\(\int \frac {\sinh ^2(a+b x)}{(c+d x)^{7/2}} \, dx\) [51]

Optimal result

Integrand size = 18, antiderivative size = 220 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}-\frac {8 b^{5/2} e^{-2 a+\frac {2 b c}{d}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}+\frac {8 b^{5/2} e^{2 a-\frac {2 b c}{d}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {32 b^2 \sinh ^2(a+b x)}{15 d^3 \sqrt {c+d x}} \] Output:

-16/15*b^2/d^3/(d*x+c)^(1/2)-8/15*b^(5/2)*exp(-2*a+2*b*c/d)*2^(1/2)*Pi^(1/ 
2)*erf(2^(1/2)*b^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^(7/2)+8/15*b^(5/2)*exp(2*a 
-2*b*c/d)*2^(1/2)*Pi^(1/2)*erfi(2^(1/2)*b^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^( 
7/2)-8/15*b*cosh(b*x+a)*sinh(b*x+a)/d^2/(d*x+c)^(3/2)-2/5*sinh(b*x+a)^2/d/ 
(d*x+c)^(5/2)-32/15*b^2*sinh(b*x+a)^2/d^3/(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.93 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\frac {6 d^2+e^{2 a} \left (-3 d^2 e^{2 b x}-4 b e^{-\frac {2 b c}{d}} (c+d x) \left (e^{\frac {2 b (c+d x)}{d}} (d+4 b (c+d x))+4 \sqrt {2} d \left (-\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 b (c+d x)}{d}\right )\right )\right )+e^{-2 (a+b x)} \left (-3 d^2+4 b (c+d x) \left (d-4 b (c+d x)+4 \sqrt {2} d e^{\frac {2 b (c+d x)}{d}} \left (\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {2 b (c+d x)}{d}\right )\right )\right )}{30 d^3 (c+d x)^{5/2}} \] Input:

Integrate[Sinh[a + b*x]^2/(c + d*x)^(7/2),x]
 

Output:

(6*d^2 + E^(2*a)*(-3*d^2*E^(2*b*x) - (4*b*(c + d*x)*(E^((2*b*(c + d*x))/d) 
*(d + 4*b*(c + d*x)) + 4*Sqrt[2]*d*(-((b*(c + d*x))/d))^(3/2)*Gamma[1/2, ( 
-2*b*(c + d*x))/d]))/E^((2*b*c)/d)) + (-3*d^2 + 4*b*(c + d*x)*(d - 4*b*(c 
+ d*x) + 4*Sqrt[2]*d*E^((2*b*(c + d*x))/d)*((b*(c + d*x))/d)^(3/2)*Gamma[1 
/2, (2*b*(c + d*x))/d]))/E^(2*(a + b*x)))/(30*d^3*(c + d*x)^(5/2))
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.80 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.10, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {3042, 25, 3795, 17, 25, 3042, 25, 3794, 27, 3042, 26, 3789, 2611, 2633, 2634}

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.

\(\displaystyle \int \frac {\sinh ^2(a+b x)}{(c+d x)^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {\sin (i a+i b x)^2}{(c+d x)^{7/2}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {\sin (i a+i b x)^2}{(c+d x)^{7/2}}dx\)

\(\Big \downarrow \) 3795

\(\displaystyle -\frac {16 b^2 \int -\frac {\sinh ^2(a+b x)}{(c+d x)^{3/2}}dx}{15 d^2}+\frac {8 b^2 \int \frac {1}{(c+d x)^{3/2}}dx}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}\)

\(\Big \downarrow \) 17

\(\displaystyle -\frac {16 b^2 \int -\frac {\sinh ^2(a+b x)}{(c+d x)^{3/2}}dx}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {16 b^2 \int \frac {\sinh ^2(a+b x)}{(c+d x)^{3/2}}dx}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {16 b^2 \int -\frac {\sin (i a+i b x)^2}{(c+d x)^{3/2}}dx}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {16 b^2 \int \frac {\sin (i a+i b x)^2}{(c+d x)^{3/2}}dx}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

