3.15 \(\int \frac{\sinh ^2(a+b x)}{(c+d x)^4} \, dx\)

Optimal. Leaf size=162 \[ \frac{2 b^3 \sinh \left (2 a-\frac{2 b c}{d}\right ) \text{Chi}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}+\frac{2 b^3 \cosh \left (2 a-\frac{2 b c}{d}\right ) \text{Shi}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}-\frac{2 b^2 \sinh ^2(a+b x)}{3 d^3 (c+d x)}-\frac{b \sinh (a+b x) \cosh (a+b x)}{3 d^2 (c+d x)^2}-\frac{\sinh ^2(a+b x)}{3 d (c+d x)^3}-\frac{b^2}{3 d^3 (c+d x)} \]

[Out]

-b^2/(3*d^3*(c + d*x)) + (2*b^3*CoshIntegral[(2*b*c)/d + 2*b*x]*Sinh[2*a - (2*b*c)/d])/(3*d^4) - (b*Cosh[a + b
*x]*Sinh[a + b*x])/(3*d^2*(c + d*x)^2) - Sinh[a + b*x]^2/(3*d*(c + d*x)^3) - (2*b^2*Sinh[a + b*x]^2)/(3*d^3*(c
+ d*x)) + (2*b^3*Cosh[2*a - (2*b*c)/d]*SinhIntegral[(2*b*c)/d + 2*b*x])/(3*d^4)

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Rubi [A]  time = 0.18726, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3314, 32, 3313, 12, 3303, 3298, 3301} \[ \frac{2 b^3 \sinh \left (2 a-\frac{2 b c}{d}\right ) \text{Chi}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}+\frac{2 b^3 \cosh \left (2 a-\frac{2 b c}{d}\right ) \text{Shi}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}-\frac{2 b^2 \sinh ^2(a+b x)}{3 d^3 (c+d x)}-\frac{b \sinh (a+b x) \cosh (a+b x)}{3 d^2 (c+d x)^2}-\frac{\sinh ^2(a+b x)}{3 d (c+d x)^3}-\frac{b^2}{3 d^3 (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^2/(c + d*x)^4,x]

[Out]

-b^2/(3*d^3*(c + d*x)) + (2*b^3*CoshIntegral[(2*b*c)/d + 2*b*x]*Sinh[2*a - (2*b*c)/d])/(3*d^4) - (b*Cosh[a + b
*x]*Sinh[a + b*x])/(3*d^2*(c + d*x)^2) - Sinh[a + b*x]^2/(3*d*(c + d*x)^3) - (2*b^2*Sinh[a + b*x]^2)/(3*d^3*(c
+ d*x)) + (2*b^3*Cosh[2*a - (2*b*c)/d]*SinhIntegral[(2*b*c)/d + 2*b*x])/(3*d^4)

