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

Optimal. Leaf size=122 \[ \frac{2 i a^2 (c+d x) \cosh (e+f x)}{f}-\frac{a^2 (c+d x) \sinh (e+f x) \cosh (e+f x)}{2 f}+\frac{a^2 (c+d x)^2}{2 d}+\frac{1}{2} a^2 c x+\frac{a^2 d \sinh ^2(e+f x)}{4 f^2}-\frac{2 i a^2 d \sinh (e+f x)}{f^2}+\frac{1}{4} a^2 d x^2 \]

[Out]

(a^2*c*x)/2 + (a^2*d*x^2)/4 + (a^2*(c + d*x)^2)/(2*d) + ((2*I)*a^2*(c + d*x)*Cosh[e + f*x])/f - ((2*I)*a^2*d*S
inh[e + f*x])/f^2 - (a^2*(c + d*x)*Cosh[e + f*x]*Sinh[e + f*x])/(2*f) + (a^2*d*Sinh[e + f*x]^2)/(4*f^2)

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Rubi [A]  time = 0.105015, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3317, 3296, 2637, 3310} \[ \frac{2 i a^2 (c+d x) \cosh (e+f x)}{f}-\frac{a^2 (c+d x) \sinh (e+f x) \cosh (e+f x)}{2 f}+\frac{a^2 (c+d x)^2}{2 d}+\frac{1}{2} a^2 c x+\frac{a^2 d \sinh ^2(e+f x)}{4 f^2}-\frac{2 i a^2 d \sinh (e+f x)}{f^2}+\frac{1}{4} a^2 d x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c*x)/2 + (a^2*d*x^2)/4 + (a^2*(c + d*x)^2)/(2*d) + ((2*I)*a^2*(c + d*x)*Cosh[e + f*x])/f - ((2*I)*a^2*d*S
inh[e + f*x])/f^2 - (a^2*(c + d*x)*Cosh[e + f*x]*Sinh[e + f*x])/(2*f) + (a^2*d*Sinh[e + f*x]^2)/(4*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 2637

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

Rule 3310

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

Rubi steps

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

Mathematica [A]  time = 1.20878, size = 86, normalized size = 0.7 \[ \frac{a^2 (-2 (3 (e+f x) (-2 c f+d e-d f x)+f (c+d x) \sinh (2 (e+f x))+8 i d \sinh (e+f x))+16 i f (c+d x) \cosh (e+f x)+d \cosh (2 (e+f x)))}{8 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*((16*I)*f*(c + d*x)*Cosh[e + f*x] + d*Cosh[2*(e + f*x)] - 2*(3*(e + f*x)*(d*e - 2*c*f - d*f*x) + (8*I)*d*
Sinh[e + f*x] + f*(c + d*x)*Sinh[2*(e + f*x)])))/(8*f^2)

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Maple [A]  time = 0.015, size = 215, normalized size = 1.8 \begin{align*}{\frac{1}{f} \left ({\frac{d{a}^{2} \left ( fx+e \right ) ^{2}}{2\,f}}+{\frac{2\,id{a}^{2} \left ( \left ( fx+e \right ) \cosh \left ( fx+e \right ) -\sinh \left ( fx+e \right ) \right ) }{f}}-{\frac{d{a}^{2}}{f} \left ({\frac{ \left ( fx+e \right ) \cosh \left ( fx+e \right ) \sinh \left ( fx+e \right ) }{2}}-{\frac{ \left ( fx+e \right ) ^{2}}{4}}-{\frac{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{4}} \right ) }-{\frac{de{a}^{2} \left ( fx+e \right ) }{f}}-{\frac{2\,ide{a}^{2}\cosh \left ( fx+e \right ) }{f}}+{\frac{de{a}^{2}}{f} \left ({\frac{\cosh \left ( fx+e \right ) \sinh \left ( fx+e \right ) }{2}}-{\frac{fx}{2}}-{\frac{e}{2}} \right ) }+c{a}^{2} \left ( fx+e \right ) +2\,ic{a}^{2}\cosh \left ( fx+e \right ) -c{a}^{2} \left ({\frac{\cosh \left ( fx+e \right ) \sinh \left ( fx+e \right ) }{2}}-{\frac{fx}{2}}-{\frac{e}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(1/2/f*d*a^2*(f*x+e)^2+2*I/f*d*a^2*((f*x+e)*cosh(f*x+e)-sinh(f*x+e))-1/f*d*a^2*(1/2*(f*x+e)*cosh(f*x+e)*si
nh(f*x+e)-1/4*(f*x+e)^2-1/4*cosh(f*x+e)^2)-d*e/f*a^2*(f*x+e)-2*I*d*e/f*a^2*cosh(f*x+e)+d*e/f*a^2*(1/2*cosh(f*x
+e)*sinh(f*x+e)-1/2*f*x-1/2*e)+c*a^2*(f*x+e)+2*I*c*a^2*cosh(f*x+e)-c*a^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 = 1.18573, size = 225, normalized size = 1.84 \begin{align*} \frac{1}{2} \, a^{2} d x^{2} + \frac{1}{16} \,{\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 )} a^{2} d + \frac{1}{8} \, a^{2} c{\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 x + i \, a^{2} 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{2 i \, a^{2} c \cosh \left (f x + e\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*d*x^2 + 1/16*(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)*a^2*
d + 1/8*a^2*c*(4*x - e^(2*f*x + 2*e)/f + e^(-2*f*x - 2*e)/f) + a^2*c*x + I*a^2*d*((f*x*e^e - e^e)*e^(f*x)/f^2
+ (f*x + 1)*e^(-f*x - e)/f^2) + 2*I*a^2*c*cosh(f*x + e)/f

