^
(2*e*x + 2*d) - 9*e^3*f^2*e^(e*x + d) + 3*I*e^3*f^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))/(f+I*f*sinh(e*x+d))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{{\left (i \, f \sinh \left (e x + d\right ) + f\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+I*f*sinh(e*x+d))^2,x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)/(I*f*sinh(e*x + d) + f)^2, x)