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 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{\sinh ^2(a+b x)}{(c+d x)^4} \, dx &=-\frac{b \cosh (a+b x) \sinh (a+b x)}{3 d^2 (c+d x)^2}-\frac{\sinh ^2(a+b x)}{3 d (c+d x)^3}+\frac{b^2 \int \frac{1}{(c+d x)^2} \, dx}{3 d^2}+\frac{\left (2 b^2\right ) \int \frac{\sinh ^2(a+b x)}{(c+d x)^2} \, dx}{3 d^2}\\ &=-\frac{b^2}{3 d^3 (c+d x)}-\frac{b \cosh (a+b x) \sinh (a+b x)}{3 d^2 (c+d x)^2}-\frac{\sinh ^2(a+b x)}{3 d (c+d x)^3}-\frac{2 b^2 \sinh ^2(a+b x)}{3 d^3 (c+d x)}-\frac{\left (4 i b^3\right ) \int \frac{i \sinh (2 a+2 b x)}{2 (c+d x)} \, dx}{3 d^3}\\ &=-\frac{b^2}{3 d^3 (c+d x)}-\frac{b \cosh (a+b x) \sinh (a+b x)}{3 d^2 (c+d x)^2}-\frac{\sinh ^2(a+b x)}{3 d (c+d x)^3}-\frac{2 b^2 \sinh ^2(a+b x)}{3 d^3 (c+d x)}+\frac{\left (2 b^3\right ) \int \frac{\sinh (2 a+2 b x)}{c+d x} \, dx}{3 d^3}\\ &=-\frac{b^2}{3 d^3 (c+d x)}-\frac{b \cosh (a+b x) \sinh (a+b x)}{3 d^2 (c+d x)^2}-\frac{\sinh ^2(a+b x)}{3 d (c+d x)^3}-\frac{2 b^2 \sinh ^2(a+b x)}{3 d^3 (c+d x)}+\frac{\left (2 b^3 \cosh \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\sinh \left (\frac{2 b c}{d}+2 b x\right )}{c+d x} \, dx}{3 d^3}+\frac{\left (2 b^3 \sinh \left (2 a-\frac{2 b c}{d}\right )\right ) \int \frac{\cosh \left (\frac{2 b c}{d}+2 b x\right )}{c+d x} \, dx}{3 d^3}\\ &=-\frac{b^2}{3 d^3 (c+d x)}+\frac{2 b^3 \text{Chi}\left (\frac{2 b c}{d}+2 b x\right ) \sinh \left (2 a-\frac{2 b c}{d}\right )}{3 d^4}-\frac{b \cosh (a+b x) \sinh (a+b x)}{3 d^2 (c+d x)^2}-\frac{\sinh ^2(a+b x)}{3 d (c+d x)^3}-\frac{2 b^2 \sinh ^2(a+b x)}{3 d^3 (c+d x)}+\frac{2 b^3 \cosh \left (2 a-\frac{2 b c}{d}\right ) \text{Shi}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}\\ \end{align*}

Mathematica [A]  time = 0.888442, size = 123, normalized size = 0.76 \[ \frac{4 b^3 \sinh \left (2 a-\frac{2 b c}{d}\right ) \text{Chi}\left (\frac{2 b (c+d x)}{d}\right )-\frac{d \left (\cosh (2 (a+b x)) \left (2 b^2 (c+d x)^2+d^2\right )+d (b (c+d x) \sinh (2 (a+b x))-d)\right )}{(c+d x)^3}+4 b^3 \cosh \left (2 a-\frac{2 b c}{d}\right ) \text{Shi}\left (\frac{2 b (c+d x)}{d}\right )}{6 d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^2/(c + d*x)^4,x]

[Out]

(4*b^3*CoshIntegral[(2*b*(c + d*x))/d]*Sinh[2*a - (2*b*c)/d] - (d*((d^2 + 2*b^2*(c + d*x)^2)*Cosh[2*(a + b*x)]
+ d*(-d + b*(c + d*x)*Sinh[2*(a + b*x)])))/(c + d*x)^3 + 4*b^3*Cosh[2*a - (2*b*c)/d]*SinhIntegral[(2*b*(c + d
*x))/d])/(6*d^4)

