Integrand size = 18, antiderivative size = 132 \[ \int F^{c (a+b x)} \sinh ^2(d+e x) \, dx=-\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (4 e^2-b^2 c^2 \log ^2(F)\right )}+\frac {2 e F^{c (a+b x)} \cosh (d+e x) \sinh (d+e x)}{4 e^2-b^2 c^2 \log ^2(F)}-\frac {b c F^{c (a+b x)} \log (F) \sinh ^2(d+e x)}{4 e^2-b^2 c^2 \log ^2(F)} \] Output:
-2*e^2*F^(c*(b*x+a))/b/c/ln(F)/(4*e^2-b^2*c^2*ln(F)^2)+2*e*F^(c*(b*x+a))*c osh(e*x+d)*sinh(e*x+d)/(4*e^2-b^2*c^2*ln(F)^2)-b*c*F^(c*(b*x+a))*ln(F)*sin h(e*x+d)^2/(4*e^2-b^2*c^2*ln(F)^2)
Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.65 \[ \int F^{c (a+b x)} \sinh ^2(d+e x) \, dx=\frac {F^{c (a+b x)} \left (4 e^2-b^2 c^2 \log ^2(F)+b^2 c^2 \cosh (2 (d+e x)) \log ^2(F)-2 b c e \log (F) \sinh (2 (d+e x))\right )}{-8 b c e^2 \log (F)+2 b^3 c^3 \log ^3(F)} \] Input:
Integrate[F^(c*(a + b*x))*Sinh[d + e*x]^2,x]
Output:
(F^(c*(a + b*x))*(4*e^2 - b^2*c^2*Log[F]^2 + b^2*c^2*Cosh[2*(d + e*x)]*Log [F]^2 - 2*b*c*e*Log[F]*Sinh[2*(d + e*x)]))/(-8*b*c*e^2*Log[F] + 2*b^3*c^3* Log[F]^3)
Time = 0.33 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5999, 2624}
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.
\(\displaystyle \int \sinh ^2(d+e x) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 5999 |
\(\displaystyle -\frac {2 e^2 \int F^{c (a+b x)}dx}{4 e^2-b^2 c^2 \log ^2(F)}-\frac {b c \log (F) \sinh ^2(d+e x) F^{c (a+b x)}}{4 e^2-b^2 c^2 \log ^2(F)}+\frac {2 e \sinh (d+e x) \cosh (d+e x) F^{c (a+b x)}}{4 e^2-b^2 c^2 \log ^2(F)}\) |
\(\Big \downarrow \) 2624 |
\(\displaystyle -\frac {b c \log (F) \sinh ^2(d+e x) F^{c (a+b x)}}{4 e^2-b^2 c^2 \log ^2(F)}+\frac {2 e \sinh (d+e x) \cosh (d+e x) F^{c (a+b x)}}{4 e^2-b^2 c^2 \log ^2(F)}-\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (4 e^2-b^2 c^2 \log ^2(F)\right )}\) |
Input:
Int[F^(c*(a + b*x))*Sinh[d + e*x]^2,x]
Output:
(-2*e^2*F^(c*(a + b*x)))/(b*c*Log[F]*(4*e^2 - b^2*c^2*Log[F]^2)) + (2*e*F^ (c*(a + b*x))*Cosh[d + e*x]*Sinh[d + e*x])/(4*e^2 - b^2*c^2*Log[F]^2) - (b *c*F^(c*(a + b*x))*Log[F]*Sinh[d + e*x]^2)/(4*e^2 - b^2*c^2*Log[F]^2)
Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; FreeQ[{F, n}, x] && LinearQ[v, x]
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sinh[(d_.) + (e_.)*(x_)]^(n_), x_Symb ol] :> Simp[(-b)*c*Log[F]*F^(c*(a + b*x))*(Sinh[d + e*x]^n/(e^2*n^2 - b^2*c ^2*Log[F]^2)), x] + (Simp[e*n*F^(c*(a + b*x))*Cosh[d + e*x]*(Sinh[d + e*x]^ (n - 1)/(e^2*n^2 - b^2*c^2*Log[F]^2)), x] - Simp[n*(n - 1)*(e^2/(e^2*n^2 - b^2*c^2*Log[F]^2)) Int[F^(c*(a + b*x))*Sinh[d + e*x]^(n - 2), x], x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 - b^2*c^2*Log[F]^2, 0] && GtQ[n , 1]
