3.164 \(\int (c+d x)^2 (a+b \sinh (e+f x))^2 \, dx\)

Optimal. Leaf size=182 \[ \frac{a^2 (c+d x)^3}{3 d}-\frac{4 a b d (c+d x) \sinh (e+f x)}{f^2}+\frac{2 a b (c+d x)^2 \cosh (e+f x)}{f}+\frac{4 a b d^2 \cosh (e+f x)}{f^3}-\frac{b^2 d (c+d x) \sinh ^2(e+f x)}{2 f^2}+\frac{b^2 (c+d x)^2 \sinh (e+f x) \cosh (e+f x)}{2 f}-\frac{b^2 (c+d x)^3}{6 d}+\frac{b^2 d^2 \sinh (e+f x) \cosh (e+f x)}{4 f^3}-\frac{b^2 d^2 x}{4 f^2} \]

[Out]

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

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Rubi [A]  time = 0.202018, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {3317, 3296, 2638, 3311, 32, 2635, 8} \[ \frac{a^2 (c+d x)^3}{3 d}-\frac{4 a b d (c+d x) \sinh (e+f x)}{f^2}+\frac{2 a b (c+d x)^2 \cosh (e+f x)}{f}+\frac{4 a b d^2 \cosh (e+f x)}{f^3}-\frac{b^2 d (c+d x) \sinh ^2(e+f x)}{2 f^2}+\frac{b^2 (c+d x)^2 \sinh (e+f x) \cosh (e+f x)}{2 f}-\frac{b^2 (c+d x)^3}{6 d}+\frac{b^2 d^2 \sinh (e+f x) \cosh (e+f x)}{4 f^3}-\frac{b^2 d^2 x}{4 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

Rule 3296

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3311

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

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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (c+d x)^2 (a+b \sinh (e+f x))^2 \, dx &=\int \left (a^2 (c+d x)^2+2 a b (c+d x)^2 \sinh (e+f x)+b^2 (c+d x)^2 \sinh ^2(e+f x)\right ) \, dx\\ &=\frac{a^2 (c+d x)^3}{3 d}+(2 a b) \int (c+d x)^2 \sinh (e+f x) \, dx+b^2 \int (c+d x)^2 \sinh ^2(e+f x) \, dx\\ &=\frac{a^2 (c+d x)^3}{3 d}+\frac{2 a b (c+d x)^2 \cosh (e+f x)}{f}+\frac{b^2 (c+d x)^2 \cosh (e+f x) \sinh (e+f x)}{2 f}-\frac{b^2 d (c+d x) \sinh ^2(e+f x)}{2 f^2}-\frac{1}{2} b^2 \int (c+d x)^2 \, dx+\frac{\left (b^2 d^2\right ) \int \sinh ^2(e+f x) \, dx}{2 f^2}-\frac{(4 a b d) \int (c+d x) \cosh (e+f x) \, dx}{f}\\ &=\frac{a^2 (c+d x)^3}{3 d}-\frac{b^2 (c+d x)^3}{6 d}+\frac{2 a b (c+d x)^2 \cosh (e+f x)}{f}-\frac{4 a b d (c+d x) \sinh (e+f x)}{f^2}+\frac{b^2 d^2 \cosh (e+f x) \sinh (e+f x)}{4 f^3}+\frac{b^2 (c+d x)^2 \cosh (e+f x) \sinh (e+f x)}{2 f}-\frac{b^2 d (c+d x) \sinh ^2(e+f x)}{2 f^2}+\frac{\left (4 a b d^2\right ) \int \sinh (e+f x) \, dx}{f^2}-\frac{\left (b^2 d^2\right ) \int 1 \, dx}{4 f^2}\\ &=-\frac{b^2 d^2 x}{4 f^2}+\frac{a^2 (c+d x)^3}{3 d}-\frac{b^2 (c+d x)^3}{6 d}+\frac{4 a b d^2 \cosh (e+f x)}{f^3}+\frac{2 a b (c+d x)^2 \cosh (e+f x)}{f}-\frac{4 a b d (c+d x) \sinh (e+f x)}{f^2}+\frac{b^2 d^2 \cosh (e+f x) \sinh (e+f x)}{4 f^3}+\frac{b^2 (c+d x)^2 \cosh (e+f x) \sinh (e+f x)}{2 f}-\frac{b^2 d (c+d x) \sinh ^2(e+f x)}{2 f^2}\\ \end{align*}