\(\Big \downarrow \) 3794

\(\displaystyle -\frac {16 b^2 \left (\frac {2 \sinh ^2(a+b x)}{d \sqrt {c+d x}}+\frac {4 i b \int \frac {i \sinh (2 a+2 b x)}{2 \sqrt {c+d x}}dx}{d}\right )}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {16 b^2 \left (\frac {2 \sinh ^2(a+b x)}{d \sqrt {c+d x}}-\frac {2 b \int \frac {\sinh (2 a+2 b x)}{\sqrt {c+d x}}dx}{d}\right )}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {16 b^2 \left (\frac {2 \sinh ^2(a+b x)}{d \sqrt {c+d x}}-\frac {2 b \int -\frac {i \sin (2 i a+2 i b x)}{\sqrt {c+d x}}dx}{d}\right )}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {16 b^2 \left (\frac {2 \sinh ^2(a+b x)}{d \sqrt {c+d x}}+\frac {2 i b \int \frac {\sin (2 i a+2 i b x)}{\sqrt {c+d x}}dx}{d}\right )}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

\(\Big \downarrow \) 3789

\(\displaystyle -\frac {16 b^2 \left (\frac {2 \sinh ^2(a+b x)}{d \sqrt {c+d x}}+\frac {2 i b \left (\frac {1}{2} i \int \frac {e^{2 (a+b x)}}{\sqrt {c+d x}}dx-\frac {1}{2} i \int \frac {e^{-2 (a+b x)}}{\sqrt {c+d x}}dx\right )}{d}\right )}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

\(\Big \downarrow \) 2611

\(\displaystyle -\frac {16 b^2 \left (\frac {2 \sinh ^2(a+b x)}{d \sqrt {c+d x}}+\frac {2 i b \left (\frac {i \int e^{2 \left (a-\frac {b c}{d}\right )+\frac {2 b (c+d x)}{d}}d\sqrt {c+d x}}{d}-\frac {i \int e^{-2 \left (a-\frac {b c}{d}\right )-\frac {2 b (c+d x)}{d}}d\sqrt {c+d x}}{d}\right )}{d}\right )}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

\(\Big \downarrow \) 2633

\(\displaystyle -\frac {16 b^2 \left (\frac {2 \sinh ^2(a+b x)}{d \sqrt {c+d x}}+\frac {2 i b \left (\frac {i \sqrt {\frac {\pi }{2}} e^{2 a-\frac {2 b c}{d}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {i \int e^{-2 \left (a-\frac {b c}{d}\right )-\frac {2 b (c+d x)}{d}}d\sqrt {c+d x}}{d}\right )}{d}\right )}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

\(\Big \downarrow \) 2634

\(\displaystyle -\frac {16 b^2 \left (\frac {2 \sinh ^2(a+b x)}{d \sqrt {c+d x}}+\frac {2 i b \left (\frac {i \sqrt {\frac {\pi }{2}} e^{2 a-\frac {2 b c}{d}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {i \sqrt {\frac {\pi }{2}} e^{\frac {2 b c}{d}-2 a} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}\right )}{d}\right )}{15 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac {16 b^2}{15 d^3 \sqrt {c+d x}}\)

Input:

Int[Sinh[a + b*x]^2/(c + d*x)^(7/2),x]
 

Output:

(-16*b^2)/(15*d^3*Sqrt[c + d*x]) - (8*b*Cosh[a + b*x]*Sinh[a + b*x])/(15*d 
^2*(c + d*x)^(3/2)) - (2*Sinh[a + b*x]^2)/(5*d*(c + d*x)^(5/2)) - (16*b^2* 
(((2*I)*b*(((-1/2*I)*E^(-2*a + (2*b*c)/d)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[b]* 
Sqrt[c + d*x])/Sqrt[d]])/(Sqrt[b]*Sqrt[d]) + ((I/2)*E^(2*a - (2*b*c)/d)*Sq 
rt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(Sqrt[b]*Sqrt[d])) 
)/d + (2*Sinh[a + b*x]^2)/(d*Sqrt[c + d*x])))/(15*d^2)
 

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 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]
 

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]

\[\int \frac {\sinh \left (b x +a \right )^{2}}{\left (d x +c \right )^{\frac {7}{2}}}d x\]

Input:

int(sinh(b*x+a)^2/(d*x+c)^(7/2),x)
 

Output:

int(sinh(b*x+a)^2/(d*x+c)^(7/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1352 vs. \(2 (172) = 344\).