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Fricas [A]  time = 2.9555, size = 389, normalized size = 3.19 \begin{align*} \frac{{\left (2 \, a^{2} d f x + 2 \, a^{2} c f + a^{2} d -{\left (2 \, a^{2} d f x + 2 \, a^{2} c f - a^{2} d\right )} e^{\left (4 \, f x + 4 \, e\right )} +{\left (16 i \, a^{2} d f x + 16 i \, a^{2} c f - 16 i \, a^{2} d\right )} e^{\left (3 \, f x + 3 \, e\right )} + 12 \,{\left (a^{2} d f^{2} x^{2} + 2 \, a^{2} c f^{2} x\right )} e^{\left (2 \, f x + 2 \, e\right )} +{\left (16 i \, a^{2} d f x + 16 i \, a^{2} c f + 16 i \, a^{2} d\right )} e^{\left (f x + e\right )}\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{16 \, f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*a^2*d*f*x + 2*a^2*c*f + a^2*d - (2*a^2*d*f*x + 2*a^2*c*f - a^2*d)*e^(4*f*x + 4*e) + (16*I*a^2*d*f*x +
16*I*a^2*c*f - 16*I*a^2*d)*e^(3*f*x + 3*e) + 12*(a^2*d*f^2*x^2 + 2*a^2*c*f^2*x)*e^(2*f*x + 2*e) + (16*I*a^2*d*
f*x + 16*I*a^2*c*f + 16*I*a^2*d)*e^(f*x + e))*e^(-2*f*x - 2*e)/f^2

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Sympy [A]  time = 2.06403, size = 364, normalized size = 2.98 \begin{align*} \frac{3 a^{2} c x}{2} + \frac{3 a^{2} d x^{2}}{4} + \begin{cases} \frac{\left (\left (32 a^{2} c f^{9} e^{2 e} + 32 a^{2} d f^{9} x e^{2 e} + 16 a^{2} d f^{8} e^{2 e}\right ) e^{- 2 f x} + \left (- 32 a^{2} c f^{9} e^{6 e} - 32 a^{2} d f^{9} x e^{6 e} + 16 a^{2} d f^{8} e^{6 e}\right ) e^{2 f x} + \left (256 i a^{2} c f^{9} e^{3 e} + 256 i a^{2} d f^{9} x e^{3 e} + 256 i a^{2} d f^{8} e^{3 e}\right ) e^{- f x} + \left (256 i a^{2} c f^{9} e^{5 e} + 256 i a^{2} d f^{9} x e^{5 e} - 256 i a^{2} d f^{8} e^{5 e}\right ) e^{f x}\right ) e^{- 4 e}}{256 f^{10}} & \text{for}\: 256 f^{10} e^{4 e} \neq 0 \\\frac{x^{2} \left (- a^{2} d e^{4 e} + 4 i a^{2} d e^{3 e} - 4 i a^{2} d e^{e} - a^{2} d\right ) e^{- 2 e}}{8} + \frac{x \left (- a^{2} c e^{4 e} + 4 i a^{2} c e^{3 e} - 4 i a^{2} c e^{e} - a^{2} c\right ) e^{- 2 e}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a**2*c*x/2 + 3*a**2*d*x**2/4 + Piecewise((((32*a**2*c*f**9*exp(2*e) + 32*a**2*d*f**9*x*exp(2*e) + 16*a**2*d*
f**8*exp(2*e))*exp(-2*f*x) + (-32*a**2*c*f**9*exp(6*e) - 32*a**2*d*f**9*x*exp(6*e) + 16*a**2*d*f**8*exp(6*e))*
exp(2*f*x) + (256*I*a**2*c*f**9*exp(3*e) + 256*I*a**2*d*f**9*x*exp(3*e) + 256*I*a**2*d*f**8*exp(3*e))*exp(-f*x
) + (256*I*a**2*c*f**9*exp(5*e) + 256*I*a**2*d*f**9*x*exp(5*e) - 256*I*a**2*d*f**8*exp(5*e))*exp(f*x))*exp(-4*
e)/(256*f**10), Ne(256*f**10*exp(4*e), 0)), (x**2*(-a**2*d*exp(4*e) + 4*I*a**2*d*exp(3*e) - 4*I*a**2*d*exp(e)
- a**2*d)*exp(-2*e)/8 + x*(-a**2*c*exp(4*e) + 4*I*a**2*c*exp(3*e) - 4*I*a**2*c*exp(e) - a**2*c)*exp(-2*e)/4, T
rue))

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Giac [A]  time = 1.24014, size = 215, normalized size = 1.76 \begin{align*} \frac{3}{4} \, a^{2} d x^{2} + \frac{3}{2} \, a^{2} c x - \frac{{\left (2 \, a^{2} d f x + 2 \, a^{2} c f - a^{2} d\right )} e^{\left (2 \, f x + 2 \, e\right )}}{16 \, f^{2}} + \frac{{\left (i \, a^{2} d f x + i \, a^{2} c f - i \, a^{2} d\right )} e^{\left (f x + e\right )}}{f^{2}} + \frac{{\left (i \, a^{2} d f x + i \, a^{2} c f + i \, a^{2} d\right )} e^{\left (-f x - e\right )}}{f^{2}} + \frac{{\left (2 \, a^{2} d f x + 2 \, a^{2} c f + a^{2} d\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{16 \, f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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