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Maple [B]  time = 0.118, size = 555, normalized size = 3.4 \begin{align*}{\frac{1}{6\,d \left ( dx+c \right ) ^{3}}}-{\frac{{b}^{5}{{\rm e}^{-2\,bx-2\,a}}{x}^{2}}{6\,d \left ({b}^{3}{d}^{3}{x}^{3}+3\,{b}^{3}c{d}^{2}{x}^{2}+3\,{b}^{3}{c}^{2}dx+{c}^{3}{b}^{3} \right ) }}-{\frac{{b}^{5}{{\rm e}^{-2\,bx-2\,a}}cx}{3\,{d}^{2} \left ({b}^{3}{d}^{3}{x}^{3}+3\,{b}^{3}c{d}^{2}{x}^{2}+3\,{b}^{3}{c}^{2}dx+{c}^{3}{b}^{3} \right ) }}-{\frac{{b}^{5}{{\rm e}^{-2\,bx-2\,a}}{c}^{2}}{6\,{d}^{3} \left ({b}^{3}{d}^{3}{x}^{3}+3\,{b}^{3}c{d}^{2}{x}^{2}+3\,{b}^{3}{c}^{2}dx+{c}^{3}{b}^{3} \right ) }}+{\frac{{b}^{4}{{\rm e}^{-2\,bx-2\,a}}x}{12\,d \left ({b}^{3}{d}^{3}{x}^{3}+3\,{b}^{3}c{d}^{2}{x}^{2}+3\,{b}^{3}{c}^{2}dx+{c}^{3}{b}^{3} \right ) }}+{\frac{{b}^{4}{{\rm e}^{-2\,bx-2\,a}}c}{12\,{d}^{2} \left ({b}^{3}{d}^{3}{x}^{3}+3\,{b}^{3}c{d}^{2}{x}^{2}+3\,{b}^{3}{c}^{2}dx+{c}^{3}{b}^{3} \right ) }}-{\frac{{b}^{3}{{\rm e}^{-2\,bx-2\,a}}}{12\,d \left ({b}^{3}{d}^{3}{x}^{3}+3\,{b}^{3}c{d}^{2}{x}^{2}+3\,{b}^{3}{c}^{2}dx+{c}^{3}{b}^{3} \right ) }}+{\frac{{b}^{3}}{3\,{d}^{4}}{{\rm e}^{-2\,{\frac{da-cb}{d}}}}{\it Ei} \left ( 1,2\,bx+2\,a-2\,{\frac{da-cb}{d}} \right ) }-{\frac{{b}^{3}{{\rm e}^{2\,bx+2\,a}}}{12\,{d}^{4}} \left ({\frac{cb}{d}}+bx \right ) ^{-3}}-{\frac{{b}^{3}{{\rm e}^{2\,bx+2\,a}}}{12\,{d}^{4}} \left ({\frac{cb}{d}}+bx \right ) ^{-2}}-{\frac{{b}^{3}{{\rm e}^{2\,bx+2\,a}}}{6\,{d}^{4}} \left ({\frac{cb}{d}}+bx \right ) ^{-1}}-{\frac{{b}^{3}}{3\,{d}^{4}}{{\rm e}^{2\,{\frac{da-cb}{d}}}}{\it Ei} \left ( 1,-2\,bx-2\,a-2\,{\frac{-da+cb}{d}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)^2/(d*x+c)^4,x)

[Out]

1/6/d/(d*x+c)^3-1/6*b^5*exp(-2*b*x-2*a)/d/(b^3*d^3*x^3+3*b^3*c*d^2*x^2+3*b^3*c^2*d*x+b^3*c^3)*x^2-1/3*b^5*exp(
-2*b*x-2*a)/d^2/(b^3*d^3*x^3+3*b^3*c*d^2*x^2+3*b^3*c^2*d*x+b^3*c^3)*c*x-1/6*b^5*exp(-2*b*x-2*a)/d^3/(b^3*d^3*x
^3+3*b^3*c*d^2*x^2+3*b^3*c^2*d*x+b^3*c^3)*c^2+1/12*b^4*exp(-2*b*x-2*a)/d/(b^3*d^3*x^3+3*b^3*c*d^2*x^2+3*b^3*c^
2*d*x+b^3*c^3)*x+1/12*b^4*exp(-2*b*x-2*a)/d^2/(b^3*d^3*x^3+3*b^3*c*d^2*x^2+3*b^3*c^2*d*x+b^3*c^3)*c-1/12*b^3*e
xp(-2*b*x-2*a)/d/(b^3*d^3*x^3+3*b^3*c*d^2*x^2+3*b^3*c^2*d*x+b^3*c^3)+1/3*b^3/d^4*exp(-2*(a*d-b*c)/d)*Ei(1,2*b*
x+2*a-2*(a*d-b*c)/d)-1/12*b^3/d^4*exp(2*b*x+2*a)/(b*c/d+b*x)^3-1/12*b^3/d^4*exp(2*b*x+2*a)/(b*c/d+b*x)^2-1/6*b
^3/d^4*exp(2*b*x+2*a)/(b*c/d+b*x)-1/3*b^3/d^4*exp(2*(a*d-b*c)/d)*Ei(1,-2*b*x-2*a-2*(-a*d+b*c)/d)

