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

Optimal. Leaf size=277 \[ \frac{\sqrt{\pi } b^{3/2} e^{\frac{b c}{d}-a} \text{Erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 d^{5/2}}-\frac{\sqrt{3 \pi } b^{3/2} e^{\frac{3 b c}{d}-3 a} \text{Erf}\left (\frac{\sqrt{3} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 d^{5/2}}-\frac{\sqrt{\pi } b^{3/2} e^{a-\frac{b c}{d}} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 d^{5/2}}+\frac{\sqrt{3 \pi } b^{3/2} e^{3 a-\frac{3 b c}{d}} \text{Erfi}\left (\frac{\sqrt{3} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 d^{5/2}}-\frac{4 b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt{c+d x}}-\frac{2 \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}} \]

[Out]

(b^(3/2)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(2*d^(5/2)) - (b^(3/2)*E^(-3*a + (3*b
*c)/d)*Sqrt[3*Pi]*Erf[(Sqrt[3]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(2*d^(5/2)) - (b^(3/2)*E^(a - (b*c)/d)*Sqrt[Pi
]*Erfi[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(2*d^(5/2)) + (b^(3/2)*E^(3*a - (3*b*c)/d)*Sqrt[3*Pi]*Erfi[(Sqrt[3]*S
qrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(2*d^(5/2)) - (4*b*Cosh[a + b*x]*Sinh[a + b*x]^2)/(d^2*Sqrt[c + d*x]) - (2*Sin
h[a + b*x]^3)/(3*d*(c + d*x)^(3/2))

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Rubi [A]  time = 0.672687, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3314, 3308, 2180, 2204, 2205, 3312} \[ \frac{\sqrt{\pi } b^{3/2} e^{\frac{b c}{d}-a} \text{Erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 d^{5/2}}-\frac{\sqrt{3 \pi } b^{3/2} e^{\frac{3 b c}{d}-3 a} \text{Erf}\left (\frac{\sqrt{3} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 d^{5/2}}-\frac{\sqrt{\pi } b^{3/2} e^{a-\frac{b c}{d}} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 d^{5/2}}+\frac{\sqrt{3 \pi } b^{3/2} e^{3 a-\frac{3 b c}{d}} \text{Erfi}\left (\frac{\sqrt{3} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{2 d^{5/2}}-\frac{4 b \sinh ^2(a+b x) \cosh (a+b x)}{d^2 \sqrt{c+d x}}-\frac{2 \sinh ^3(a+b x)}{3 d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^(3/2)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(2*d^(5/2)) - (b^(3/2)*E^(-3*a + (3*b
*c)/d)*Sqrt[3*Pi]*Erf[(Sqrt[3]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(2*d^(5/2)) - (b^(3/2)*E^(a - (b*c)/d)*Sqrt[Pi
]*Erfi[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(2*d^(5/2)) + (b^(3/2)*E^(3*a - (3*b*c)/d)*Sqrt[3*Pi]*Erfi[(Sqrt[3]*S
qrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(2*d^(5/2)) - (4*b*Cosh[a + b*x]*Sinh[a + b*x]^2)/(d^2*Sqrt[c + d*x]) - (2*Sin
h[a + b*x]^3)/(3*d*(c + d*x)^(3/2))

Rule 3314

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

Rule 3308

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

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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] &&  !$UseGamma === True

Rule 2204

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 2205

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] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[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])
)

Rubi steps

\begin{align*} \int \frac{\sinh ^3(a+b x)}{(c+d x)^{5/2}} \, dx &=-\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}}+\frac{\left (8 b^2\right ) \int \frac{\sinh (a+b x)}{\sqrt{c+d x}} \, dx}{d^2}+\frac{\left (12 b^2\right ) \int \frac{\sinh ^3(a+b x)}{\sqrt{c+d x}} \, dx}{d^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}}+\frac{\left (12 i b^2\right ) \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{\left (4 b^2\right ) \int \frac{e^{-i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{d^2}-\frac{\left (4 b^2\right ) \int \frac{e^{i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{d^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}}-\frac{\left (8 b^2\right ) \operatorname{Subst}\left (\int e^{i \left (i a-\frac{i b c}{d}\right )-\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{d^3}+\frac{\left (8 b^2\right ) \operatorname{Subst}\left (\int e^{-i \left (i a-\frac{i b c}{d}\right )+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{d^3}+\frac{\left (3 b^2\right ) \int \frac{\sinh (3 a+3 b x)}{\sqrt{c+d x}} \, dx}{d^2}-\frac{\left (9 b^2\right ) \int \frac{\sinh (a+b x)}{\sqrt{c+d x}} \, dx}{d^2}\\ &=-\frac{4 b^{3/2} e^{-a+\frac{b c}{d}} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}+\frac{4 b^{3/2} e^{a-\frac{b c}{d}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{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}}+\frac{\left (3 b^2\right ) \int \frac{e^{-i (3 i a+3 i b x)}}{\sqrt{c+d x}} \, dx}{2 d^2}-\frac{\left (3 b^2\right ) \int \frac{e^{i (3 i a+3 i b x)}}{\sqrt{c+d x}} \, dx}{2 d^2}-\frac{\left (9 b^2\right ) \int \frac{e^{-i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{2 d^2}+\frac{\left (9 b^2\right ) \int \frac{e^{i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{2 d^2}\\ &=-\frac{4 b^{3/2} e^{-a+\frac{b c}{d}} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{5/2}}+\frac{4 b^{3/2} e^{a-\frac{b c}{d}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{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}}-\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int e^{i \left (3 i a-\frac{3 i b c}{d}\right )-\frac{3 b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{d^3}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int e^{-i \left (3 i a-\frac{3 i b c}{d}\right )+\frac{3 b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{d^3}+\frac{\left (9 b^2\right ) \operatorname{Subst}\left (\int e^{i \left (i a-\frac{i b c}{d}\right )-\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{d^3}-\frac{\left (9 b^2\right ) \operatorname{Subst}\left (\int e^{-i \left (i a-\frac{i b c}{d}\right )+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{d^3}\\ &=\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}}\\ \end{align*}

