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

Optimal. Leaf size=74 \[ -\frac{2 i a d (c+d x) \sinh (e+f x)}{f^2}+\frac{i a (c+d x)^2 \cosh (e+f x)}{f}+\frac{a (c+d x)^3}{3 d}+\frac{2 i a d^2 \cosh (e+f x)}{f^3} \]

[Out]

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

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Rubi [A]  time = 0.0970964, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3317, 3296, 2638} \[ -\frac{2 i a d (c+d x) \sinh (e+f x)}{f^2}+\frac{i a (c+d x)^2 \cosh (e+f x)}{f}+\frac{a (c+d x)^3}{3 d}+\frac{2 i a d^2 \cosh (e+f x)}{f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.472278, size = 88, normalized size = 1.19 \[ \frac{a \left (3 i \left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2+2\right )\right ) \cosh (e+f x)+f^3 x \left (3 c^2+3 c d x+d^2 x^2\right )-6 i d f (c+d x) \sinh (e+f x)\right )}{3 f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.01, size = 249, normalized size = 3.4 \begin{align*}{\frac{1}{f} \left ({\frac{a{d}^{2} \left ( fx+e \right ) ^{3}}{3\,{f}^{2}}}+{\frac{i{d}^{2}a \left ( \left ( fx+e \right ) ^{2}\cosh \left ( fx+e \right ) -2\, \left ( fx+e \right ) \sinh \left ( fx+e \right ) +2\,\cosh \left ( fx+e \right ) \right ) }{{f}^{2}}}-{\frac{{d}^{2}ea \left ( fx+e \right ) ^{2}}{{f}^{2}}}-{\frac{2\,i{d}^{2}ea \left ( \left ( fx+e \right ) \cosh \left ( fx+e \right ) -\sinh \left ( fx+e \right ) \right ) }{{f}^{2}}}+{\frac{cda \left ( fx+e \right ) ^{2}}{f}}+{\frac{2\,idca \left ( \left ( fx+e \right ) \cosh \left ( fx+e \right ) -\sinh \left ( fx+e \right ) \right ) }{f}}+{\frac{{d}^{2}{e}^{2}a \left ( fx+e \right ) }{{f}^{2}}}+{\frac{i{d}^{2}{e}^{2}a\cosh \left ( fx+e \right ) }{{f}^{2}}}-2\,{\frac{deca \left ( fx+e \right ) }{f}}-{\frac{2\,ideca\cosh \left ( fx+e \right ) }{f}}+{c}^{2}a \left ( fx+e \right ) +i{c}^{2}a\cosh \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.06462, size = 190, normalized size = 2.57 \begin{align*} \frac{1}{3} \, a d^{2} x^{3} + a c d x^{2} + a c^{2} x + i \, a 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 )} + \frac{1}{2} i \, a 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{i \, a 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+I*a*sinh(f*x+e)),x, algorithm="maxima")

[Out]

1/3*a*d^2*x^3 + a*c*d*x^2 + a*c^2*x + I*a*c*d*((f*x*e^e - e^e)*e^(f*x)/f^2 + (f*x + 1)*e^(-f*x - e)/f^2) + 1/2
*I*a*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) + I*a*c^2*co
sh(f*x + e)/f