Time = 0.13 (sec) , antiderivative size = 1352, normalized size of antiderivative = 6.15 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate(sinh(b*x+a)^2/(d*x+c)^(7/2),x, algorithm="fricas")
 

Output:

-1/30*(16*sqrt(2)*sqrt(pi)*((b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x 
 + b^2*c^3)*cosh(b*x + a)^2*cosh(-2*(b*c - a*d)/d) - (b^2*d^3*x^3 + 3*b^2* 
c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)^2*sinh(-2*(b*c - a*d)/d 
) + ((b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(-2*(b* 
c - a*d)/d) - (b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*si 
nh(-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*((b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 
+ 3*b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)*cosh(-2*(b*c - a*d)/d) - (b^2*d^3 
*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)*sinh(-2*(b 
*c - a*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(2)*sqrt(d*x + c)*sqrt(b/d) 
) + 16*sqrt(2)*sqrt(pi)*((b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + 
b^2*c^3)*cosh(b*x + a)^2*cosh(-2*(b*c - a*d)/d) + (b^2*d^3*x^3 + 3*b^2*c*d 
^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)^2*sinh(-2*(b*c - a*d)/d) + 
 ((b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(-2*(b*c - 
 a*d)/d) + (b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*sinh( 
-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*((b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3 
*b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)*cosh(-2*(b*c - a*d)/d) + (b^2*d^3*x^ 
3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)*sinh(-2*(b*c 
- a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(2)*sqrt(d*x + c)*sqrt(-b/d)) 
 + (16*b^2*d^2*x^2 + (16*b^2*d^2*x^2 + 16*b^2*c^2 + 4*b*c*d + 3*d^2 + 4*(8 
*b^2*c*d + b*d^2)*x)*cosh(b*x + a)^4 + 4*(16*b^2*d^2*x^2 + 16*b^2*c^2 +...
 

Sympy [F]

\[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {\sinh ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{\frac {7}{2}}}\, dx \] Input:

integrate(sinh(b*x+a)**2/(d*x+c)**(7/2),x)
 

Output:

Integral(sinh(a + b*x)**2/(c + d*x)**(7/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.54 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=-\frac {\frac {5 \, \sqrt {2} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {5}{2}} e^{\left (\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {5}{2}, \frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {5}{2}}} + \frac {5 \, \sqrt {2} \left (-\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {5}{2}} e^{\left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {5}{2}, -\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {5}{2}}} - \frac {1}{{\left (d x + c\right )}^{\frac {5}{2}}}}{5 \, d} \] Input:

integrate(sinh(b*x+a)^2/(d*x+c)^(7/2),x, algorithm="maxima")
 

Output:

-1/5*(5*sqrt(2)*((d*x + c)*b/d)^(5/2)*e^(2*(b*c - a*d)/d)*gamma(-5/2, 2*(d 
*x + c)*b/d)/(d*x + c)^(5/2) + 5*sqrt(2)*(-(d*x + c)*b/d)^(5/2)*e^(-2*(b*c 
 - a*d)/d)*gamma(-5/2, -2*(d*x + c)*b/d)/(d*x + c)^(5/2) - 1/(d*x + c)^(5/ 
2))/d
 

Giac [F]

\[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int { \frac {\sinh \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate(sinh(b*x+a)^2/(d*x+c)^(7/2),x, algorithm="giac")
 

Output:

integrate(sinh(b*x + a)^2/(d*x + c)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {{\mathrm {sinh}\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{7/2}} \,d x \] Input:

int(sinh(a + b*x)^2/(c + d*x)^(7/2),x)
 

Output:

int(sinh(a + b*x)^2/(c + d*x)^(7/2), x)
 

Reduce [F]

\[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {\sinh \left (b x +a \right )^{2}}{\sqrt {d x +c}\, c^{3}+3 \sqrt {d x +c}\, c^{2} d x +3 \sqrt {d x +c}\, c \,d^{2} x^{2}+\sqrt {d x +c}\, d^{3} x^{3}}d x \] Input:

int(sinh(b*x+a)^2/(d*x+c)^(7/2),x)
 

Output:

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