\(\int \frac {\sinh ^3(a+b x)}{(c+d x)^2} \, dx\) [21]

Optimal result

Integrand size = 16, antiderivative size = 145 \[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^2} \, dx=-\frac {3 b \cosh \left (a-\frac {b c}{d}\right ) \text {Chi}\left (\frac {b c}{d}+b x\right )}{4 d^2}+\frac {3 b \cosh \left (3 a-\frac {3 b c}{d}\right ) \text {Chi}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2}-\frac {\sinh ^3(a+b x)}{d (c+d x)}-\frac {3 b \sinh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (\frac {b c}{d}+b x\right )}{4 d^2}+\frac {3 b \sinh \left (3 a-\frac {3 b c}{d}\right ) \text {Shi}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2} \] Output:

-3/4*b*cosh(a-b*c/d)*Chi(b*c/d+b*x)/d^2+3/4*b*cosh(3*a-3*b*c/d)*Chi(3*b*c/ 
d+3*b*x)/d^2-sinh(b*x+a)^3/d/(d*x+c)-3/4*b*sinh(a-b*c/d)*Shi(b*c/d+b*x)/d^ 
2+3/4*b*sinh(3*a-3*b*c/d)*Shi(3*b*c/d+3*b*x)/d^2
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.10 \[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^2} \, dx=\frac {6 d \cosh (b x) \sinh (a)-2 d \cosh (3 b x) \sinh (3 a)+6 d \cosh (a) \sinh (b x)-2 d \cosh (3 a) \sinh (3 b x)+6 b (c+d x) \left (-\cosh \left (a-\frac {b c}{d}\right ) \text {Chi}\left (b \left (\frac {c}{d}+x\right )\right )+\cosh \left (3 a-\frac {3 b c}{d}\right ) \text {Chi}\left (\frac {3 b (c+d x)}{d}\right )-\sinh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (b \left (\frac {c}{d}+x\right )\right )+\sinh \left (3 a-\frac {3 b c}{d}\right ) \text {Shi}\left (\frac {3 b (c+d x)}{d}\right )\right )}{8 d^2 (c+d x)} \] Input:

Integrate[Sinh[a + b*x]^3/(c + d*x)^2,x]
 

Output:

(6*d*Cosh[b*x]*Sinh[a] - 2*d*Cosh[3*b*x]*Sinh[3*a] + 6*d*Cosh[a]*Sinh[b*x] 
 - 2*d*Cosh[3*a]*Sinh[3*b*x] + 6*b*(c + d*x)*(-(Cosh[a - (b*c)/d]*CoshInte 
gral[b*(c/d + x)]) + Cosh[3*a - (3*b*c)/d]*CoshIntegral[(3*b*(c + d*x))/d] 
 - Sinh[a - (b*c)/d]*SinhIntegral[b*(c/d + x)] + Sinh[3*a - (3*b*c)/d]*Sin 
hIntegral[(3*b*(c + d*x))/d]))/(8*d^2*(c + d*x))
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.49 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 26, 3794, 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.

\(\displaystyle \int \frac {\sinh ^3(a+b x)}{(c+d x)^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i \sin (i a+i b x)^3}{(c+d x)^2}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {\sin (i a+i b x)^3}{(c+d x)^2}dx\)

\(\Big \downarrow \) 3794

\(\displaystyle i \left (\frac {3 i b \int \left (\frac {\cosh (a+b x)}{4 (c+d x)}-\frac {\cosh (3 a+3 b x)}{4 (c+d x)}\right )dx}{d}+\frac {i \sinh ^3(a+b x)}{d (c+d x)}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle i \left (\frac {3 i b \left (\frac {\cosh \left (a-\frac {b c}{d}\right ) \text {Chi}\left (\frac {b c}{d}+b x\right )}{4 d}-\frac {\cosh \left (3 a-\frac {3 b c}{d}\right ) \text {Chi}\left (\frac {3 b c}{d}+3 b x\right )}{4 d}+\frac {\sinh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (\frac {b c}{d}+b x\right )}{4 d}-\frac {\sinh \left (3 a-\frac {3 b c}{d}\right ) \text {Shi}\left (\frac {3 b c}{d}+3 b x\right )}{4 d}\right )}{d}+\frac {i \sinh ^3(a+b x)}{d (c+d x)}\right )\)

