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

Optimal. Leaf size=220 \[ -\frac{8 \sqrt{2 \pi } b^{5/2} e^{\frac{2 b c}{d}-2 a} \text{Erf}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}+\frac{8 \sqrt{2 \pi } b^{5/2} e^{2 a-\frac{2 b c}{d}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}-\frac{32 b^2 \sinh ^2(a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac{16 b^2}{15 d^3 \sqrt{c+d x}} \]

[Out]

(-16*b^2)/(15*d^3*Sqrt[c + d*x]) - (8*b^(5/2)*E^(-2*a + (2*b*c)/d)*Sqrt[2*Pi]*Erf[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*
x])/Sqrt[d]])/(15*d^(7/2)) + (8*b^(5/2)*E^(2*a - (2*b*c)/d)*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sq
rt[d]])/(15*d^(7/2)) - (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)) - (32*b^2*Sinh[a + b*x]^2)/(15*d^3*Sqrt[c + d*x])

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Rubi [A]  time = 0.321564, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3314, 32, 3313, 12, 3308, 2180, 2204, 2205} \[ -\frac{8 \sqrt{2 \pi } b^{5/2} e^{\frac{2 b c}{d}-2 a} \text{Erf}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}+\frac{8 \sqrt{2 \pi } b^{5/2} e^{2 a-\frac{2 b c}{d}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}-\frac{32 b^2 \sinh ^2(a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{8 b \sinh (a+b x) \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \sinh ^2(a+b x)}{5 d (c+d x)^{5/2}}-\frac{16 b^2}{15 d^3 \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b^2)/(15*d^3*Sqrt[c + d*x]) - (8*b^(5/2)*E^(-2*a + (2*b*c)/d)*Sqrt[2*Pi]*Erf[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*
x])/Sqrt[d]])/(15*d^(7/2)) + (8*b^(5/2)*E^(2*a - (2*b*c)/d)*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sq
rt[d]])/(15*d^(7/2)) - (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)) - (32*b^2*Sinh[a + b*x]^2)/(15*d^3*Sqrt[c + d*x])

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 32

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

Rule 3313

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x]^
n)/(d*(m + 1)), x] - Dist[(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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rubi steps

\begin{align*} \int \frac{\sinh ^2(a+b x)}{(c+d x)^{7/2}} \, dx &=-\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{\left (8 b^2\right ) \int \frac{1}{(c+d x)^{3/2}} \, dx}{15 d^2}+\frac{\left (16 b^2\right ) \int \frac{\sinh ^2(a+b x)}{(c+d x)^{3/2}} \, dx}{15 d^2}\\ &=-\frac{16 b^2}{15 d^3 \sqrt{c+d x}}-\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}}-\frac{\left (64 i b^3\right ) \int \frac{i \sinh (2 a+2 b x)}{2 \sqrt{c+d x}} \, dx}{15 d^3}\\ &=-\frac{16 b^2}{15 d^3 \sqrt{c+d x}}-\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}}+\frac{\left (32 b^3\right ) \int \frac{\sinh (2 a+2 b x)}{\sqrt{c+d x}} \, dx}{15 d^3}\\ &=-\frac{16 b^2}{15 d^3 \sqrt{c+d x}}-\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}}+\frac{\left (16 b^3\right ) \int \frac{e^{-i (2 i a+2 i b x)}}{\sqrt{c+d x}} \, dx}{15 d^3}-\frac{\left (16 b^3\right ) \int \frac{e^{i (2 i a+2 i b x)}}{\sqrt{c+d x}} \, dx}{15 d^3}\\ &=-\frac{16 b^2}{15 d^3 \sqrt{c+d x}}-\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}}-\frac{\left (32 b^3\right ) \operatorname{Subst}\left (\int e^{i \left (2 i a-\frac{2 i b c}{d}\right )-\frac{2 b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{15 d^4}+\frac{\left (32 b^3\right ) \operatorname{Subst}\left (\int e^{-i \left (2 i a-\frac{2 i b c}{d}\right )+\frac{2 b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{15 d^4}\\ &=-\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}}\\ \end{align*}

