3.1.38 \(\int (c+d x)^{5/2} \sinh (a+b x) \, dx\) [38]

3.1.38.1 Optimal result

Integrand size = 16, antiderivative size = 171 \[ \int (c+d x)^{5/2} \sinh (a+b x) \, dx=\frac {15 d^2 \sqrt {c+d x} \cosh (a+b x)}{4 b^3}+\frac {(c+d x)^{5/2} \cosh (a+b x)}{b}-\frac {15 d^{5/2} e^{-a+\frac {b c}{d}} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{7/2}}-\frac {15 d^{5/2} e^{a-\frac {b c}{d}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{7/2}}-\frac {5 d (c+d x)^{3/2} \sinh (a+b x)}{2 b^2} \]

output

(d*x+c)^(5/2)*cosh(b*x+a)/b-5/2*d*(d*x+c)^(3/2)*sinh(b*x+a)/b^2-15/16*d^(5 
/2)*exp(-a+b*c/d)*erf(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))*Pi^(1/2)/b^(7/2)-15/1 
6*d^(5/2)*exp(a-b*c/d)*erfi(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))*Pi^(1/2)/b^(7/2 
)+15/4*d^2*cosh(b*x+a)*(d*x+c)^(1/2)/b^3
 

3.1.38.2 Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.63 \[ \int (c+d x)^{5/2} \sinh (a+b x) \, dx=\frac {d^3 e^{-a-\frac {b c}{d}} \left (-e^{2 a} \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {7}{2},-\frac {b (c+d x)}{d}\right )+e^{\frac {2 b c}{d}} \sqrt {\frac {b (c+d x)}{d}} \Gamma \left (\frac {7}{2},\frac {b (c+d x)}{d}\right )\right )}{2 b^4 \sqrt {c+d x}} \]

input

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

output

(d^3*E^(-a - (b*c)/d)*(-(E^(2*a)*Sqrt[-((b*(c + d*x))/d)]*Gamma[7/2, -((b* 
(c + d*x))/d)]) + E^((2*b*c)/d)*Sqrt[(b*(c + d*x))/d]*Gamma[7/2, (b*(c + d 
*x))/d]))/(2*b^4*Sqrt[c + d*x])
 

3.1.38.3 Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.83 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.19, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.938, Rules used = {3042, 26, 3777, 3042, 3777, 26, 3042, 26, 3777, 3042, 3788, 26, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3788

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 2611

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

\(\Big \downarrow \) 2633

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

\(\Big \downarrow \) 2634

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

input

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

output

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

3.1.38.3.1 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 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 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 3788

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

3.1.38.4 Maple [F]

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

input

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

output

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

3.1.38.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 521 vs. \(2 (131) = 262\).

Time = 0.26 (sec) , antiderivative size = 521, normalized size of antiderivative = 3.05 \[ \int (c+d x)^{5/2} \sinh (a+b x) \, dx=-\frac {15 \, \sqrt {\pi } {\left (d^{3} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) - d^{3} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left (d^{3} \cosh \left (-\frac {b c - a d}{d}\right ) - d^{3} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) - 15 \, \sqrt {\pi } {\left (d^{3} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) + d^{3} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left (d^{3} \cosh \left (-\frac {b c - a d}{d}\right ) + d^{3} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {-\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) - 2 \, {\left (4 \, b^{3} d^{2} x^{2} + 4 \, b^{3} c^{2} + 10 \, b^{2} c d + 15 \, b d^{2} + {\left (4 \, b^{3} d^{2} x^{2} + 4 \, b^{3} c^{2} - 10 \, b^{2} c d + 15 \, b d^{2} + 2 \, {\left (4 \, b^{3} c d - 5 \, b^{2} d^{2}\right )} x\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (4 \, b^{3} d^{2} x^{2} + 4 \, b^{3} c^{2} - 10 \, b^{2} c d + 15 \, b d^{2} + 2 \, {\left (4 \, b^{3} c d - 5 \, b^{2} d^{2}\right )} x\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (4 \, b^{3} d^{2} x^{2} + 4 \, b^{3} c^{2} - 10 \, b^{2} c d + 15 \, b d^{2} + 2 \, {\left (4 \, b^{3} c d - 5 \, b^{2} d^{2}\right )} x\right )} \sinh \left (b x + a\right )^{2} + 2 \, {\left (4 \, b^{3} c d + 5 \, b^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{16 \, {\left (b^{4} \cosh \left (b x + a\right ) + b^{4} \sinh \left (b x + a\right )\right )}} \]

input

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

output

-1/16*(15*sqrt(pi)*(d^3*cosh(b*x + a)*cosh(-(b*c - a*d)/d) - d^3*cosh(b*x 
+ a)*sinh(-(b*c - a*d)/d) + (d^3*cosh(-(b*c - a*d)/d) - d^3*sinh(-(b*c - a 
*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(d*x + c)*sqrt(b/d)) - 15*sqrt(pi 
)*(d^3*cosh(b*x + a)*cosh(-(b*c - a*d)/d) + d^3*cosh(b*x + a)*sinh(-(b*c - 
 a*d)/d) + (d^3*cosh(-(b*c - a*d)/d) + d^3*sinh(-(b*c - a*d)/d))*sinh(b*x 
+ a))*sqrt(-b/d)*erf(sqrt(d*x + c)*sqrt(-b/d)) - 2*(4*b^3*d^2*x^2 + 4*b^3* 
c^2 + 10*b^2*c*d + 15*b*d^2 + (4*b^3*d^2*x^2 + 4*b^3*c^2 - 10*b^2*c*d + 15 
*b*d^2 + 2*(4*b^3*c*d - 5*b^2*d^2)*x)*cosh(b*x + a)^2 + 2*(4*b^3*d^2*x^2 + 
 4*b^3*c^2 - 10*b^2*c*d + 15*b*d^2 + 2*(4*b^3*c*d - 5*b^2*d^2)*x)*cosh(b*x 
 + a)*sinh(b*x + a) + (4*b^3*d^2*x^2 + 4*b^3*c^2 - 10*b^2*c*d + 15*b*d^2 + 
 2*(4*b^3*c*d - 5*b^2*d^2)*x)*sinh(b*x + a)^2 + 2*(4*b^3*c*d + 5*b^2*d^2)* 
x)*sqrt(d*x + c))/(b^4*cosh(b*x + a) + b^4*sinh(b*x + a))
 

