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

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

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3317, 3296, 2637, 3311, 32, 2635, 8} \[ \frac {a^2 (c+d x)^3}{3 d}-\frac {4 a b d (c+d x) \cosh (e+f x)}{f^2}+\frac {2 a b (c+d x)^2 \sinh (e+f x)}{f}+\frac {4 a b d^2 \sinh (e+f x)}{f^3}-\frac {b^2 d (c+d x) \cosh ^2(e+f x)}{2 f^2}+\frac {b^2 (c+d x)^2 \sinh (e+f x) \cosh (e+f x)}{2 f}+\frac {b^2 (c+d x)^3}{6 d}+\frac {b^2 d^2 \sinh (e+f x) \cosh (e+f x)}{4 f^3}+\frac {b^2 d^2 x}{4 f^2} \]

Antiderivative was successfully verified.

[In]

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

Rule 8

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

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 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 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])

Rubi steps

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

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Mathematica [A]  time = 1.05, size = 252, normalized size = 1.38 \[ \frac {1}{24} \left (24 a^2 c^2 x+24 a^2 c d x^2+8 a^2 d^2 x^3+\frac {48 a b c^2 \sinh (e+f x)}{f}-\frac {96 a b d (c+d x) \cosh (e+f x)}{f^2}+\frac {96 a b c d x \sinh (e+f x)}{f}+\frac {96 a b d^2 \sinh (e+f x)}{f^3}+\frac {48 a b d^2 x^2 \sinh (e+f x)}{f}+\frac {6 b^2 c^2 \sinh (2 (e+f x))}{f}+12 b^2 c^2 x-\frac {6 b^2 d (c+d x) \cosh (2 (e+f x))}{f^2}+\frac {12 b^2 c d x \sinh (2 (e+f x))}{f}+12 b^2 c d x^2+\frac {3 b^2 d^2 \sinh (2 (e+f x))}{f^3}+\frac {6 b^2 d^2 x^2 \sinh (2 (e+f x))}{f}+4 b^2 d^2 x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 240, normalized size = 1.32 \[ \frac {2 \, {\left (2 \, a^{2} + b^{2}\right )} d^{2} f^{3} x^{3} + 6 \, {\left (2 \, a^{2} + b^{2}\right )} c d f^{3} x^{2} + 6 \, {\left (2 \, a^{2} + b^{2}\right )} c^{2} f^{3} x - 3 \, {\left (b^{2} d^{2} f x + b^{2} c d f\right )} \cosh \left (f x + e\right )^{2} - 3 \, {\left (b^{2} d^{2} f x + b^{2} c d f\right )} \sinh \left (f x + e\right )^{2} - 48 \, {\left (a b d^{2} f x + a b c d f\right )} \cosh \left (f x + e\right ) + 3 \, {\left (8 \, a b d^{2} f^{2} x^{2} + 16 \, a b c d f^{2} x + 8 \, a b c^{2} f^{2} + 16 \, a b d^{2} + {\left (2 \, b^{2} d^{2} f^{2} x^{2} + 4 \, b^{2} c d f^{2} x + 2 \, b^{2} c^{2} f^{2} + b^{2} d^{2}\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )}{12 \, f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*(a+b*cosh(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 - 48*(a*b*d^2*f*x + a*b*c*d*f)*cosh(
f*x + e) + 3*(8*a*b*d^2*f^2*x^2 + 16*a*b*c*d*f^2*x + 8*a*b*c^2*f^2 + 16*a*b*d^2 + (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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giac [B]  time = 0.15, size = 349, normalized size = 1.92 \[ \frac {1}{3} \, a^{2} d^{2} x^{3} + \frac {1}{6} \, b^{2} d^{2} x^{3} + a^{2} c d x^{2} + \frac {1}{2} \, b^{2} c d x^{2} + a^{2} c^{2} x + \frac {1}{2} \, b^{2} c^{2} x + \frac {{\left (2 \, b^{2} d^{2} f^{2} x^{2} + 4 \, b^{2} c d f^{2} x + 2 \, b^{2} c^{2} f^{2} - 2 \, b^{2} d^{2} f x - 2 \, b^{2} c d f + b^{2} d^{2}\right )} e^{\left (2 \, f x + 2 \, e\right )}}{16 \, f^{3}} + \frac {{\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2} - 2 \, a b d^{2} f x - 2 \, a b c d f + 2 \, a b d^{2}\right )} e^{\left (f x + e\right )}}{f^{3}} - \frac {{\left (a b d^{2} f^{2} x^{2} + 2 \, a b c d f^{2} x + a b c^{2} f^{2} + 2 \, a b d^{2} f x + 2 \, a b c d f + 2 \, a b d^{2}\right )} e^{\left (-f x - e\right )}}{f^{3}} - \frac {{\left (2 \, b^{2} d^{2} f^{2} x^{2} + 4 \, b^{2} c d f^{2} x + 2 \, b^{2} c^{2} f^{2} + 2 \, b^{2} d^{2} f x + 2 \, b^{2} c d f + b^{2} d^{2}\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{16 \, f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*(a+b*cosh(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