Mathematica [A]  time = 0.752984, size = 249, normalized size = 1.37 \[ \frac{24 a^2 c^2 f^3 x+24 a^2 c d f^3 x^2+8 a^2 d^2 f^3 x^3+48 a b \left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2+2\right )\right ) \cosh (e+f x)-96 a b c d f \sinh (e+f x)-96 a b d^2 f x \sinh (e+f x)+6 b^2 c^2 f^2 \sinh (2 (e+f x))-12 b^2 c^2 f^3 x+12 b^2 c d f^2 x \sinh (2 (e+f x))-6 b^2 d f (c+d x) \cosh (2 (e+f x))-12 b^2 c d f^3 x^2+6 b^2 d^2 f^2 x^2 \sinh (2 (e+f x))+3 b^2 d^2 \sinh (2 (e+f x))-4 b^2 d^2 f^3 x^3}{24 f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*a^2*c^2*f^3*x - 12*b^2*c^2*f^3*x + 24*a^2*c*d*f^3*x^2 - 12*b^2*c*d*f^3*x^2 + 8*a^2*d^2*f^3*x^3 - 4*b^2*d^2
*f^3*x^3 + 48*a*b*(c^2*f^2 + 2*c*d*f^2*x + d^2*(2 + f^2*x^2))*Cosh[e + f*x] - 6*b^2*d*f*(c + d*x)*Cosh[2*(e +
f*x)] - 96*a*b*c*d*f*Sinh[e + f*x] - 96*a*b*d^2*f*x*Sinh[e + f*x] + 3*b^2*d^2*Sinh[2*(e + f*x)] + 6*b^2*c^2*f^
2*Sinh[2*(e + f*x)] + 12*b^2*c*d*f^2*x*Sinh[2*(e + f*x)] + 6*b^2*d^2*f^2*x^2*Sinh[2*(e + f*x)])/(24*f^3)

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Maple [B]  time = 0.016, size = 535, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*(a+b*sinh(f*x+e))^2,x)

[Out]

1/f*(1/3/f^2*d^2*a^2*(f*x+e)^3+2/f^2*d^2*a*b*((f*x+e)^2*cosh(f*x+e)-2*(f*x+e)*sinh(f*x+e)+2*cosh(f*x+e))+1/f^2
*d^2*b^2*(1/2*(f*x+e)^2*cosh(f*x+e)*sinh(f*x+e)-1/6*(f*x+e)^3-1/2*(f*x+e)*cosh(f*x+e)^2+1/4*cosh(f*x+e)*sinh(f
*x+e)+1/4*f*x+1/4*e)-1/f^2*d^2*e*a^2*(f*x+e)^2-4/f^2*d^2*e*a*b*((f*x+e)*cosh(f*x+e)-sinh(f*x+e))-2/f^2*d^2*e*b
^2*(1/2*(f*x+e)*cosh(f*x+e)*sinh(f*x+e)-1/4*(f*x+e)^2-1/4*cosh(f*x+e)^2)+1/f^2*d^2*e^2*a^2*(f*x+e)+2/f^2*d^2*e
^2*a*b*cosh(f*x+e)+1/f^2*d^2*e^2*b^2*(1/2*cosh(f*x+e)*sinh(f*x+e)-1/2*f*x-1/2*e)+1/f*c*d*a^2*(f*x+e)^2+4/f*c*d
*a*b*((f*x+e)*cosh(f*x+e)-sinh(f*x+e))+2/f*c*d*b^2*(1/2*(f*x+e)*cosh(f*x+e)*sinh(f*x+e)-1/4*(f*x+e)^2-1/4*cosh
(f*x+e)^2)-2/f*c*d*e*a^2*(f*x+e)-4/f*c*d*e*a*b*cosh(f*x+e)-2/f*c*d*e*b^2*(1/2*cosh(f*x+e)*sinh(f*x+e)-1/2*f*x-
1/2*e)+c^2*a^2*(f*x+e)+2*c^2*a*b*cosh(f*x+e)+c^2*b^2*(1/2*cosh(f*x+e)*sinh(f*x+e)-1/2*f*x-1/2*e))

