\(\int \frac {\sinh ^3(a+b x)}{(c+d x)^{5/2}} \, dx\) [58]

Optimal result

Integrand size = 18, antiderivative size = 277 \[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^{5/2}} \, dx=\frac {b^{3/2} e^{-a+\frac {b c}{d}} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 d^{5/2}}-\frac {b^{3/2} e^{-3 a+\frac {3 b c}{d}} \sqrt {3 \pi } \text {erf}\left (\frac {\sqrt {3} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 d^{5/2}}-\frac {b^{3/2} e^{a-\frac {b c}{d}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {b^{3/2} e^{3 a-\frac {3 b c}{d}} \sqrt {3 \pi } \text {erfi}\left (\frac {\sqrt {3} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 d^{5/2}}-\frac {4 b \cosh (a+b x) \sinh ^2(a+b x)}{d^2 \sqrt {c+d x}}-\frac {2 \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}} \] Output:

1/2*b^(3/2)*exp(-a+b*c/d)*Pi^(1/2)*erf(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^(5 
/2)-1/2*b^(3/2)*exp(-3*a+3*b*c/d)*3^(1/2)*Pi^(1/2)*erf(3^(1/2)*b^(1/2)*(d* 
x+c)^(1/2)/d^(1/2))/d^(5/2)-1/2*b^(3/2)*exp(a-b*c/d)*Pi^(1/2)*erfi(b^(1/2) 
*(d*x+c)^(1/2)/d^(1/2))/d^(5/2)+1/2*b^(3/2)*exp(3*a-3*b*c/d)*3^(1/2)*Pi^(1 
/2)*erfi(3^(1/2)*b^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^(5/2)-4*b*cosh(b*x+a)*si 
nh(b*x+a)^2/d^2/(d*x+c)^(1/2)-2/3*sinh(b*x+a)^3/d/(d*x+c)^(3/2)
 

Mathematica [A] (verified)

Time = 1.86 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.91 \[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^{5/2}} \, dx=\frac {e^{-3 \left (a+\frac {b c}{d}\right )} \left (-3 \sqrt {3} d e^{6 a} \left (-\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {3 b (c+d x)}{d}\right )+3 d e^{4 a+\frac {2 b c}{d}} \left (-\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {b (c+d x)}{d}\right )-3 d e^{2 a+\frac {4 b c}{d}} \left (\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {b (c+d x)}{d}\right )+3 \sqrt {3} d e^{\frac {6 b c}{d}} \left (\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {3 b (c+d x)}{d}\right )-4 e^{3 \left (a+\frac {b c}{d}\right )} \sinh ^2(a+b x) (6 b (c+d x) \cosh (a+b x)+d \sinh (a+b x))\right )}{6 d^2 (c+d x)^{3/2}} \] Input:

Integrate[Sinh[a + b*x]^3/(c + d*x)^(5/2),x]
 

Output:

(-3*Sqrt[3]*d*E^(6*a)*(-((b*(c + d*x))/d))^(3/2)*Gamma[1/2, (-3*b*(c + d*x 
))/d] + 3*d*E^(4*a + (2*b*c)/d)*(-((b*(c + d*x))/d))^(3/2)*Gamma[1/2, -((b 
*(c + d*x))/d)] - 3*d*E^(2*a + (4*b*c)/d)*((b*(c + d*x))/d)^(3/2)*Gamma[1/ 
2, (b*(c + d*x))/d] + 3*Sqrt[3]*d*E^((6*b*c)/d)*((b*(c + d*x))/d)^(3/2)*Ga 
mma[1/2, (3*b*(c + d*x))/d] - 4*E^(3*(a + (b*c)/d))*Sinh[a + b*x]^2*(6*b*( 
c + d*x)*Cosh[a + b*x] + d*Sinh[a + b*x]))/(6*d^2*E^(3*(a + (b*c)/d))*(c + 
 d*x)^(3/2))
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 1.26 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.52, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 26, 3795, 26, 3042, 26, 3789, 2611, 2633, 2634, 3793, 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.

\(\displaystyle \int \frac {\sinh ^3(a+b x)}{(c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {i \sin (i a+i b x)^3}{(c+d x)^{5/2}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int \frac {\sin (i a+i b x)^3}{(c+d x)^{5/2}}dx\)

\(\Big \downarrow \) 3795

\(\displaystyle i \left (\frac {12 b^2 \int -\frac {i \sinh ^3(a+b x)}{\sqrt {c+d x}}dx}{d^2}-\frac {8 b^2 \int \frac {i \sinh (a+b x)}{\sqrt {c+d x}}dx}{d^2}+\frac {4 i b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt {c+d x}}+\frac {2 i \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle i \left (-\frac {12 i b^2 \int \frac {\sinh ^3(a+b x)}{\sqrt {c+d x}}dx}{d^2}-\frac {8 i b^2 \int \frac {\sinh (a+b x)}{\sqrt {c+d x}}dx}{d^2}+\frac {4 i b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt {c+d x}}+\frac {2 i \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle i \left (-\frac {8 i b^2 \int -\frac {i \sin (i a+i b x)}{\sqrt {c+d x}}dx}{d^2}-\frac {12 i b^2 \int \frac {i \sin (i a+i b x)^3}{\sqrt {c+d x}}dx}{d^2}+\frac {4 i b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt {c+d x}}+\frac {2 i \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle i \left (-\frac {8 b^2 \int \frac {\sin (i a+i b x)}{\sqrt {c+d x}}dx}{d^2}+\frac {12 b^2 \int \frac {\sin (i a+i b x)^3}{\sqrt {c+d x}}dx}{d^2}+\frac {4 i b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt {c+d x}}+\frac {2 i \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 3789

