Integrand size = 25, antiderivative size = 192 \[ \int F^{c (a+b x)} (f+i f \sinh (d+e x))^2 \, dx=\frac {f^2 F^{a c+b c x}}{b c \log (F)}+\frac {2 e^2 f^2 F^{a c+b c x}}{b c \log (F) \left (4 e^2-b^2 c^2 \log ^2(F)\right )}+\frac {2 i f^2 F^{a c+b c x} (e \cosh (d+e x)-b c \log (F) \sinh (d+e x))}{e^2-b^2 c^2 \log ^2(F)}-\frac {f^2 F^{a c+b c x} \sinh (d+e x) (2 e \cosh (d+e x)-b c \log (F) \sinh (d+e x))}{4 e^2-b^2 c^2 \log ^2(F)} \] Output:
f^2*F^(b*c*x+a*c)/b/c/ln(F)+2*e^2*f^2*F^(b*c*x+a*c)/b/c/ln(F)/(4*e^2-b^2*c ^2*ln(F)^2)+2*I*f^2*F^(b*c*x+a*c)*(e*cosh(e*x+d)-b*c*ln(F)*sinh(e*x+d))/(e ^2-b^2*c^2*ln(F)^2)-f^2*F^(b*c*x+a*c)*sinh(e*x+d)*(2*e*cosh(e*x+d)-b*c*ln( F)*sinh(e*x+d))/(4*e^2-b^2*c^2*ln(F)^2)
Time = 7.60 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.02 \[ \int F^{c (a+b x)} (f+i f \sinh (d+e x))^2 \, dx=\frac {F^{c (a+b x)} (f+i f \sinh (d+e x))^2 \left (\frac {3}{b c \log (F)}+\frac {4 i e \cosh (d+e x)}{(e-b c \log (F)) (e+b c \log (F))}-\frac {b c \cosh (2 (d+e x)) \log (F)}{-4 e^2+b^2 c^2 \log ^2(F)}+\frac {4 i b c \log (F) \sinh (d+e x)}{(-e+b c \log (F)) (e+b c \log (F))}-\frac {2 e \sinh (2 (d+e x))}{4 e^2-b^2 c^2 \log ^2(F)}\right )}{2 \left (\cosh \left (\frac {1}{2} (d+e x)\right )+i \sinh \left (\frac {1}{2} (d+e x)\right )\right )^4} \] Input:
Integrate[F^(c*(a + b*x))*(f + I*f*Sinh[d + e*x])^2,x]
Output:
(F^(c*(a + b*x))*(f + I*f*Sinh[d + e*x])^2*(3/(b*c*Log[F]) + ((4*I)*e*Cosh [d + e*x])/((e - b*c*Log[F])*(e + b*c*Log[F])) - (b*c*Cosh[2*(d + e*x)]*Lo g[F])/(-4*e^2 + b^2*c^2*Log[F]^2) + ((4*I)*b*c*Log[F]*Sinh[d + e*x])/((-e + b*c*Log[F])*(e + b*c*Log[F])) - (2*e*Sinh[2*(d + e*x)])/(4*e^2 - b^2*c^2 *Log[F]^2)))/(2*(Cosh[(d + e*x)/2] + I*Sinh[(d + e*x)/2])^4)
Time = 0.67 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {7292, 27, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int F^{c (a+b x)} (f+i f \sinh (d+e x))^2 \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int f^2 (1+i \sinh (d+e x))^2 F^{a c+b c x}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle f^2 \int F^{a c+b x c} (i \sinh (d+e x)+1)^2dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle f^2 \int \left (-\sinh ^2(d+e x) F^{a c+b x c}+2 i \sinh (d+e x) F^{a c+b x c}+F^{a c+b x c}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle f^2 \left (\frac {b c \log (F) \sinh ^2(d+e x) F^{a c+b c x}}{4 e^2-b^2 c^2 \log ^2(F)}-\frac {2 i b c \log (F) \sinh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}+\frac {2 i e \cosh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}-\frac {2 e \sinh (d+e x) \cosh (d+e x) F^{a c+b c x}}{4 e^2-b^2 c^2 \log ^2(F)}+\frac {2 e^2 F^{a c+b c x}}{b c \log (F) \left (4 e^2-b^2 c^2 \log ^2(F)\right )}+\frac {F^{a c+b c x}}{b c \log (F)}\right )\) |
Input:
Int[F^(c*(a + b*x))*(f + I*f*Sinh[d + e*x])^2,x]
Output:
f^2*(F^(a*c + b*c*x)/(b*c*Log[F]) + ((2*I)*e*F^(a*c + b*c*x)*Cosh[d + e*x] )/(e^2 - b^2*c^2*Log[F]^2) + (2*e^2*F^(a*c + b*c*x))/(b*c*Log[F]*(4*e^2 - b^2*c^2*Log[F]^2)) - ((2*I)*b*c*F^(a*c + b*c*x)*Log[F]*Sinh[d + e*x])/(e^2 - b^2*c^2*Log[F]^2) - (2*e*F^(a*c + b*c*x)*Cosh[d + e*x]*Sinh[d + e*x])/( 4*e^2 - b^2*c^2*Log[F]^2) + (b*c*F^(a*c + b*c*x)*Log[F]*Sinh[d + e*x]^2)/( 4*e^2 - b^2*c^2*Log[F]^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.84 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.02