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Maxima [A]  time = 1.27115, size = 435, normalized size = 2.39 \begin{align*} \frac{1}{3} \, a^{2} d^{2} x^{3} + a^{2} c d x^{2} - \frac{1}{8} \,{\left (4 \, x^{2} - \frac{{\left (2 \, f x e^{\left (2 \, e\right )} - e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f^{2}} + \frac{{\left (2 \, f x + 1\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{f^{2}}\right )} b^{2} c d - \frac{1}{48} \,{\left (8 \, x^{3} - \frac{3 \,{\left (2 \, f^{2} x^{2} e^{\left (2 \, e\right )} - 2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f^{3}} + \frac{3 \,{\left (2 \, f^{2} x^{2} + 2 \, f x + 1\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{f^{3}}\right )} b^{2} d^{2} - \frac{1}{8} \, b^{2} c^{2}{\left (4 \, x - \frac{e^{\left (2 \, f x + 2 \, e\right )}}{f} + \frac{e^{\left (-2 \, f x - 2 \, e\right )}}{f}\right )} + a^{2} c^{2} x + 2 \, a b c d{\left (\frac{{\left (f x e^{e} - e^{e}\right )} e^{\left (f x\right )}}{f^{2}} + \frac{{\left (f x + 1\right )} e^{\left (-f x - e\right )}}{f^{2}}\right )} + a b d^{2}{\left (\frac{{\left (f^{2} x^{2} e^{e} - 2 \, f x e^{e} + 2 \, e^{e}\right )} e^{\left (f x\right )}}{f^{3}} + \frac{{\left (f^{2} x^{2} + 2 \, f x + 2\right )} e^{\left (-f x - e\right )}}{f^{3}}\right )} + \frac{2 \, a b c^{2} \cosh \left (f x + e\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*d^2*x^3 + a^2*c*d*x^2 - 1/8*(4*x^2 - (2*f*x*e^(2*e) - e^(2*e))*e^(2*f*x)/f^2 + (2*f*x + 1)*e^(-2*f*x -
2*e)/f^2)*b^2*c*d - 1/48*(8*x^3 - 3*(2*f^2*x^2*e^(2*e) - 2*f*x*e^(2*e) + e^(2*e))*e^(2*f*x)/f^3 + 3*(2*f^2*x^
2 + 2*f*x + 1)*e^(-2*f*x - 2*e)/f^3)*b^2*d^2 - 1/8*b^2*c^2*(4*x - e^(2*f*x + 2*e)/f + e^(-2*f*x - 2*e)/f) + a^
2*c^2*x + 2*a*b*c*d*((f*x*e^e - e^e)*e^(f*x)/f^2 + (f*x + 1)*e^(-f*x - e)/f^2) + a*b*d^2*((f^2*x^2*e^e - 2*f*x
*e^e + 2*e^e)*e^(f*x)/f^3 + (f^2*x^2 + 2*f*x + 2)*e^(-f*x - e)/f^3) + 2*a*b*c^2*cosh(f*x + e)/f