\(\displaystyle i \left (-\frac {8 b^2 \left (\frac {1}{2} i \int \frac {e^{a+b x}}{\sqrt {c+d x}}dx-\frac {1}{2} i \int \frac {e^{-a-b x}}{\sqrt {c+d x}}dx\right )}{d^2}+\frac {12 b^2 \int \frac {\sin (i a+i b x)^3}{\sqrt {c+d x}}dx}{d^2}+\frac {4 i b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt {c+d x}}+\frac {2 i \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 2611

\(\displaystyle i \left (-\frac {8 b^2 \left (\frac {i \int e^{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{d}-\frac {i \int e^{-a-\frac {b (c+d x)}{d}+\frac {b c}{d}}d\sqrt {c+d x}}{d}\right )}{d^2}+\frac {12 b^2 \int \frac {\sin (i a+i b x)^3}{\sqrt {c+d x}}dx}{d^2}+\frac {4 i b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt {c+d x}}+\frac {2 i \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 2633

\(\displaystyle i \left (-\frac {8 b^2 \left (\frac {i \sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {i \int e^{-a-\frac {b (c+d x)}{d}+\frac {b c}{d}}d\sqrt {c+d x}}{d}\right )}{d^2}+\frac {12 b^2 \int \frac {\sin (i a+i b x)^3}{\sqrt {c+d x}}dx}{d^2}+\frac {4 i b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt {c+d x}}+\frac {2 i \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 2634

\(\displaystyle i \left (\frac {12 b^2 \int \frac {\sin (i a+i b x)^3}{\sqrt {c+d x}}dx}{d^2}-\frac {8 b^2 \left (\frac {i \sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {i \sqrt {\pi } e^{\frac {b c}{d}-a} \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}\right )}{d^2}+\frac {4 i b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt {c+d x}}+\frac {2 i \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 3793

\(\displaystyle i \left (\frac {12 b^2 \int \left (\frac {3 i \sinh (a+b x)}{4 \sqrt {c+d x}}-\frac {i \sinh (3 a+3 b x)}{4 \sqrt {c+d x}}\right )dx}{d^2}-\frac {8 b^2 \left (\frac {i \sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {i \sqrt {\pi } e^{\frac {b c}{d}-a} \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}\right )}{d^2}+\frac {4 i b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt {c+d x}}+\frac {2 i \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle i \left (-\frac {8 b^2 \left (\frac {i \sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {i \sqrt {\pi } e^{\frac {b c}{d}-a} \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}\right )}{d^2}+\frac {12 b^2 \left (-\frac {3 i \sqrt {\pi } e^{\frac {b c}{d}-a} \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{8 \sqrt {b} \sqrt {d}}+\frac {i \sqrt {\frac {\pi }{3}} e^{\frac {3 b c}{d}-3 a} \text {erf}\left (\frac {\sqrt {3} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{8 \sqrt {b} \sqrt {d}}+\frac {3 i \sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{8 \sqrt {b} \sqrt {d}}-\frac {i \sqrt {\frac {\pi }{3}} e^{3 a-\frac {3 b c}{d}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{8 \sqrt {b} \sqrt {d}}\right )}{d^2}+\frac {4 i b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt {c+d x}}+\frac {2 i \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}}\right )\)

Input:

Int[Sinh[a + b*x]^3/(c + d*x)^(5/2),x]
 

Output:

I*((-8*b^2*(((-1/2*I)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x] 
)/Sqrt[d]])/(Sqrt[b]*Sqrt[d]) + ((I/2)*E^(a - (b*c)/d)*Sqrt[Pi]*Erfi[(Sqrt 
[b]*Sqrt[c + d*x])/Sqrt[d]])/(Sqrt[b]*Sqrt[d])))/d^2 + (12*b^2*((((-3*I)/8 
)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(Sqrt[b] 
*Sqrt[d]) + ((I/8)*E^(-3*a + (3*b*c)/d)*Sqrt[Pi/3]*Erf[(Sqrt[3]*Sqrt[b]*Sq 
rt[c + d*x])/Sqrt[d]])/(Sqrt[b]*Sqrt[d]) + (((3*I)/8)*E^(a - (b*c)/d)*Sqrt 
[Pi]*Erfi[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(Sqrt[b]*Sqrt[d]) - ((I/8)*E^( 
3*a - (3*b*c)/d)*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]]) 
/(Sqrt[b]*Sqrt[d])))/d^2 + ((4*I)*b*Cosh[a + b*x]*Sinh[a + b*x]^2)/(d^2*Sq 
rt[c + d*x]) + (((2*I)/3)*Sinh[a + b*x]^3)/(d*(c + d*x)^(3/2)))
 