| method | result | size |
| parallelrisch | \(\frac {2 f^{2} F^{c \left (b x +a \right )} \left (\frac {\left (\ln \left (F \right )^{4} b^{4} c^{4}-\ln \left (F \right )^{2} b^{2} c^{2} e^{2}\right ) \cosh \left (2 e x +2 d \right )}{4}+\frac {\left (-\ln \left (F \right )^{3} b^{3} c^{3} e +\ln \left (F \right ) b c \,e^{3}\right ) \sinh \left (2 e x +2 d \right )}{2}+\left (-i \sinh \left (e x +d \right ) b^{2} c^{2} \ln \left (F \right )^{2}+i \cosh \left (e x +d \right ) b c e \ln \left (F \right )-\frac {3 b^{2} c^{2} \ln \left (F \right )^{2}}{4}+\frac {3 e^{2}}{4}\right ) \left (2 e +b c \ln \left (F \right )\right ) \left (b c \ln \left (F \right )-2 e \right )\right )}{-c^{5} b^{5} \ln \left (F \right )^{5}+5 c^{3} b^{3} \ln \left (F \right )^{3} e^{2}-4 c b \ln \left (F \right ) e^{4}}\) | \(196\) |
| risch | \(\frac {f^{2} \left (-4 i \ln \left (F \right )^{3} b^{3} c^{3} e \,{\mathrm e}^{3 e x +3 d}-\ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{4 e x +4 d}+16 i \ln \left (F \right ) b c \,e^{3} {\mathrm e}^{3 e x +3 d}+6 \ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{2 e x +2 d}-4 i \ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{e x +d}+2 \ln \left (F \right )^{3} b^{3} c^{3} e \,{\mathrm e}^{4 e x +4 d}-\ln \left (F \right )^{4} b^{4} c^{4}-16 i \ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{3 e x +3 d}-4 i \ln \left (F \right )^{3} b^{3} c^{3} e \,{\mathrm e}^{e x +d}+\ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{4 e x +4 d}-2 \ln \left (F \right )^{3} b^{3} c^{3} e +16 i \ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{e x +d}-30 \ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{2 e x +2 d}+4 i \ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{3 e x +3 d}-2 \ln \left (F \right ) b c \,e^{3} {\mathrm e}^{4 e x +4 d}+\ln \left (F \right )^{2} b^{2} c^{2} e^{2}+16 i \ln \left (F \right ) b c \,e^{3} {\mathrm e}^{e x +d}+2 \ln \left (F \right ) b c \,e^{3}+24 e^{4} {\mathrm e}^{2 e x +2 d}\right ) {\mathrm e}^{-2 e x -2 d} F^{c \left (b x +a \right )}}{4 b c \ln \left (F \right ) \left (-b c \ln \left (F \right )+e \right ) \left (-b c \ln \left (F \right )+2 e \right ) \left (e +b c \ln \left (F \right )\right ) \left (2 e +b c \ln \left (F \right )\right )}\) | \(434\) |
| orering | \(\text {Expression too large to display}\) | \(1011\) |
Input:
int(F^(c*(b*x+a))*(f+I*f*sinh(e*x+d))^2,x,method=_RETURNVERBOSE)
Output:
2/(-c^5*b^5*ln(F)^5+5*c^3*b^3*ln(F)^3*e^2-4*c*b*ln(F)*e^4)*f^2*F^(c*(b*x+a ))*(1/4*(ln(F)^4*b^4*c^4-ln(F)^2*b^2*c^2*e^2)*cosh(2*e*x+2*d)+1/2*(-ln(F)^ 3*b^3*c^3*e+ln(F)*b*c*e^3)*sinh(2*e*x+2*d)+(-I*sinh(e*x+d)*b^2*c^2*ln(F)^2 +I*cosh(e*x+d)*b*c*e*ln(F)-3/4*b^2*c^2*ln(F)^2+3/4*e^2)*(2*e+b*c*ln(F))*(b *c*ln(F)-2*e))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 441 vs. \(2 (188) = 376\).