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Fricas [B]  time = 2.64756, size = 402, normalized size = 5.43 \begin{align*} \frac{{\left (3 i \, a d^{2} f^{2} x^{2} + 3 i \, a c^{2} f^{2} + 6 i \, a c d f + 6 i \, a d^{2} +{\left (6 i \, a c d f^{2} + 6 i \, a d^{2} f\right )} x +{\left (3 i \, a d^{2} f^{2} x^{2} + 3 i \, a c^{2} f^{2} - 6 i \, a c d f + 6 i \, a d^{2} +{\left (6 i \, a c d f^{2} - 6 i \, a d^{2} f\right )} x\right )} e^{\left (2 \, f x + 2 \, e\right )} + 2 \,{\left (a d^{2} f^{3} x^{3} + 3 \, a c d f^{3} x^{2} + 3 \, a c^{2} f^{3} x\right )} e^{\left (f x + e\right )}\right )} e^{\left (-f x - e\right )}}{6 \, f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.53159, size = 350, normalized size = 4.73 \begin{align*} a c^{2} x + a c d x^{2} + \frac{a d^{2} x^{3}}{3} + \begin{cases} \frac{\left (\left (2 i a c^{2} f^{11} e^{2 e} + 4 i a c d f^{11} x e^{2 e} + 4 i a c d f^{10} e^{2 e} + 2 i a d^{2} f^{11} x^{2} e^{2 e} + 4 i a d^{2} f^{10} x e^{2 e} + 4 i a d^{2} f^{9} e^{2 e}\right ) e^{- f x} + \left (2 i a c^{2} f^{11} e^{4 e} + 4 i a c d f^{11} x e^{4 e} - 4 i a c d f^{10} e^{4 e} + 2 i a d^{2} f^{11} x^{2} e^{4 e} - 4 i a d^{2} f^{10} x e^{4 e} + 4 i a d^{2} f^{9} e^{4 e}\right ) e^{f x}\right ) e^{- 3 e}}{4 f^{12}} & \text{for}\: 4 f^{12} e^{3 e} \neq 0 \\\frac{x^{3} \left (i a d^{2} e^{2 e} - i a d^{2}\right ) e^{- e}}{6} + \frac{x^{2} \left (i a c d e^{2 e} - i a c d\right ) e^{- e}}{2} + \frac{x \left (i a c^{2} e^{2 e} - i a c^{2}\right ) e^{- e}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c**2*x + a*c*d*x**2 + a*d**2*x**3/3 + Piecewise((((2*I*a*c**2*f**11*exp(2*e) + 4*I*a*c*d*f**11*x*exp(2*e) +
4*I*a*c*d*f**10*exp(2*e) + 2*I*a*d**2*f**11*x**2*exp(2*e) + 4*I*a*d**2*f**10*x*exp(2*e) + 4*I*a*d**2*f**9*exp(
2*e))*exp(-f*x) + (2*I*a*c**2*f**11*exp(4*e) + 4*I*a*c*d*f**11*x*exp(4*e) - 4*I*a*c*d*f**10*exp(4*e) + 2*I*a*d
**2*f**11*x**2*exp(4*e) - 4*I*a*d**2*f**10*x*exp(4*e) + 4*I*a*d**2*f**9*exp(4*e))*exp(f*x))*exp(-3*e)/(4*f**12
), Ne(4*f**12*exp(3*e), 0)), (x**3*(I*a*d**2*exp(2*e) - I*a*d**2)*exp(-e)/6 + x**2*(I*a*c*d*exp(2*e) - I*a*c*d
)*exp(-e)/2 + x*(I*a*c**2*exp(2*e) - I*a*c**2)*exp(-e)/2, True))

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Giac [B]  time = 1.26303, size = 205, normalized size = 2.77 \begin{align*} \frac{1}{3} \, a d^{2} x^{3} + a c d x^{2} + a c^{2} x - \frac{{\left (-i \, a d^{2} f^{2} x^{2} - 2 i \, a c d f^{2} x - i \, a c^{2} f^{2} + 2 i \, a d^{2} f x + 2 i \, a c d f - 2 i \, a d^{2}\right )} e^{\left (f x + e\right )}}{2 \, f^{3}} + \frac{{\left (i \, a d^{2} f^{2} x^{2} + 2 i \, a c d f^{2} x + i \, a c^{2} f^{2} + 2 i \, a d^{2} f x + 2 i \, a c d f + 2 i \, a d^{2}\right )} e^{\left (-f x - e\right )}}{2 \, f^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*d^2*x^3 + a*c*d*x^2 + a*c^2*x - 1/2*(-I*a*d^2*f^2*x^2 - 2*I*a*c*d*f^2*x - I*a*c^2*f^2 + 2*I*a*d^2*f*x +
2*I*a*c*d*f - 2*I*a*d^2)*e^(f*x + e)/f^3 + 1/2*(I*a*d^2*f^2*x^2 + 2*I*a*c*d*f^2*x + I*a*c^2*f^2 + 2*I*a*d^2*f*
x + 2*I*a*c*d*f + 2*I*a*d^2)*e^(-f*x - e)/f^3