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Maxima [A]  time = 1.51533, size = 149, normalized size = 0.92 \begin{align*} \frac{1}{6 \,{\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )}} - \frac{e^{\left (-2 \, a + \frac{2 \, b c}{d}\right )} E_{4}\left (\frac{2 \,{\left (d x + c\right )} b}{d}\right )}{4 \,{\left (d x + c\right )}^{3} d} - \frac{e^{\left (2 \, a - \frac{2 \, b c}{d}\right )} E_{4}\left (-\frac{2 \,{\left (d x + c\right )} b}{d}\right )}{4 \,{\left (d x + c\right )}^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^2/(d*x+c)^4,x, algorithm="maxima")

[Out]

1/6/(d^4*x^3 + 3*c*d^3*x^2 + 3*c^2*d^2*x + c^3*d) - 1/4*e^(-2*a + 2*b*c/d)*exp_integral_e(4, 2*(d*x + c)*b/d)/
((d*x + c)^3*d) - 1/4*e^(2*a - 2*b*c/d)*exp_integral_e(4, -2*(d*x + c)*b/d)/((d*x + c)^3*d)

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Fricas [B]  time = 2.73919, size = 857, normalized size = 5.29 \begin{align*} \frac{d^{3} -{\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d + d^{3}\right )} \cosh \left (b x + a\right )^{2} - 2 \,{\left (b d^{3} x + b c d^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) -{\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d + d^{3}\right )} \sinh \left (b x + a\right )^{2} + 2 \,{\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )}{\rm Ei}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) -{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )}{\rm Ei}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right )\right )} \cosh \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) + 2 \,{\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )}{\rm Ei}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )}{\rm Ei}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right )\right )} \sinh \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right )}{6 \,{\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^2/(d*x+c)^4,x, algorithm="fricas")

[Out]

1/6*(d^3 - (2*b^2*d^3*x^2 + 4*b^2*c*d^2*x + 2*b^2*c^2*d + d^3)*cosh(b*x + a)^2 - 2*(b*d^3*x + b*c*d^2)*cosh(b*
x + a)*sinh(b*x + a) - (2*b^2*d^3*x^2 + 4*b^2*c*d^2*x + 2*b^2*c^2*d + d^3)*sinh(b*x + a)^2 + 2*((b^3*d^3*x^3 +
3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*Ei(2*(b*d*x + b*c)/d) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2
*d*x + b^3*c^3)*Ei(-2*(b*d*x + b*c)/d))*cosh(-2*(b*c - a*d)/d) + 2*((b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2
*d*x + b^3*c^3)*Ei(2*(b*d*x + b*c)/d) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*Ei(-2*(b*d*x
+ b*c)/d))*sinh(-2*(b*c - a*d)/d))/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2*d^5*x + c^3*d^4)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)**2/(d*x+c)**4,x)

[Out]

Integral(sinh(a + b*x)**2/(c + d*x)**4, x)