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maple [B]  time = 0.07, size = 535, normalized size = 2.94 \[ \frac {\frac {d^{2} a^{2} \left (f x +e \right )^{3}}{3 f^{2}}+\frac {2 d^{2} a b \left (\left (f x +e \right )^{2} \sinh \left (f x +e \right )-2 \left (f x +e \right ) \cosh \left (f x +e \right )+2 \sinh \left (f x +e \right )\right )}{f^{2}}+\frac {d^{2} b^{2} \left (\frac {\left (f x +e \right )^{2} \cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}+\frac {\left (f x +e \right )^{3}}{6}-\frac {\left (f x +e \right ) \left (\cosh ^{2}\left (f x +e \right )\right )}{2}+\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{4}+\frac {f x}{4}+\frac {e}{4}\right )}{f^{2}}-\frac {d^{2} e \,a^{2} \left (f x +e \right )^{2}}{f^{2}}-\frac {4 d^{2} e a b \left (\left (f x +e \right ) \sinh \left (f x +e \right )-\cosh \left (f x +e \right )\right )}{f^{2}}-\frac {2 d^{2} e \,b^{2} \left (\frac {\left (f x +e \right ) \cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}+\frac {\left (f x +e \right )^{2}}{4}-\frac {\left (\cosh ^{2}\left (f x +e \right )\right )}{4}\right )}{f^{2}}+\frac {d^{2} e^{2} a^{2} \left (f x +e \right )}{f^{2}}+\frac {2 d^{2} e^{2} a b \sinh \left (f x +e \right )}{f^{2}}+\frac {d^{2} e^{2} b^{2} \left (\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f^{2}}+\frac {c d \,a^{2} \left (f x +e \right )^{2}}{f}+\frac {4 c d a b \left (\left (f x +e \right ) \sinh \left (f x +e \right )-\cosh \left (f x +e \right )\right )}{f}+\frac {2 c d \,b^{2} \left (\frac {\left (f x +e \right ) \cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}+\frac {\left (f x +e \right )^{2}}{4}-\frac {\left (\cosh ^{2}\left (f x +e \right )\right )}{4}\right )}{f}-\frac {2 c d e \,a^{2} \left (f x +e \right )}{f}-\frac {4 c d e a b \sinh \left (f x +e \right )}{f}-\frac {2 c d e \,b^{2} \left (\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}+c^{2} a^{2} \left (f x +e \right )+2 c^{2} a b \sinh \left (f x +e \right )+c^{2} b^{2} \left (\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*(a+b*cosh(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*sinh(f*x+e)-2*(f*x+e)*cosh(f*x+e)+2*sinh(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)*sinh(f*x+e)-cosh(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*sinh(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)*sinh(f*x+e)-cosh(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*sinh(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*sinh(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 = 0.37, size = 324, normalized size = 1.78 \[ \frac {1}{3} \, a^{2} d^{2} x^{3} + a^{2} c d x^{2} + \frac {1}{8} \, {\left (4 \, x^{2} + \frac {{\left (2 \, f x e^{\left (2 \, e\right )} - e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f^{2}} - \frac {{\left (2 \, f x + 1\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{f^{2}}\right )} b^{2} c d + \frac {1}{48} \, {\left (8 \, x^{3} + \frac {3 \, {\left (2 \, f^{2} x^{2} e^{\left (2 \, e\right )} - 2 \, f x e^{\left (2 \, e\right )} + e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f^{3}} - \frac {3 \, {\left (2 \, f^{2} x^{2} + 2 \, f x + 1\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{f^{3}}\right )} b^{2} d^{2} + \frac {1}{8} \, b^{2} c^{2} {\left (4 \, x + \frac {e^{\left (2 \, f x + 2 \, e\right )}}{f} - \frac {e^{\left (-2 \, f x - 2 \, e\right )}}{f}\right )} + a^{2} c^{2} x + 2 \, a b c d {\left (\frac {{\left (f x e^{e} - e^{e}\right )} e^{\left (f x\right )}}{f^{2}} - \frac {{\left (f x + 1\right )} e^{\left (-f x - e\right )}}{f^{2}}\right )} + a b d^{2} {\left (\frac {{\left (f^{2} x^{2} e^{e} - 2 \, f x e^{e} + 2 \, e^{e}\right )} e^{\left (f x\right )}}{f^{3}} - \frac {{\left (f^{2} x^{2} + 2 \, f x + 2\right )} e^{\left (-f x - e\right )}}{f^{3}}\right )} + \frac {2 \, a b c^{2} \sinh \left (f x + e\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*(a+b*cosh(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*sinh(f*x + e)/f