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 2611

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] : 
> Simp[2/d   Subst[Int[F^(g*(e - c*(f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d 
*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

rule 2633

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ 
F, a, b, c, d}, x] && PosQ[b]
 

rule 2634

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F], 2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; Fr 
eeQ[{F, a, b, c, d}, x] && NegQ[b]
 

rule 3042

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

rule 3789

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
/2   Int[(c + d*x)^m/E^(I*(e + f*x)), x], x] - Simp[I/2   Int[(c + d*x)^m*E 
^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]
 

rule 3793

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

rule 3795

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo 
l] :> Simp[(c + d*x)^(m + 1)*((b*Sin[e + f*x])^n/(d*(m + 1))), x] + (-Simp[ 
b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(d^2*(m + 1) 
*(m + 2))), x] + Simp[b^2*f^2*n*((n - 1)/(d^2*(m + 1)*(m + 2)))   Int[(c + 
d*x)^(m + 2)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[f^2*(n^2/(d^2*(m + 1)* 
(m + 2)))   Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x], x]) /; FreeQ[{b, 
c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]
 

Maple [F]

\[\int \frac {\sinh \left (b x +a \right )^{3}}{\left (d x +c \right )^{\frac {5}{2}}}d x\]

Input:

int(sinh(b*x+a)^3/(d*x+c)^(5/2),x)
 

Output:

int(sinh(b*x+a)^3/(d*x+c)^(5/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2059 vs. \(2 (209) = 418\).

Time = 0.13 (sec) , antiderivative size = 2059, normalized size of antiderivative = 7.43 \[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(sinh(b*x+a)^3/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

-1/12*(6*sqrt(3)*sqrt(pi)*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)^3 
*cosh(-3*(b*c - a*d)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)^3* 
sinh(-3*(b*c - a*d)/d) + ((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(-3*(b*c - a 
*d)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*sinh(-3*(b*c - a*d)/d))*sinh(b*x 
+ a)^3 + 3*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*cosh(-3*(b*c - a 
*d)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*sinh(-3*(b*c - a*d) 
/d))*sinh(b*x + a)^2 + 3*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)^2* 
cosh(-3*(b*c - a*d)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)^2*s 
inh(-3*(b*c - a*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(3)*sqrt(d*x + c)* 
sqrt(b/d)) + 6*sqrt(3)*sqrt(pi)*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x 
+ a)^3*cosh(-3*(b*c - a*d)/d) + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + 
 a)^3*sinh(-3*(b*c - a*d)/d) + ((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(-3*(b 
*c - a*d)/d) + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*sinh(-3*(b*c - a*d)/d))*sin 
h(b*x + a)^3 + 3*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*cosh(-3*(b 
*c - a*d)/d) + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*sinh(-3*(b*c 
- a*d)/d))*sinh(b*x + a)^2 + 3*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + 
 a)^2*cosh(-3*(b*c - a*d)/d) + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + 
a)^2*sinh(-3*(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(3)*sqrt(d* 
x + c)*sqrt(-b/d)) - 6*sqrt(pi)*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x 
+ a)^3*cosh(-(b*c - a*d)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x ...
 

Sympy [F]

\[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^{5/2}} \, dx=\int \frac {\sinh ^{3}{\left (a + b x \right )}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(sinh(b*x+a)**3/(d*x+c)**(5/2),x)
 

Output:

Integral(sinh(a + b*x)**3/(c + d*x)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.71 \[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^{5/2}} \, dx=\frac {3 \, {\left (\frac {\sqrt {3} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (\frac {3 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {3}{2}, \frac {3 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}} - \frac {\sqrt {3} \left (-\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {3}{2}, -\frac {3 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}} - \frac {\left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (-a + \frac {b c}{d}\right )} \Gamma \left (-\frac {3}{2}, \frac {{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}} + \frac {\left (-\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (a - \frac {b c}{d}\right )} \Gamma \left (-\frac {3}{2}, -\frac {{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}}\right )}}{8 \, d} \] Input:

integrate(sinh(b*x+a)^3/(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^{5/2}} \, dx=\int { \frac {\sinh \left (b x + a\right )^{3}}{{\left (d x + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(sinh(b*x+a)^3/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

integrate(sinh(b*x + a)^3/(d*x + c)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^{5/2}} \, dx=\int \frac {{\mathrm {sinh}\left (a+b\,x\right )}^3}{{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(sinh(a + b*x)^3/(c + d*x)^(5/2),x)
 

Output:

int(sinh(a + b*x)^3/(c + d*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\sinh ^3(a+b x)}{(c+d x)^{5/2}} \, dx=\int \frac {\sinh \left (b x +a \right )^{3}}{\sqrt {d x +c}\, c^{2}+2 \sqrt {d x +c}\, c d x +\sqrt {d x +c}\, d^{2} x^{2}}d x \] Input:

int(sinh(b*x+a)^3/(d*x+c)^(5/2),x)
 

Output:

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