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Fricas [A]  time = 2.47877, size = 541, normalized size = 2.97 \begin{align*} \frac{2 \,{\left (2 \, a^{2} - b^{2}\right )} d^{2} f^{3} x^{3} + 6 \,{\left (2 \, a^{2} - b^{2}\right )} c d f^{3} x^{2} + 6 \,{\left (2 \, a^{2} - b^{2}\right )} c^{2} f^{3} x - 3 \,{\left (b^{2} d^{2} f x + b^{2} c d f\right )} \cosh \left (f x + e\right )^{2} - 3 \,{\left (b^{2} d^{2} f x + b^{2} c d f\right )} \sinh \left (f x + e\right )^{2} + 24 \,{\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2} + 2 \, a b d^{2}\right )} \cosh \left (f x + e\right ) - 3 \,{\left (16 \, a b d^{2} f x + 16 \, a b c d f -{\left (2 \, b^{2} d^{2} f^{2} x^{2} + 4 \, b^{2} c d f^{2} x + 2 \, b^{2} c^{2} f^{2} + b^{2} d^{2}\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )}{12 \, f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*(2*a^2 - b^2)*d^2*f^3*x^3 + 6*(2*a^2 - b^2)*c*d*f^3*x^2 + 6*(2*a^2 - b^2)*c^2*f^3*x - 3*(b^2*d^2*f*x +
b^2*c*d*f)*cosh(f*x + e)^2 - 3*(b^2*d^2*f*x + b^2*c*d*f)*sinh(f*x + e)^2 + 24*(a*b*d^2*f^2*x^2 + 2*a*b*c*d*f^
2*x + a*b*c^2*f^2 + 2*a*b*d^2)*cosh(f*x + e) - 3*(16*a*b*d^2*f*x + 16*a*b*c*d*f - (2*b^2*d^2*f^2*x^2 + 4*b^2*c
*d*f^2*x + 2*b^2*c^2*f^2 + b^2*d^2)*cosh(f*x + e))*sinh(f*x + e))/f^3

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Sympy [A]  time = 2.35746, size = 456, normalized size = 2.51 \begin{align*} \begin{cases} a^{2} c^{2} x + a^{2} c d x^{2} + \frac{a^{2} d^{2} x^{3}}{3} + \frac{2 a b c^{2} \cosh{\left (e + f x \right )}}{f} + \frac{4 a b c d x \cosh{\left (e + f x \right )}}{f} - \frac{4 a b c d \sinh{\left (e + f x \right )}}{f^{2}} + \frac{2 a b d^{2} x^{2} \cosh{\left (e + f x \right )}}{f} - \frac{4 a b d^{2} x \sinh{\left (e + f x \right )}}{f^{2}} + \frac{4 a b d^{2} \cosh{\left (e + f x \right )}}{f^{3}} + \frac{b^{2} c^{2} x \sinh ^{2}{\left (e + f x \right )}}{2} - \frac{b^{2} c^{2} x \cosh ^{2}{\left (e + f x \right )}}{2} + \frac{b^{2} c^{2} \sinh{\left (e + f x \right )} \cosh{\left (e + f x \right )}}{2 f} + \frac{b^{2} c d x^{2} \sinh ^{2}{\left (e + f x \right )}}{2} - \frac{b^{2} c d x^{2} \cosh ^{2}{\left (e + f x \right )}}{2} + \frac{b^{2} c d x \sinh{\left (e + f x \right )} \cosh{\left (e + f x \right )}}{f} - \frac{b^{2} c d \sinh ^{2}{\left (e + f x \right )}}{2 f^{2}} + \frac{b^{2} d^{2} x^{3} \sinh ^{2}{\left (e + f x \right )}}{6} - \frac{b^{2} d^{2} x^{3} \cosh ^{2}{\left (e + f x \right )}}{6} + \frac{b^{2} d^{2} x^{2} \sinh{\left (e + f x \right )} \cosh{\left (e + f x \right )}}{2 f} - \frac{b^{2} d^{2} x \sinh ^{2}{\left (e + f x \right )}}{4 f^{2}} - \frac{b^{2} d^{2} x \cosh ^{2}{\left (e + f x \right )}}{4 f^{2}} + \frac{b^{2} d^{2} \sinh{\left (e + f x \right )} \cosh{\left (e + f x \right )}}{4 f^{3}} & \text{for}\: f \neq 0 \\\left (a + b \sinh{\left (e \right )}\right )^{2} \left (c^{2} x + c d x^{2} + \frac{d^{2} x^{3}}{3}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*(a+b*sinh(f*x+e))**2,x)

[Out]