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mupad [B]  time = 1.33, size = 281, normalized size = 1.54 \[ a^2\,c^2\,x+\frac {b^2\,c^2\,x}{2}+\frac {a^2\,d^2\,x^3}{3}+\frac {b^2\,d^2\,x^3}{6}+\frac {b^2\,c^2\,\mathrm {sinh}\left (2\,e+2\,f\,x\right )}{4\,f}+\frac {b^2\,d^2\,\mathrm {sinh}\left (2\,e+2\,f\,x\right )}{8\,f^3}+a^2\,c\,d\,x^2+\frac {b^2\,c\,d\,x^2}{2}+\frac {2\,a\,b\,c^2\,\mathrm {sinh}\left (e+f\,x\right )}{f}+\frac {4\,a\,b\,d^2\,\mathrm {sinh}\left (e+f\,x\right )}{f^3}+\frac {b^2\,d^2\,x^2\,\mathrm {sinh}\left (2\,e+2\,f\,x\right )}{4\,f}-\frac {b^2\,c\,d\,\mathrm {cosh}\left (2\,e+2\,f\,x\right )}{4\,f^2}-\frac {b^2\,d^2\,x\,\mathrm {cosh}\left (2\,e+2\,f\,x\right )}{4\,f^2}-\frac {4\,a\,b\,c\,d\,\mathrm {cosh}\left (e+f\,x\right )}{f^2}-\frac {4\,a\,b\,d^2\,x\,\mathrm {cosh}\left (e+f\,x\right )}{f^2}+\frac {2\,a\,b\,d^2\,x^2\,\mathrm {sinh}\left (e+f\,x\right )}{f}+\frac {b^2\,c\,d\,x\,\mathrm {sinh}\left (2\,e+2\,f\,x\right )}{2\,f}+\frac {4\,a\,b\,c\,d\,x\,\mathrm {sinh}\left (e+f\,x\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*c^2*x + (b^2*c^2*x)/2 + (a^2*d^2*x^3)/3 + (b^2*d^2*x^3)/6 + (b^2*c^2*sinh(2*e + 2*f*x))/(4*f) + (b^2*d^2*s
inh(2*e + 2*f*x))/(8*f^3) + a^2*c*d*x^2 + (b^2*c*d*x^2)/2 + (2*a*b*c^2*sinh(e + f*x))/f + (4*a*b*d^2*sinh(e +
f*x))/f^3 + (b^2*d^2*x^2*sinh(2*e + 2*f*x))/(4*f) - (b^2*c*d*cosh(2*e + 2*f*x))/(4*f^2) - (b^2*d^2*x*cosh(2*e
+ 2*f*x))/(4*f^2) - (4*a*b*c*d*cosh(e + f*x))/f^2 - (4*a*b*d^2*x*cosh(e + f*x))/f^2 + (2*a*b*d^2*x^2*sinh(e +
f*x))/f + (b^2*c*d*x*sinh(2*e + 2*f*x))/(2*f) + (4*a*b*c*d*x*sinh(e + f*x))/f

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sympy [A]  time = 1.65, size = 456, normalized size = 2.51 \[ \begin {cases} a^{2} c^{2} x + a^{2} c d x^{2} + \frac {a^{2} d^{2} x^{3}}{3} + \frac {2 a b c^{2} \sinh {\left (e + f x \right )}}{f} + \frac {4 a b c d x \sinh {\left (e + f x \right )}}{f} - \frac {4 a b c d \cosh {\left (e + f x \right )}}{f^{2}} + \frac {2 a b d^{2} x^{2} \sinh {\left (e + f x \right )}}{f} - \frac {4 a b d^{2} x \cosh {\left (e + f x \right )}}{f^{2}} + \frac {4 a b d^{2} \sinh {\left (e + f x \right )}}{f^{3}} - \frac {b^{2} c^{2} x \sinh ^{2}{\left (e + f x \right )}}{2} + \frac {b^{2} c^{2} x \cosh ^{2}{\left (e + f x \right )}}{2} + \frac {b^{2} c^{2} \sinh {\left (e + f x \right )} \cosh {\left (e + f x \right )}}{2 f} - \frac {b^{2} c d x^{2} \sinh ^{2}{\left (e + f x \right )}}{2} + \frac {b^{2} c d x^{2} \cosh ^{2}{\left (e + f x \right )}}{2} + \frac {b^{2} c d x \sinh {\left (e + f x \right )} \cosh {\left (e + f x \right )}}{f} - \frac {b^{2} c d \sinh ^{2}{\left (e + f x \right )}}{2 f^{2}} - \frac {b^{2} d^{2} x^{3} \sinh ^{2}{\left (e + f x \right )}}{6} + \frac {b^{2} d^{2} x^{3} \cosh ^{2}{\left (e + f x \right )}}{6} + \frac {b^{2} d^{2} x^{2} \sinh {\left (e + f x \right )} \cosh {\left (e + f x \right )}}{2 f} - \frac {b^{2} d^{2} x \sinh ^{2}{\left (e + f x \right )}}{4 f^{2}} - \frac {b^{2} d^{2} x \cosh ^{2}{\left (e + f x \right )}}{4 f^{2}} + \frac {b^{2} d^{2} \sinh {\left (e + f x \right )} \cosh {\left (e + f x \right )}}{4 f^{3}} & \text {for}\: f \neq 0 \\\left (a + b \cosh {\relax (e )}\right )^{2} \left (c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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