3.1.38.6 Sympy [F]

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

input

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

output

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

3.1.38.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (131) = 262\).

Time = 0.22 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.80 \[ \int (c+d x)^{5/2} \sinh (a+b x) \, dx=\frac {32 \, {\left (d x + c\right )}^{\frac {7}{2}} \sinh \left (b x + a\right ) - \frac {{\left (\frac {105 \, \sqrt {\pi } d^{4} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) e^{\left (a - \frac {b c}{d}\right )}}{b^{4} \sqrt {-\frac {b}{d}}} + \frac {105 \, \sqrt {\pi } d^{4} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) e^{\left (-a + \frac {b c}{d}\right )}}{b^{4} \sqrt {\frac {b}{d}}} - \frac {2 \, {\left (8 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{3} d e^{\left (\frac {b c}{d}\right )} + 28 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{2} e^{\left (\frac {b c}{d}\right )} + 70 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{3} e^{\left (\frac {b c}{d}\right )} + 105 \, \sqrt {d x + c} d^{4} e^{\left (\frac {b c}{d}\right )}\right )} e^{\left (-a - \frac {{\left (d x + c\right )} b}{d}\right )}}{b^{4}} + \frac {2 \, {\left (8 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{3} d e^{a} - 28 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{2} e^{a} + 70 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{3} e^{a} - 105 \, \sqrt {d x + c} d^{4} e^{a}\right )} e^{\left (\frac {{\left (d x + c\right )} b}{d} - \frac {b c}{d}\right )}}{b^{4}}\right )} b}{d}}{112 \, d} \]

input

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

output

1/112*(32*(d*x + c)^(7/2)*sinh(b*x + a) - (105*sqrt(pi)*d^4*erf(sqrt(d*x + 
 c)*sqrt(-b/d))*e^(a - b*c/d)/(b^4*sqrt(-b/d)) + 105*sqrt(pi)*d^4*erf(sqrt 
(d*x + c)*sqrt(b/d))*e^(-a + b*c/d)/(b^4*sqrt(b/d)) - 2*(8*(d*x + c)^(7/2) 
*b^3*d*e^(b*c/d) + 28*(d*x + c)^(5/2)*b^2*d^2*e^(b*c/d) + 70*(d*x + c)^(3/ 
2)*b*d^3*e^(b*c/d) + 105*sqrt(d*x + c)*d^4*e^(b*c/d))*e^(-a - (d*x + c)*b/ 
d)/b^4 + 2*(8*(d*x + c)^(7/2)*b^3*d*e^a - 28*(d*x + c)^(5/2)*b^2*d^2*e^a + 
 70*(d*x + c)^(3/2)*b*d^3*e^a - 105*sqrt(d*x + c)*d^4*e^a)*e^((d*x + c)*b/ 
d - b*c/d)/b^4)*b/d)/d
 

3.1.38.8 Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.36 \[ \int (c+d x)^{5/2} \sinh (a+b x) \, dx=\frac {\frac {15 \, \sqrt {\pi } d^{4} \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c}}{d}\right ) e^{\left (\frac {b c - a d}{d}\right )}}{\sqrt {b d} b^{3}} + \frac {15 \, \sqrt {\pi } d^{4} \operatorname {erf}\left (-\frac {\sqrt {-b d} \sqrt {d x + c}}{d}\right ) e^{\left (-\frac {b c - a d}{d}\right )}}{\sqrt {-b d} b^{3}} + \frac {2 \, {\left (4 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} + 15 \, \sqrt {d x + c} d^{3}\right )} e^{\left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right )}}{b^{3}} + \frac {2 \, {\left (4 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d + 10 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} + 15 \, \sqrt {d x + c} d^{3}\right )} e^{\left (-\frac {{\left (d x + c\right )} b - b c + a d}{d}\right )}}{b^{3}}}{16 \, d} \]

input

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

output

1/16*(15*sqrt(pi)*d^4*erf(-sqrt(b*d)*sqrt(d*x + c)/d)*e^((b*c - a*d)/d)/(s 
qrt(b*d)*b^3) + 15*sqrt(pi)*d^4*erf(-sqrt(-b*d)*sqrt(d*x + c)/d)*e^(-(b*c 
- a*d)/d)/(sqrt(-b*d)*b^3) + 2*(4*(d*x + c)^(5/2)*b^2*d - 10*(d*x + c)^(3/ 
2)*b*d^2 + 15*sqrt(d*x + c)*d^3)*e^(((d*x + c)*b - b*c + a*d)/d)/b^3 + 2*( 
4*(d*x + c)^(5/2)*b^2*d + 10*(d*x + c)^(3/2)*b*d^2 + 15*sqrt(d*x + c)*d^3) 
*e^(-((d*x + c)*b - b*c + a*d)/d)/b^3)/d
 

3.1.38.9 Mupad [F(-1)]

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

input

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

output

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