Input:

Int[Sinh[a + b*x]^3/(c + d*x)^2,x]
 

Output:

I*((I*Sinh[a + b*x]^3)/(d*(c + d*x)) + ((3*I)*b*((Cosh[a - (b*c)/d]*CoshIn 
tegral[(b*c)/d + b*x])/(4*d) - (Cosh[3*a - (3*b*c)/d]*CoshIntegral[(3*b*c) 
/d + 3*b*x])/(4*d) + (Sinh[a - (b*c)/d]*SinhIntegral[(b*c)/d + b*x])/(4*d) 
 - (Sinh[3*a - (3*b*c)/d]*SinhIntegral[(3*b*c)/d + 3*b*x])/(4*d)))/d)
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3794

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Si 
mp[(c + d*x)^(m + 1)*(Sin[e + f*x]^n/(d*(m + 1))), x] - Simp[f*(n/(d*(m + 1 
)))   Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]^(n 
- 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] & 
& LtQ[m, -1]
 

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.87

method	result	size
risch	\(\frac {b \,{\mathrm e}^{-3 b x -3 a}}{8 d \left (d x b +b c \right )}-\frac {3 b \,{\mathrm e}^{-\frac {3 \left (d a -b c \right )}{d}} \operatorname {expIntegral}_{1}\left (3 b x +3 a -\frac {3 \left (d a -b c \right )}{d}\right )}{8 d^{2}}-\frac {3 b \,{\mathrm e}^{-b x -a}}{8 d \left (d x b +b c \right )}+\frac {3 b \,{\mathrm e}^{-\frac {d a -b c}{d}} \operatorname {expIntegral}_{1}\left (b x +a -\frac {d a -b c}{d}\right )}{8 d^{2}}+\frac {3 b \,{\mathrm e}^{b x +a}}{8 d^{2} \left (\frac {b c}{d}+b x \right )}+\frac {3 b \,{\mathrm e}^{\frac {d a -b c}{d}} \operatorname {expIntegral}_{1}\left (-b x -a -\frac {-d a +b c}{d}\right )}{8 d^{2}}-\frac {b \,{\mathrm e}^{3 b x +3 a}}{8 d^{2} \left (\frac {b c}{d}+b x \right )}-\frac {3 b \,{\mathrm e}^{\frac {3 d a -3 b c}{d}} \operatorname {expIntegral}_{1}\left (-3 b x -3 a -\frac {3 \left (-d a +b c \right )}{d}\right )}{8 d^{2}}\)	\(271\)

Input:

int(sinh(b*x+a)^3/(d*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

1/8*b*exp(-3*b*x-3*a)/d/(b*d*x+b*c)-3/8*b/d^2*exp(-3*(a*d-b*c)/d)*Ei(1,3*b 
*x+3*a-3*(a*d-b*c)/d)-3/8*b*exp(-b*x-a)/d/(b*d*x+b*c)+3/8*b/d^2*exp(-(a*d- 
b*c)/d)*Ei(1,b*x+a-(a*d-b*c)/d)+3/8*b/d^2*exp(b*x+a)/(b*c/d+b*x)+3/8*b/d^2 
*exp((a*d-b*c)/d)*Ei(1,-b*x-a-(-a*d+b*c)/d)-1/8*b/d^2*exp(3*b*x+3*a)/(b*c/ 
d+b*x)-3/8*b/d^2*exp(3*(a*d-b*c)/d)*Ei(1,-3*b*x-3*a-3*(-a*d+b*c)/d)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (137) = 274\).