Piecewise((a**2*c**2*x + a**2*c*d*x**2 + a**2*d**2*x**3/3 + 2*a*b*c**2*cosh(e + f*x)/f + 4*a*b*c*d*x*cosh(e +
f*x)/f - 4*a*b*c*d*sinh(e + f*x)/f**2 + 2*a*b*d**2*x**2*cosh(e + f*x)/f - 4*a*b*d**2*x*sinh(e + f*x)/f**2 + 4*
a*b*d**2*cosh(e + f*x)/f**3 + b**2*c**2*x*sinh(e + f*x)**2/2 - b**2*c**2*x*cosh(e + f*x)**2/2 + b**2*c**2*sinh
(e + f*x)*cosh(e + f*x)/(2*f) + b**2*c*d*x**2*sinh(e + f*x)**2/2 - b**2*c*d*x**2*cosh(e + f*x)**2/2 + b**2*c*d
*x*sinh(e + f*x)*cosh(e + f*x)/f - b**2*c*d*sinh(e + f*x)**2/(2*f**2) + b**2*d**2*x**3*sinh(e + f*x)**2/6 - b*
*2*d**2*x**3*cosh(e + f*x)**2/6 + b**2*d**2*x**2*sinh(e + f*x)*cosh(e + f*x)/(2*f) - b**2*d**2*x*sinh(e + f*x)
**2/(4*f**2) - b**2*d**2*x*cosh(e + f*x)**2/(4*f**2) + b**2*d**2*sinh(e + f*x)*cosh(e + f*x)/(4*f**3), Ne(f, 0
)), ((a + b*sinh(e))**2*(c**2*x + c*d*x**2 + d**2*x**3/3), True))

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Giac [B]  time = 1.2962, size = 470, normalized size = 2.58 \begin{align*} \frac{1}{3} \, a^{2} d^{2} x^{3} - \frac{1}{6} \, b^{2} d^{2} x^{3} + a^{2} c d x^{2} - \frac{1}{2} \, b^{2} c d x^{2} + a^{2} c^{2} x - \frac{1}{2} \, b^{2} c^{2} x + \frac{{\left (2 \, b^{2} d^{2} f^{2} x^{2} + 4 \, b^{2} c d f^{2} x + 2 \, b^{2} c^{2} f^{2} - 2 \, b^{2} d^{2} f x - 2 \, b^{2} c d f + b^{2} d^{2}\right )} e^{\left (2 \, f x + 2 \, e\right )}}{16 \, f^{3}} + \frac{{\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2} - 2 \, a b d^{2} f x - 2 \, a b c d f + 2 \, a b d^{2}\right )} e^{\left (f x + e\right )}}{f^{3}} + \frac{{\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2} + 2 \, a b d^{2} f x + 2 \, a b c d f + 2 \, a b d^{2}\right )} e^{\left (-f x - e\right )}}{f^{3}} - \frac{{\left (2 \, b^{2} d^{2} f^{2} x^{2} + 4 \, b^{2} c d f^{2} x + 2 \, b^{2} c^{2} f^{2} + 2 \, b^{2} d^{2} f x + 2 \, b^{2} c d f + b^{2} d^{2}\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{16 \, f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*d^2*x^3 - 1/6*b^2*d^2*x^3 + a^2*c*d*x^2 - 1/2*b^2*c*d*x^2 + a^2*c^2*x - 1/2*b^2*c^2*x + 1/16*(2*b^2*d^
2*f^2*x^2 + 4*b^2*c*d*f^2*x + 2*b^2*c^2*f^2 - 2*b^2*d^2*f*x - 2*b^2*c*d*f + b^2*d^2)*e^(2*f*x + 2*e)/f^3 + (a*
b*d^2*f^2*x^2 + 2*a*b*c*d*f^2*x + a*b*c^2*f^2 - 2*a*b*d^2*f*x - 2*a*b*c*d*f + 2*a*b*d^2)*e^(f*x + e)/f^3 + (a*
b*d^2*f^2*x^2 + 2*a*b*c*d*f^2*x + a*b*c^2*f^2 + 2*a*b*d^2*f*x + 2*a*b*c*d*f + 2*a*b*d^2)*e^(-f*x - e)/f^3 - 1/
16*(2*b^2*d^2*f^2*x^2 + 4*b^2*c*d*f^2*x + 2*b^2*c^2*f^2 + 2*b^2*d^2*f*x + 2*b^2*c*d*f + b^2*d^2)*e^(-2*f*x - 2
*e)/f^3