Mathematica [B]  time = 9.18661, size = 825, normalized size = 3.75 \[ \frac{e^{-\frac{2 b (c+d x)}{d}} \left (16 \sqrt{2} d^2 e^{\frac{2 b (c+d x)}{d}} \text{Gamma}\left (\frac{1}{2},-\frac{2 b (c+d x)}{d}\right ) \left (\cosh \left (2 a-\frac{2 b c}{d}\right )+\sinh \left (2 a-\frac{2 b c}{d}\right )\right ) \left (-\frac{b (c+d x)}{d}\right )^{5/2}+6 d^2 e^{\frac{2 b (c+d x)}{d}}-16 b^2 c^2 e^{\frac{4 b (c+d x)}{d}} \cosh \left (2 a-\frac{2 b c}{d}\right )-3 d^2 e^{\frac{4 b (c+d x)}{d}} \cosh \left (2 a-\frac{2 b c}{d}\right )-4 b c d e^{\frac{4 b (c+d x)}{d}} \cosh \left (2 a-\frac{2 b c}{d}\right )-16 b^2 c^2 \cosh \left (2 a-\frac{2 b c}{d}\right )-3 d^2 \cosh \left (2 a-\frac{2 b c}{d}\right )-16 b^2 d^2 e^{\frac{4 b (c+d x)}{d}} x^2 \cosh \left (2 a-\frac{2 b c}{d}\right )-16 b^2 d^2 x^2 \cosh \left (2 a-\frac{2 b c}{d}\right )+4 b c d \cosh \left (2 a-\frac{2 b c}{d}\right )-4 b d^2 e^{\frac{4 b (c+d x)}{d}} x \cosh \left (2 a-\frac{2 b c}{d}\right )-32 b^2 c d e^{\frac{4 b (c+d x)}{d}} x \cosh \left (2 a-\frac{2 b c}{d}\right )+4 b d^2 x \cosh \left (2 a-\frac{2 b c}{d}\right )-32 b^2 c d x \cosh \left (2 a-\frac{2 b c}{d}\right )+16 \sqrt{2} d^2 e^{\frac{2 b (c+d x)}{d}} \left (\frac{b (c+d x)}{d}\right )^{5/2} \text{Gamma}\left (\frac{1}{2},\frac{2 b (c+d x)}{d}\right ) \left (\cosh \left (2 a-\frac{2 b c}{d}\right )-\sinh \left (2 a-\frac{2 b c}{d}\right )\right )-16 b^2 c^2 e^{\frac{4 b (c+d x)}{d}} \sinh \left (2 a-\frac{2 b c}{d}\right )-3 d^2 e^{\frac{4 b (c+d x)}{d}} \sinh \left (2 a-\frac{2 b c}{d}\right )-4 b c d e^{\frac{4 b (c+d x)}{d}} \sinh \left (2 a-\frac{2 b c}{d}\right )+16 b^2 c^2 \sinh \left (2 a-\frac{2 b c}{d}\right )+3 d^2 \sinh \left (2 a-\frac{2 b c}{d}\right )-16 b^2 d^2 e^{\frac{4 b (c+d x)}{d}} x^2 \sinh \left (2 a-\frac{2 b c}{d}\right )+16 b^2 d^2 x^2 \sinh \left (2 a-\frac{2 b c}{d}\right )-4 b c d \sinh \left (2 a-\frac{2 b c}{d}\right )-4 b d^2 e^{\frac{4 b (c+d x)}{d}} x \sinh \left (2 a-\frac{2 b c}{d}\right )-32 b^2 c d e^{\frac{4 b (c+d x)}{d}} x \sinh \left (2 a-\frac{2 b c}{d}\right )-4 b d^2 x \sinh \left (2 a-\frac{2 b c}{d}\right )+32 b^2 c d x \sinh \left (2 a-\frac{2 b c}{d}\right )\right )}{30 d^3 (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*d^2*E^((2*b*(c + d*x))/d) - 16*b^2*c^2*Cosh[2*a - (2*b*c)/d] + 4*b*c*d*Cosh[2*a - (2*b*c)/d] - 3*d^2*Cosh[2
*a - (2*b*c)/d] - 16*b^2*c^2*E^((4*b*(c + d*x))/d)*Cosh[2*a - (2*b*c)/d] - 4*b*c*d*E^((4*b*(c + d*x))/d)*Cosh[
2*a - (2*b*c)/d] - 3*d^2*E^((4*b*(c + d*x))/d)*Cosh[2*a - (2*b*c)/d] - 32*b^2*c*d*x*Cosh[2*a - (2*b*c)/d] + 4*
b*d^2*x*Cosh[2*a - (2*b*c)/d] - 32*b^2*c*d*E^((4*b*(c + d*x))/d)*x*Cosh[2*a - (2*b*c)/d] - 4*b*d^2*E^((4*b*(c