Time = 0.08 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.08 \[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^2} \, dx=-\frac {2 \, d \sinh \left (b x + a\right )^{3} + 3 \, {\left ({\left (b d x + b c\right )} {\rm Ei}\left (\frac {b d x + b c}{d}\right ) + {\left (b d x + b c\right )} {\rm Ei}\left (-\frac {b d x + b c}{d}\right )\right )} \cosh \left (-\frac {b c - a d}{d}\right ) - 3 \, {\left ({\left (b d x + b c\right )} {\rm Ei}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b d x + b c\right )} {\rm Ei}\left (-\frac {3 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cosh \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + 6 \, {\left (d \cosh \left (b x + a\right )^{2} - d\right )} \sinh \left (b x + a\right ) + 3 \, {\left ({\left (b d x + b c\right )} {\rm Ei}\left (\frac {b d x + b c}{d}\right ) - {\left (b d x + b c\right )} {\rm Ei}\left (-\frac {b d x + b c}{d}\right )\right )} \sinh \left (-\frac {b c - a d}{d}\right ) - 3 \, {\left ({\left (b d x + b c\right )} {\rm Ei}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) - {\left (b d x + b c\right )} {\rm Ei}\left (-\frac {3 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sinh \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{8 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:

integrate(sinh(b*x+a)^3/(d*x+c)^2,x, algorithm="fricas")
 

Output:

-1/8*(2*d*sinh(b*x + a)^3 + 3*((b*d*x + b*c)*Ei((b*d*x + b*c)/d) + (b*d*x 
+ b*c)*Ei(-(b*d*x + b*c)/d))*cosh(-(b*c - a*d)/d) - 3*((b*d*x + b*c)*Ei(3* 
(b*d*x + b*c)/d) + (b*d*x + b*c)*Ei(-3*(b*d*x + b*c)/d))*cosh(-3*(b*c - a* 
d)/d) + 6*(d*cosh(b*x + a)^2 - d)*sinh(b*x + a) + 3*((b*d*x + b*c)*Ei((b*d 
*x + b*c)/d) - (b*d*x + b*c)*Ei(-(b*d*x + b*c)/d))*sinh(-(b*c - a*d)/d) - 
3*((b*d*x + b*c)*Ei(3*(b*d*x + b*c)/d) - (b*d*x + b*c)*Ei(-3*(b*d*x + b*c) 
/d))*sinh(-3*(b*c - a*d)/d))/(d^3*x + c*d^2)
 

Sympy [F]

\[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^2} \, dx=\int \frac {\sinh ^{3}{\left (a + b x \right )}}{\left (c + d x\right )^{2}}\, dx \] Input:

integrate(sinh(b*x+a)**3/(d*x+c)**2,x)
 

Output:

Integral(sinh(a + b*x)**3/(c + d*x)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00 \[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^2} \, dx=\frac {e^{\left (-3 \, a + \frac {3 \, b c}{d}\right )} E_{2}\left (\frac {3 \, {\left (d x + c\right )} b}{d}\right )}{8 \, {\left (d x + c\right )} d} - \frac {3 \, e^{\left (-a + \frac {b c}{d}\right )} E_{2}\left (\frac {{\left (d x + c\right )} b}{d}\right )}{8 \, {\left (d x + c\right )} d} + \frac {3 \, e^{\left (a - \frac {b c}{d}\right )} E_{2}\left (-\frac {{\left (d x + c\right )} b}{d}\right )}{8 \, {\left (d x + c\right )} d} - \frac {e^{\left (3 \, a - \frac {3 \, b c}{d}\right )} E_{2}\left (-\frac {3 \, {\left (d x + c\right )} b}{d}\right )}{8 \, {\left (d x + c\right )} d} \] Input:

integrate(sinh(b*x+a)^3/(d*x+c)^2,x, algorithm="maxima")
 

Output:

1/8*e^(-3*a + 3*b*c/d)*exp_integral_e(2, 3*(d*x + c)*b/d)/((d*x + c)*d) - 
3/8*e^(-a + b*c/d)*exp_integral_e(2, (d*x + c)*b/d)/((d*x + c)*d) + 3/8*e^ 
(a - b*c/d)*exp_integral_e(2, -(d*x + c)*b/d)/((d*x + c)*d) - 1/8*e^(3*a - 
 3*b*c/d)*exp_integral_e(2, -3*(d*x + c)*b/d)/((d*x + c)*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1076 vs. \(2 (137) = 274\).