Time = 0.09 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.30 \[ \int F^{c (a+b x)} (f+i f \sinh (d+e x))^2 \, dx=\frac {{\left (24 \, e^{4} f^{2} e^{\left (2 \, e x + 2 \, d\right )} - {\left (b^{4} c^{4} f^{2} e^{\left (4 \, e x + 4 \, d\right )} - 4 i \, b^{4} c^{4} f^{2} e^{\left (3 \, e x + 3 \, d\right )} - 6 \, b^{4} c^{4} f^{2} e^{\left (2 \, e x + 2 \, d\right )} + 4 i \, b^{4} c^{4} f^{2} e^{\left (e x + d\right )} + b^{4} c^{4} f^{2}\right )} \log \left (F\right )^{4} + 2 \, {\left (b^{3} c^{3} e f^{2} e^{\left (4 \, e x + 4 \, d\right )} - 2 i \, b^{3} c^{3} e f^{2} e^{\left (3 \, e x + 3 \, d\right )} - 2 i \, b^{3} c^{3} e f^{2} e^{\left (e x + d\right )} - b^{3} c^{3} e f^{2}\right )} \log \left (F\right )^{3} + {\left (b^{2} c^{2} e^{2} f^{2} e^{\left (4 \, e x + 4 \, d\right )} - 16 i \, b^{2} c^{2} e^{2} f^{2} e^{\left (3 \, e x + 3 \, d\right )} - 30 \, b^{2} c^{2} e^{2} f^{2} e^{\left (2 \, e x + 2 \, d\right )} + 16 i \, b^{2} c^{2} e^{2} f^{2} e^{\left (e x + d\right )} + b^{2} c^{2} e^{2} f^{2}\right )} \log \left (F\right )^{2} - 2 \, {\left (b c e^{3} f^{2} e^{\left (4 \, e x + 4 \, d\right )} - 8 i \, b c e^{3} f^{2} e^{\left (3 \, e x + 3 \, d\right )} - 8 i \, b c e^{3} f^{2} e^{\left (e x + d\right )} - b c e^{3} f^{2}\right )} \log \left (F\right )\right )} F^{b c x + a c}}{4 \, {\left (b^{5} c^{5} e^{\left (2 \, e x + 2 \, d\right )} \log \left (F\right )^{5} - 5 \, b^{3} c^{3} e^{2} e^{\left (2 \, e x + 2 \, d\right )} \log \left (F\right )^{3} + 4 \, b c e^{4} e^{\left (2 \, e x + 2 \, d\right )} \log \left (F\right )\right )}} \] Input:
integrate(F^(c*(b*x+a))*(f+I*f*sinh(e*x+d))^2,x, algorithm="fricas")
Output:
1/4*(24*e^4*f^2*e^(2*e*x + 2*d) - (b^4*c^4*f^2*e^(4*e*x + 4*d) - 4*I*b^4*c ^4*f^2*e^(3*e*x + 3*d) - 6*b^4*c^4*f^2*e^(2*e*x + 2*d) + 4*I*b^4*c^4*f^2*e ^(e*x + d) + b^4*c^4*f^2)*log(F)^4 + 2*(b^3*c^3*e*f^2*e^(4*e*x + 4*d) - 2* I*b^3*c^3*e*f^2*e^(3*e*x + 3*d) - 2*I*b^3*c^3*e*f^2*e^(e*x + d) - b^3*c^3* e*f^2)*log(F)^3 + (b^2*c^2*e^2*f^2*e^(4*e*x + 4*d) - 16*I*b^2*c^2*e^2*f^2* e^(3*e*x + 3*d) - 30*b^2*c^2*e^2*f^2*e^(2*e*x + 2*d) + 16*I*b^2*c^2*e^2*f^ 2*e^(e*x + d) + b^2*c^2*e^2*f^2)*log(F)^2 - 2*(b*c*e^3*f^2*e^(4*e*x + 4*d) - 8*I*b*c*e^3*f^2*e^(3*e*x + 3*d) - 8*I*b*c*e^3*f^2*e^(e*x + d) - b*c*e^3 *f^2)*log(F))*F^(b*c*x + a*c)/(b^5*c^5*e^(2*e*x + 2*d)*log(F)^5 - 5*b^3*c^ 3*e^2*e^(2*e*x + 2*d)*log(F)^3 + 4*b*c*e^4*e^(2*e*x + 2*d)*log(F))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2414 vs. \(2 (177) = 354\).