Time = 0.54 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.68
method | result | size |
parallelrisch | \(-\frac {2 F^{c \left (b x +a \right )} \left (-\frac {b^{2} c^{2} \ln \left (F \right )^{2} \cosh \left (2 e x +2 d \right )}{2}+\frac {b^{2} c^{2} \ln \left (F \right )^{2}}{2}+e b c \ln \left (F \right ) \sinh \left (2 e x +2 d \right )-2 e^{2}\right )}{2 \ln \left (F \right )^{3} b^{3} c^{3}-8 \ln \left (F \right ) b c \,e^{2}}\) | \(90\) |
risch | \(\frac {\left (\ln \left (F \right )^{2} b^{2} c^{2} {\mathrm e}^{4 e x +4 d}-2 \ln \left (F \right )^{2} b^{2} c^{2} {\mathrm e}^{2 e x +2 d}-2 \ln \left (F \right ) b c e \,{\mathrm e}^{4 e x +4 d}+b^{2} c^{2} \ln \left (F \right )^{2}+2 e b c \ln \left (F \right )+8 e^{2} {\mathrm e}^{2 e x +2 d}\right ) {\mathrm e}^{-2 e x -2 d} F^{c \left (b x +a \right )}}{4 \ln \left (F \right ) b c \left (b c \ln \left (F \right )-2 e \right ) \left (2 e +b c \ln \left (F \right )\right )}\) | \(143\) |
orering | \(\frac {\left (3 b^{2} c^{2} \ln \left (F \right )^{2}-4 e^{2}\right ) F^{c \left (b x +a \right )} \sinh \left (e x +d \right )^{2}}{\ln \left (F \right ) b c \left (b^{2} c^{2} \ln \left (F \right )^{2}-4 e^{2}\right )}-\frac {3 \left (F^{c \left (b x +a \right )} b c \ln \left (F \right ) \sinh \left (e x +d \right )^{2}+2 F^{c \left (b x +a \right )} \sinh \left (e x +d \right ) e \cosh \left (e x +d \right )\right )}{b^{2} c^{2} \ln \left (F \right )^{2}-4 e^{2}}+\frac {F^{c \left (b x +a \right )} b^{2} c^{2} \ln \left (F \right )^{2} \sinh \left (e x +d \right )^{2}+4 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \sinh \left (e x +d \right ) e \cosh \left (e x +d \right )+2 F^{c \left (b x +a \right )} e^{2} \cosh \left (e x +d \right )^{2}+2 F^{c \left (b x +a \right )} \sinh \left (e x +d \right )^{2} e^{2}}{\ln \left (F \right ) b c \left (b^{2} c^{2} \ln \left (F \right )^{2}-4 e^{2}\right )}\) | \(266\) |
Input:
int(F^(c*(b*x+a))*sinh(e*x+d)^2,x,method=_RETURNVERBOSE)
Output:
-2*F^(c*(b*x+a))*(-1/2*b^2*c^2*ln(F)^2*cosh(2*e*x+2*d)+1/2*b^2*c^2*ln(F)^2 +e*b*c*ln(F)*sinh(2*e*x+2*d)-2*e^2)/(2*ln(F)^3*b^3*c^3-8*ln(F)*b*c*e^2)
Leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (128) = 256\).