Mathematica [A]  time = 3.0608, size = 253, normalized size = 0.91 \[ \frac{e^{-3 \left (a+\frac{b c}{d}\right )} \left (-3 \sqrt{3} e^{6 a} d \left (-\frac{b (c+d x)}{d}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{3 b (c+d x)}{d}\right )+3 d e^{4 a+\frac{2 b c}{d}} \left (-\frac{b (c+d x)}{d}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{b (c+d x)}{d}\right )-3 d e^{2 a+\frac{4 b c}{d}} \left (\frac{b (c+d x)}{d}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},\frac{b (c+d x)}{d}\right )+3 \sqrt{3} d e^{\frac{6 b c}{d}} \left (\frac{b (c+d x)}{d}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},\frac{3 b (c+d x)}{d}\right )-4 e^{3 \left (a+\frac{b c}{d}\right )} \sinh ^2(a+b x) (6 b (c+d x) \cosh (a+b x)+d \sinh (a+b x))\right )}{6 d^2 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-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)*Gamma[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))

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Maple [F]  time = 0.084, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sinh \left ( bx+a \right ) \right ) ^{3} \left ( dx+c \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.42798, size = 265, normalized size = 0.96 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 3.15274, size = 4826, normalized size = 17.42 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*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(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)*co
sh(-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)*sin
h(-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(sqr
t(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 + a)^3*sinh(-(b*c - a*d)/d) + ((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*
cosh(-(b*c - a*d)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*sinh(-(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(-(b*c - a*d)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*sinh(-(
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(-(b*c - a*d)/d) - (b*
d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)^2*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(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 + a)^3*sinh(-(b*c - a*d)/d) + ((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(-(b*c - a*d)/d)
+ (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*sinh(-(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(-(b*c - a*d)/d) + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*sinh(-(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(-(b*c - a*d)/d) + (b*d^2*x^2 + 2*b*c*d*x
+ b*c^2)*cosh(b*x + a)^2*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(d*x + c)*sqrt(-b/d)) + ((6*b
*d*x + 6*b*c + d)*cosh(b*x + a)^6 + 6*(6*b*d*x + 6*b*c + d)*cosh(b*x + a)*sinh(b*x + a)^5 + (6*b*d*x + 6*b*c +
d)*sinh(b*x + a)^6 - 3*(2*b*d*x + 2*b*c + d)*cosh(b*x + a)^4 - 3*(2*b*d*x - 5*(6*b*d*x + 6*b*c + d)*cosh(b*x
+ a)^2 + 2*b*c + d)*sinh(b*x + a)^4 + 4*(5*(6*b*d*x + 6*b*c + d)*cosh(b*x + a)^3 - 3*(2*b*d*x + 2*b*c + d)*cos
h(b*x + a))*sinh(b*x + a)^3 + 6*b*d*x - 3*(2*b*d*x + 2*b*c - d)*cosh(b*x + a)^2 + 3*(5*(6*b*d*x + 6*b*c + d)*c
osh(b*x + a)^4 - 2*b*d*x - 6*(2*b*d*x + 2*b*c + d)*cosh(b*x + a)^2 - 2*b*c + d)*sinh(b*x + a)^2 + 6*b*c + 6*((
6*b*d*x + 6*b*c + d)*cosh(b*x + a)^5 - 2*(2*b*d*x + 2*b*c + d)*cosh(b*x + a)^3 - (2*b*d*x + 2*b*c - d)*cosh(b*
x + a))*sinh(b*x + a) - d)*sqrt(d*x + c))/((d^4*x^2 + 2*c*d^3*x + c^2*d^2)*cosh(b*x + a)^3 + 3*(d^4*x^2 + 2*c*
d^3*x + c^2*d^2)*cosh(b*x + a)^2*sinh(b*x + a) + 3*(d^4*x^2 + 2*c*d^3*x + c^2*d^2)*cosh(b*x + a)*sinh(b*x + a)
^2 + (d^4*x^2 + 2*c*d^3*x + c^2*d^2)*sinh(b*x + a)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{3}{\left (a + b x \right )}}{\left (c + d x\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (b x + a\right )^{3}}{{\left (d x + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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