Time = 39.81 (sec) , antiderivative size = 2414, normalized size of antiderivative = 12.57 \[ \int F^{c (a+b x)} (f+i f \sinh (d+e x))^2 \, dx=\text {Too large to display} \] Input:
integrate(F**(c*(b*x+a))*(f+I*f*sinh(e*x+d))**2,x)
Output:
Piecewise((x*(I*f*sinh(d) + f)**2, Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(e, 0)), (-f**2*x*sinh(d + e*x)**2/2 + f**2*x*cosh(d + e*x)**2/2 + f**2*x - f* *2*sinh(d + e*x)*cosh(d + e*x)/(2*e) + 2*I*f**2*cosh(d + e*x)/e, Eq(F, 1)) , (F**(a*c)*(-f**2*x*sinh(d + e*x)**2/2 + f**2*x*cosh(d + e*x)**2/2 + f**2 *x - f**2*sinh(d + e*x)*cosh(d + e*x)/(2*e) + 2*I*f**2*cosh(d + e*x)/e), E q(b, 0)), (-f**2*x*sinh(d + e*x)**2/2 + f**2*x*cosh(d + e*x)**2/2 + f**2*x - f**2*sinh(d + e*x)*cosh(d + e*x)/(2*e) + 2*I*f**2*cosh(d + e*x)/e, Eq(c , 0)), (-I*F**(a*c + b*c*x)*f**2*x*sinh(b*c*x*log(F) - d) + I*F**(a*c + b* c*x)*f**2*x*cosh(b*c*x*log(F) - d) - F**(a*c + b*c*x)*f**2*sinh(b*c*x*log( F) - d)**2/(3*b*c*log(F)) - 2*F**(a*c + b*c*x)*f**2*sinh(b*c*x*log(F) - d) *cosh(b*c*x*log(F) - d)/(3*b*c*log(F)) + I*F**(a*c + b*c*x)*f**2*sinh(b*c* x*log(F) - d)/(b*c*log(F)) + 2*F**(a*c + b*c*x)*f**2*cosh(b*c*x*log(F) - d )**2/(3*b*c*log(F)) - 2*I*F**(a*c + b*c*x)*f**2*cosh(b*c*x*log(F) - d)/(b* c*log(F)) + F**(a*c + b*c*x)*f**2/(b*c*log(F)), Eq(e, -b*c*log(F))), (-F** (a*c + b*c*x)*f**2*x*sinh(b*c*x*log(F)/2 - d)**2/4 + F**(a*c + b*c*x)*f**2 *x*sinh(b*c*x*log(F)/2 - d)*cosh(b*c*x*log(F)/2 - d)/2 - F**(a*c + b*c*x)* f**2*x*cosh(b*c*x*log(F)/2 - d)**2/4 - F**(a*c + b*c*x)*f**2*sinh(b*c*x*lo g(F)/2 - d)**2/(b*c*log(F)) + F**(a*c + b*c*x)*f**2*sinh(b*c*x*log(F)/2 - d)*cosh(b*c*x*log(F)/2 - d)/(2*b*c*log(F)) - 8*I*F**(a*c + b*c*x)*f**2*sin h(b*c*x*log(F)/2 - d)/(3*b*c*log(F)) + 4*I*F**(a*c + b*c*x)*f**2*cosh(b...