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Giac [B]  time = 1.15627, size = 725, normalized size = 4.48 \begin{align*} \frac{4 \, b^{3} d^{3} x^{3}{\rm Ei}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) e^{\left (2 \, a - \frac{2 \, b c}{d}\right )} - 4 \, b^{3} d^{3} x^{3}{\rm Ei}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) e^{\left (-2 \, a + \frac{2 \, b c}{d}\right )} + 12 \, b^{3} c d^{2} x^{2}{\rm Ei}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) e^{\left (2 \, a - \frac{2 \, b c}{d}\right )} - 12 \, b^{3} c d^{2} x^{2}{\rm Ei}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) e^{\left (-2 \, a + \frac{2 \, b c}{d}\right )} + 12 \, b^{3} c^{2} d x{\rm Ei}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) e^{\left (2 \, a - \frac{2 \, b c}{d}\right )} - 12 \, b^{3} c^{2} d x{\rm Ei}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) e^{\left (-2 \, a + \frac{2 \, b c}{d}\right )} - 2 \, b^{2} d^{3} x^{2} e^{\left (2 \, b x + 2 \, a\right )} - 2 \, b^{2} d^{3} x^{2} e^{\left (-2 \, b x - 2 \, a\right )} + 4 \, b^{3} c^{3}{\rm Ei}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) e^{\left (2 \, a - \frac{2 \, b c}{d}\right )} - 4 \, b^{3} c^{3}{\rm Ei}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) e^{\left (-2 \, a + \frac{2 \, b c}{d}\right )} - 4 \, b^{2} c d^{2} x e^{\left (2 \, b x + 2 \, a\right )} - 4 \, b^{2} c d^{2} x e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, b^{2} c^{2} d e^{\left (2 \, b x + 2 \, a\right )} - b d^{3} x e^{\left (2 \, b x + 2 \, a\right )} - 2 \, b^{2} c^{2} d e^{\left (-2 \, b x - 2 \, a\right )} + b d^{3} x e^{\left (-2 \, b x - 2 \, a\right )} - b c d^{2} e^{\left (2 \, b x + 2 \, a\right )} + b c d^{2} e^{\left (-2 \, b x - 2 \, a\right )} - d^{3} e^{\left (2 \, b x + 2 \, a\right )} - d^{3} e^{\left (-2 \, b x - 2 \, a\right )} + 2 \, d^{3}}{12 \,{\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^2/(d*x+c)^4,x, algorithm="giac")

[Out]

1/12*(4*b^3*d^3*x^3*Ei(2*(b*d*x + b*c)/d)*e^(2*a - 2*b*c/d) - 4*b^3*d^3*x^3*Ei(-2*(b*d*x + b*c)/d)*e^(-2*a + 2
*b*c/d) + 12*b^3*c*d^2*x^2*Ei(2*(b*d*x + b*c)/d)*e^(2*a - 2*b*c/d) - 12*b^3*c*d^2*x^2*Ei(-2*(b*d*x + b*c)/d)*e
^(-2*a + 2*b*c/d) + 12*b^3*c^2*d*x*Ei(2*(b*d*x + b*c)/d)*e^(2*a - 2*b*c/d) - 12*b^3*c^2*d*x*Ei(-2*(b*d*x + b*c
)/d)*e^(-2*a + 2*b*c/d) - 2*b^2*d^3*x^2*e^(2*b*x + 2*a) - 2*b^2*d^3*x^2*e^(-2*b*x - 2*a) + 4*b^3*c^3*Ei(2*(b*d
*x + b*c)/d)*e^(2*a - 2*b*c/d) - 4*b^3*c^3*Ei(-2*(b*d*x + b*c)/d)*e^(-2*a + 2*b*c/d) - 4*b^2*c*d^2*x*e^(2*b*x
+ 2*a) - 4*b^2*c*d^2*x*e^(-2*b*x - 2*a) - 2*b^2*c^2*d*e^(2*b*x + 2*a) - b*d^3*x*e^(2*b*x + 2*a) - 2*b^2*c^2*d*
e^(-2*b*x - 2*a) + b*d^3*x*e^(-2*b*x - 2*a) - b*c*d^2*e^(2*b*x + 2*a) + b*c*d^2*e^(-2*b*x - 2*a) - d^3*e^(2*b*
x + 2*a) - d^3*e^(-2*b*x - 2*a) + 2*d^3)/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2*d^5*x + c^3*d^4)