Time = 0.18 (sec) , antiderivative size = 1076, normalized size of antiderivative = 7.42 \[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^2} \, dx=\text {Too large to display} \] Input:

integrate(sinh(b*x+a)^3/(d*x+c)^2,x, algorithm="giac")
 

Output:

1/8*(3*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*Ei(-3*((d*x + c)* 
(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d)*e^(3*(b*c - a*d)/d) + 
3*b^3*c*Ei(-3*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/ 
d)*e^(3*(b*c - a*d)/d) - 3*a*b^2*d*Ei(-3*((d*x + c)*(b - b*c/(d*x + c) + a 
*d/(d*x + c)) + b*c - a*d)/d)*e^(3*(b*c - a*d)/d) - 3*(d*x + c)*(b - b*c/( 
d*x + c) + a*d/(d*x + c))*b^2*Ei(-((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x 
 + c)) + b*c - a*d)/d)*e^((b*c - a*d)/d) - 3*b^3*c*Ei(-((d*x + c)*(b - b*c 
/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d)*e^((b*c - a*d)/d) + 3*a*b^2*d* 
Ei(-((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d)*e^((b*c 
 - a*d)/d) - 3*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*Ei(((d*x 
+ c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d)*e^(-(b*c - a*d)/d 
) - 3*b^3*c*Ei(((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d) 
/d)*e^(-(b*c - a*d)/d) + 3*a*b^2*d*Ei(((d*x + c)*(b - b*c/(d*x + c) + a*d/ 
(d*x + c)) + b*c - a*d)/d)*e^(-(b*c - a*d)/d) + 3*(d*x + c)*(b - b*c/(d*x 
+ c) + a*d/(d*x + c))*b^2*Ei(3*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + 
c)) + b*c - a*d)/d)*e^(-3*(b*c - a*d)/d) + 3*b^3*c*Ei(3*((d*x + c)*(b - b* 
c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d)*e^(-3*(b*c - a*d)/d) - 3*a*b^ 
2*d*Ei(3*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d)*e^ 
(-3*(b*c - a*d)/d) - b^2*d*e^(3*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + 
c))/d) + 3*b^2*d*e^((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))/d) - ...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^2} \, dx=\int \frac {{\mathrm {sinh}\left (a+b\,x\right )}^3}{{\left (c+d\,x\right )}^2} \,d x \] Input:

int(sinh(a + b*x)^3/(c + d*x)^2,x)
 

Output:

int(sinh(a + b*x)^3/(c + d*x)^2, x)
 

Reduce [F]

\[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^2} \, dx=\frac {e^{4 a} \left (\int \frac {e^{3 b x}}{d^{2} x^{2}+2 c d x +c^{2}}d x \right )-3 e^{2 a} \left (\int \frac {e^{b x}}{d^{2} x^{2}+2 c d x +c^{2}}d x \right )-e^{a} \left (\int \frac {1}{e^{3 b x +3 a} c^{2}+2 e^{3 b x +3 a} c d x +e^{3 b x +3 a} d^{2} x^{2}}d x \right )+3 \left (\int \frac {1}{e^{b x} c^{2}+2 e^{b x} c d x +e^{b x} d^{2} x^{2}}d x \right )}{8 e^{a}} \] Input:

int(sinh(b*x+a)^3/(d*x+c)^2,x)
 

Output:

(e**(4*a)*int(e**(3*b*x)/(c**2 + 2*c*d*x + d**2*x**2),x) - 3*e**(2*a)*int( 
e**(b*x)/(c**2 + 2*c*d*x + d**2*x**2),x) - e**a*int(1/(e**(3*a + 3*b*x)*c* 
*2 + 2*e**(3*a + 3*b*x)*c*d*x + e**(3*a + 3*b*x)*d**2*x**2),x) + 3*int(1/( 
e**(b*x)*c**2 + 2*e**(b*x)*c*d*x + e**(b*x)*d**2*x**2),x))/(8*e**a)