Time = 0.05 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.98 \[ \int F^{c (a+b x)} (f+i f \sinh (d+e x))^2 \, dx=-\frac {1}{4} \, f^{2} {\left (\frac {F^{a c} e^{\left (b c x \log \left (F\right ) + 2 \, e x + 2 \, d\right )}}{b c \log \left (F\right ) + 2 \, e} + \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - 2 \, e x\right )}}{b c e^{\left (2 \, d\right )} \log \left (F\right ) - 2 \, e e^{\left (2 \, d\right )}} - \frac {2 \, F^{b c x + a c}}{b c \log \left (F\right )}\right )} + i \, f^{2} {\left (\frac {F^{a c} e^{\left (b c x \log \left (F\right ) + e x + d\right )}}{b c \log \left (F\right ) + e} - \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - e x\right )}}{b c e^{d} \log \left (F\right ) - e e^{d}}\right )} + \frac {F^{b c x + a c} f^{2}}{b c \log \left (F\right )} \] Input:
integrate(F^(c*(b*x+a))*(f+I*f*sinh(e*x+d))^2,x, algorithm="maxima")
Output:
-1/4*f^2*(F^(a*c)*e^(b*c*x*log(F) + 2*e*x + 2*d)/(b*c*log(F) + 2*e) + F^(a *c)*e^(b*c*x*log(F) - 2*e*x)/(b*c*e^(2*d)*log(F) - 2*e*e^(2*d)) - 2*F^(b*c *x + a*c)/(b*c*log(F))) + I*f^2*(F^(a*c)*e^(b*c*x*log(F) + e*x + d)/(b*c*l og(F) + e) - F^(a*c)*e^(b*c*x*log(F) - e*x)/(b*c*e^d*log(F) - e*e^d)) + F^ (b*c*x + a*c)*f^2/(b*c*log(F))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1546 vs. \(2 (188) = 376\).
Time = 0.17 (sec) , antiderivative size = 1546, normalized size of antiderivative = 8.05 \[ \int F^{c (a+b x)} (f+i f \sinh (d+e x))^2 \, dx=\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*(f+I*f*sinh(e*x+d))^2,x, algorithm="giac")
Output:
3*(2*b*c*f^2*cos(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b* c)^2) - (pi*b*c*sgn(F) - pi*b*c)*f^2*sin(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c *x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sg n(F) - pi*b*c)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 3*I*(I*f^2*e^ (1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a *c)/(2*I*pi*b*c*sgn(F) - 2*I*pi*b*c + 4*b*c*log(abs(F))) - I*f^2*e^(-1/2*I *pi*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c)/(- 2*I*pi*b*c*sgn(F) + 2*I*pi*b*c + 4*b*c*log(abs(F))))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) - 1/2*(2*(b*c*log(abs(F)) + 2*e)*f^2*cos(-1/2*pi*b*c*x* sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)/((pi*b*c*sgn(F) - pi*b*c)^2 + 4*(b*c*log(abs(F)) + 2*e)^2) - (pi*b*c*sgn(F) - pi*b*c)*f^2*si n(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)/(( pi*b*c*sgn(F) - pi*b*c)^2 + 4*(b*c*log(abs(F)) + 2*e)^2))*e^(a*c*log(abs(F )) + (b*c*log(abs(F)) + 2*e)*x + 2*d) + I*(-I*f^2*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*sgn(F) - 1/2*I*pi*a*c)/(4*I*pi*b*c*sgn(F) - 4*I*pi*b*c + 8*b*c*log(abs(F)) + 16*e) + I*f^2*e^(-1/2*I*pi*b*c*x*sgn(F ) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*sgn(F) + 1/2*I*pi*a*c)/(-4*I*pi*b*c*sgn( F) + 4*I*pi*b*c + 8*b*c*log(abs(F)) + 16*e))*e^(a*c*log(abs(F)) + (b*c*log (abs(F)) + 2*e)*x + 2*d) - 2*((pi*b*c*sgn(F) - pi*b*c)*f^2*cos(-1/2*pi*...