Time = 0.09 (sec) , antiderivative size = 703, normalized size of antiderivative = 5.33 \[ \int F^{c (a+b x)} \sinh ^2(d+e x) \, dx =\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*sinh(e*x+d)^2,x, algorithm="fricas")
Output:
1/4*(((b^2*c^2*log(F)^2 - 2*b*c*e*log(F))*sinh(e*x + d)^4 + 8*e^2*cosh(e*x + d)^2 + 4*(b^2*c^2*cosh(e*x + d)*log(F)^2 - 2*b*c*e*cosh(e*x + d)*log(F) )*sinh(e*x + d)^3 + (b^2*c^2*cosh(e*x + d)^4 - 2*b^2*c^2*cosh(e*x + d)^2 + b^2*c^2)*log(F)^2 - 2*(6*b*c*e*cosh(e*x + d)^2*log(F) - (3*b^2*c^2*cosh(e *x + d)^2 - b^2*c^2)*log(F)^2 - 4*e^2)*sinh(e*x + d)^2 - 2*(b*c*e*cosh(e*x + d)^4 - b*c*e)*log(F) - 4*(2*b*c*e*cosh(e*x + d)^3*log(F) - 4*e^2*cosh(e *x + d) - (b^2*c^2*cosh(e*x + d)^3 - b^2*c^2*cosh(e*x + d))*log(F)^2)*sinh (e*x + d))*cosh((b*c*x + a*c)*log(F)) + ((b^2*c^2*log(F)^2 - 2*b*c*e*log(F ))*sinh(e*x + d)^4 + 8*e^2*cosh(e*x + d)^2 + 4*(b^2*c^2*cosh(e*x + d)*log( F)^2 - 2*b*c*e*cosh(e*x + d)*log(F))*sinh(e*x + d)^3 + (b^2*c^2*cosh(e*x + d)^4 - 2*b^2*c^2*cosh(e*x + d)^2 + b^2*c^2)*log(F)^2 - 2*(6*b*c*e*cosh(e* x + d)^2*log(F) - (3*b^2*c^2*cosh(e*x + d)^2 - b^2*c^2)*log(F)^2 - 4*e^2)* sinh(e*x + d)^2 - 2*(b*c*e*cosh(e*x + d)^4 - b*c*e)*log(F) - 4*(2*b*c*e*co sh(e*x + d)^3*log(F) - 4*e^2*cosh(e*x + d) - (b^2*c^2*cosh(e*x + d)^3 - b^ 2*c^2*cosh(e*x + d))*log(F)^2)*sinh(e*x + d))*sinh((b*c*x + a*c)*log(F)))/ (b^3*c^3*cosh(e*x + d)^2*log(F)^3 - 4*b*c*e^2*cosh(e*x + d)^2*log(F) + (b^ 3*c^3*log(F)^3 - 4*b*c*e^2*log(F))*sinh(e*x + d)^2 + 2*(b^3*c^3*cosh(e*x + d)*log(F)^3 - 4*b*c*e^2*cosh(e*x + d)*log(F))*sinh(e*x + d))
Leaf count of result is larger than twice the leaf count of optimal. 707 vs. \(2 (119) = 238\).
Time = 1.26 (sec) , antiderivative size = 707, normalized size of antiderivative = 5.36 \[ \int F^{c (a+b x)} \sinh ^2(d+e x) \, dx =\text {Too large to display} \] Input:
integrate(F**(c*(b*x+a))*sinh(e*x+d)**2,x)
Output:
Piecewise((x*sinh(d)**2, Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(e, 0)), (x*si nh(d + e*x)**2/2 - x*cosh(d + e*x)**2/2 + sinh(d + e*x)*cosh(d + e*x)/(2*e ), Eq(F, 1)), (F**(a*c)*(x*sinh(d + e*x)**2/2 - x*cosh(d + e*x)**2/2 + sin h(d + e*x)*cosh(d + e*x)/(2*e)), Eq(b, 0)), (x*sinh(d + e*x)**2/2 - x*cosh (d + e*x)**2/2 + sinh(d + e*x)*cosh(d + e*x)/(2*e), Eq(c, 0)), (F**(a*c + b*c*x)*x*sinh(b*c*x*log(F)/2 - d)**2/4 - F**(a*c + b*c*x)*x*sinh(b*c*x*log (F)/2 - d)*cosh(b*c*x*log(F)/2 - d)/2 + F**(a*c + b*c*x)*x*cosh(b*c*x*log( F)/2 - d)**2/4 + 3*F**(a*c + b*c*x)*sinh(b*c*x*log(F)/2 - d)*cosh(b*c*x*lo g(F)/2 - d)/(2*b*c*log(F)) - F**(a*c + b*c*x)*cosh(b*c*x*log(F)/2 - d)**2/ (b*c*log(F)), Eq(e, -b*c*log(F)/2)), (F**(a*c + b*c*x)*x*sinh(b*c*x*log(F) /2 + d)**2/4 - F**(a*c + b*c*x)*x*sinh(b*c*x*log(F)/2 + d)*cosh(b*c*x*log( F)/2 + d)/2 + F**(a*c + b*c*x)*x*cosh(b*c*x*log(F)/2 + d)**2/4 + F**(a*c + b*c*x)*sinh(b*c*x*log(F)/2 + d)**2/(b*c*log(F)) - F**(a*c + b*c*x)*sinh(b *c*x*log(F)/2 + d)*cosh(b*c*x*log(F)/2 + d)/(2*b*c*log(F)), Eq(e, b*c*log( F)/2)), (F**(a*c + b*c*x)*b**2*c**2*log(F)**2*sinh(d + e*x)**2/(b**3*c**3* log(F)**3 - 4*b*c*e**2*log(F)) - 2*F**(a*c + b*c*x)*b*c*e*log(F)*sinh(d + e*x)*cosh(d + e*x)/(b**3*c**3*log(F)**3 - 4*b*c*e**2*log(F)) - 2*F**(a*c + b*c*x)*e**2*sinh(d + e*x)**2/(b**3*c**3*log(F)**3 - 4*b*c*e**2*log(F)) + 2*F**(a*c + b*c*x)*e**2*cosh(d + e*x)**2/(b**3*c**3*log(F)**3 - 4*b*c*e**2 *log(F)), True))
Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.71 \[ \int F^{c (a+b x)} \sinh ^2(d+e x) \, dx=\frac {F^{a c} e^{\left (b c x \log \left (F\right ) + 2 \, e x + 2 \, d\right )}}{4 \, {\left (b c \log \left (F\right ) + 2 \, e\right )}} + \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - 2 \, e x\right )}}{4 \, {\left (b c e^{\left (2 \, d\right )} \log \left (F\right ) - 2 \, e e^{\left (2 \, d\right )}\right )}} - \frac {F^{b c x + a c}}{2 \, b c \log \left (F\right )} \] Input:
integrate(F^(c*(b*x+a))*sinh(e*x+d)^2,x, algorithm="maxima")
Output:
1/4*F^(a*c)*e^(b*c*x*log(F) + 2*e*x + 2*d)/(b*c*log(F) + 2*e) + 1/4*F^(a*c )*e^(b*c*x*log(F) - 2*e*x)/(b*c*e^(2*d)*log(F) - 2*e*e^(2*d)) - 1/2*F^(b*c *x + a*c)/(b*c*log(F))
Result contains complex when optimal does not.
Time = 0.15 (sec) , antiderivative size = 890, normalized size of antiderivative = 6.74 \[ \int F^{c (a+b x)} \sinh ^2(d+e x) \, dx=\text {Too large to display} \] Input:
integrate(F^(c*(b*x+a))*sinh(e*x+d)^2,x, algorithm="giac")
Output:
-(2*b*c*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)*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*sgn(F) - pi *b*c)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + I*(-I*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*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)*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)*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)/((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*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*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c*s gn(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) + 1/2*(2*( b*c*log(abs(F)) - 2*e)*cos(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi...
Time = 1.91 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.73 \[ \int F^{c (a+b x)} \sinh ^2(d+e x) \, dx=-\frac {F^{a\,c+b\,c\,x}\,\left (2\,e^2-\frac {b^2\,c^2\,{\ln \left (F\right )}^2}{2}+\frac {b^2\,c^2\,{\ln \left (F\right )}^2\,\mathrm {cosh}\left (2\,d+2\,e\,x\right )}{2}-b\,c\,e\,\ln \left (F\right )\,\mathrm {sinh}\left (2\,d+2\,e\,x\right )\right )}{b\,c\,\ln \left (F\right )\,\left (4\,e^2-b^2\,c^2\,{\ln \left (F\right )}^2\right )} \] Input:
int(F^(c*(a + b*x))*sinh(d + e*x)^2,x)
Output:
-(F^(a*c + b*c*x)*(2*e^2 - (b^2*c^2*log(F)^2)/2 + (b^2*c^2*log(F)^2*cosh(2 *d + 2*e*x))/2 - b*c*e*log(F)*sinh(2*d + 2*e*x)))/(b*c*log(F)*(4*e^2 - b^2 *c^2*log(F)^2))
\[ \int F^{c (a+b x)} \sinh ^2(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \sinh \left (e x +d \right )^{2}d x \right ) \] Input:
int(F^(c*(b*x+a))*sinh(e*x+d)^2,x)
Output:
f**(a*c)*int(f**(b*c*x)*sinh(d + e*x)**2,x)