+ d*x))/d)*x*Cosh[2*a - (2*b*c)/d] - 16*b^2*d^2*x^2*Cosh[2*a - (2*b*c)/d] - 16*b^2*d^2*E^((4*b*(c + d*x))/d)*x
^2*Cosh[2*a - (2*b*c)/d] + 16*Sqrt[2]*d^2*E^((2*b*(c + d*x))/d)*((b*(c + d*x))/d)^(5/2)*Gamma[1/2, (2*b*(c + d
*x))/d]*(Cosh[2*a - (2*b*c)/d] - Sinh[2*a - (2*b*c)/d]) + 16*b^2*c^2*Sinh[2*a - (2*b*c)/d] - 4*b*c*d*Sinh[2*a
- (2*b*c)/d] + 3*d^2*Sinh[2*a - (2*b*c)/d] - 16*b^2*c^2*E^((4*b*(c + d*x))/d)*Sinh[2*a - (2*b*c)/d] - 4*b*c*d*
E^((4*b*(c + d*x))/d)*Sinh[2*a - (2*b*c)/d] - 3*d^2*E^((4*b*(c + d*x))/d)*Sinh[2*a - (2*b*c)/d] + 32*b^2*c*d*x
*Sinh[2*a - (2*b*c)/d] - 4*b*d^2*x*Sinh[2*a - (2*b*c)/d] - 32*b^2*c*d*E^((4*b*(c + d*x))/d)*x*Sinh[2*a - (2*b*
c)/d] - 4*b*d^2*E^((4*b*(c + d*x))/d)*x*Sinh[2*a - (2*b*c)/d] + 16*b^2*d^2*x^2*Sinh[2*a - (2*b*c)/d] - 16*b^2*
d^2*E^((4*b*(c + d*x))/d)*x^2*Sinh[2*a - (2*b*c)/d] + 16*Sqrt[2]*d^2*E^((2*b*(c + d*x))/d)*(-((b*(c + d*x))/d)
)^(5/2)*Gamma[1/2, (-2*b*(c + d*x))/d]*(Cosh[2*a - (2*b*c)/d] + Sinh[2*a - (2*b*c)/d]))/(30*d^3*E^((2*b*(c + d
*x))/d)*(c + d*x)^(5/2))

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Maple [F]  time = 0.067, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( dx+c \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.31562, size = 159, normalized size = 0.72 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*sqr
t(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

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Fricas [B]  time = 3.08432, size = 2974, normalized size = 13.52 \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)^2/(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-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)*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)*sqr
t(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)*cos
h(b*x + a)^4 + 4*(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)*sinh(
b*x + a)^3 + (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)*sinh(b*x + a)^4 + 16*b^
2*c^2 - 6*d^2*cosh(b*x + a)^2 - 4*b*c*d + 6*((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)^2 - d^2)*sinh(b*x + a)^2 + 3*d^2 + 4*(8*b^2*c*d - b*d^2)*x + 4*((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)^3 - 3*d^2*cosh(b*x + a))*sinh(b*x + a))*sqrt(
d*x + c))/((d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)*cosh(b*x + a)^2 + 2*(d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*
d^4*x + c^3*d^3)*cosh(b*x + a)*sinh(b*x + a) + (d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)*sinh(b*x + a)^2
)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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