Time = 3.59 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.31 \[ \int F^{c (a+b x)} (f+i f \sinh (d+e x))^2 \, dx=\frac {F^{c\,\left (a+b\,x\right )}\,f^2\,\left (12\,e^4+3\,b^4\,c^4\,{\ln \left (F\right )}^4+b^4\,c^4\,\mathrm {sinh}\left (d+e\,x\right )\,{\ln \left (F\right )}^4\,4{}\mathrm {i}-b^4\,c^4\,{\ln \left (F\right )}^4\,\mathrm {cosh}\left (2\,d+2\,e\,x\right )-15\,b^2\,c^2\,e^2\,{\ln \left (F\right )}^2+2\,b^3\,c^3\,e\,{\ln \left (F\right )}^3\,\mathrm {sinh}\left (2\,d+2\,e\,x\right )-b^2\,c^2\,e^2\,\mathrm {sinh}\left (d+e\,x\right )\,{\ln \left (F\right )}^2\,16{}\mathrm {i}-2\,b\,c\,e^3\,\ln \left (F\right )\,\mathrm {sinh}\left (2\,d+2\,e\,x\right )+b^2\,c^2\,e^2\,{\ln \left (F\right )}^2\,\mathrm {cosh}\left (2\,d+2\,e\,x\right )-b^3\,c^3\,e\,\mathrm {cosh}\left (d+e\,x\right )\,{\ln \left (F\right )}^3\,4{}\mathrm {i}+b\,c\,e^3\,\mathrm {cosh}\left (d+e\,x\right )\,\ln \left (F\right )\,16{}\mathrm {i}\right )}{2\,b\,c\,\ln \left (F\right )\,\left (b^4\,c^4\,{\ln \left (F\right )}^4-5\,b^2\,c^2\,e^2\,{\ln \left (F\right )}^2+4\,e^4\right )} \] Input:
int(F^(c*(a + b*x))*(f + f*sinh(d + e*x)*1i)^2,x)
Output:
(F^(c*(a + b*x))*f^2*(12*e^4 + 3*b^4*c^4*log(F)^4 + b^4*c^4*sinh(d + e*x)* log(F)^4*4i - b^4*c^4*log(F)^4*cosh(2*d + 2*e*x) - 15*b^2*c^2*e^2*log(F)^2 + 2*b^3*c^3*e*log(F)^3*sinh(2*d + 2*e*x) - b^2*c^2*e^2*sinh(d + e*x)*log( F)^2*16i - 2*b*c*e^3*log(F)*sinh(2*d + 2*e*x) + b^2*c^2*e^2*log(F)^2*cosh( 2*d + 2*e*x) - b^3*c^3*e*cosh(d + e*x)*log(F)^3*4i + b*c*e^3*cosh(d + e*x) *log(F)*16i))/(2*b*c*log(F)*(4*e^4 + b^4*c^4*log(F)^4 - 5*b^2*c^2*e^2*log( F)^2))
\[ \int F^{c (a+b x)} (f+i f \sinh (d+e x))^2 \, dx=\frac {f^{a c} f^{2} \left (-2 f^{b c x} \cosh \left (e x +d \right ) \mathrm {log}\left (f \right ) b c e i +2 f^{b c x} \mathrm {log}\left (f \right )^{2} \sinh \left (e x +d \right ) b^{2} c^{2} i +f^{b c x} \mathrm {log}\left (f \right )^{2} b^{2} c^{2}-f^{b c x} e^{2}-\left (\int f^{b c x} \sinh \left (e x +d \right )^{2}d x \right ) \mathrm {log}\left (f \right )^{3} b^{3} c^{3}+\left (\int f^{b c x} \sinh \left (e x +d \right )^{2}d x \right ) \mathrm {log}\left (f \right ) b c \,e^{2}\right )}{\mathrm {log}\left (f \right ) b c \left (\mathrm {log}\left (f \right )^{2} b^{2} c^{2}-e^{2}\right )} \] Input:
int(F^(c*(b*x+a))*(f+I*f*sinh(e*x+d))^2,x)
Output:
(f**(a*c)*f**2*( - 2*f**(b*c*x)*cosh(d + e*x)*log(f)*b*c*e*i + 2*f**(b*c*x )*log(f)**2*sinh(d + e*x)*b**2*c**2*i + f**(b*c*x)*log(f)**2*b**2*c**2 - f **(b*c*x)*e**2 - int(f**(b*c*x)*sinh(d + e*x)**2,x)*log(f)**3*b**3*c**3 + int(f**(b*c*x)*sinh(d + e*x)**2,x)*log(f)*b*c*e**2))/(log(f)*b*c*(log(f)** 2*b**2*c**2 - e**2))