3.98 \(\int (c+d x) (a+i a \sinh (e+f x)) \, dx\)

Optimal. Leaf size=50 \[ \frac{i a (c+d x) \cosh (e+f x)}{f}+\frac{a (c+d x)^2}{2 d}-\frac{i a d \sinh (e+f x)}{f^2} \]

[Out]

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

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Rubi [A]  time = 0.0533599, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3317, 3296, 2637} \[ \frac{i a (c+d x) \cosh (e+f x)}{f}+\frac{a (c+d x)^2}{2 d}-\frac{i a d \sinh (e+f x)}{f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Rubi steps

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

Mathematica [A]  time = 0.0842476, size = 48, normalized size = 0.96 \[ \frac{a \left (2 i f (c+d x) \cosh (e+f x)+f^2 x (2 c+d x)-2 i d \sinh (e+f x)\right )}{2 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.009, size = 96, normalized size = 1.9 \begin{align*}{\frac{1}{f} \left ({\frac{da \left ( fx+e \right ) ^{2}}{2\,f}}+{\frac{ida \left ( \left ( fx+e \right ) \cosh \left ( fx+e \right ) -\sinh \left ( fx+e \right ) \right ) }{f}}-{\frac{dea \left ( fx+e \right ) }{f}}-{\frac{idea\cosh \left ( fx+e \right ) }{f}}+ca \left ( fx+e \right ) +ica\cosh \left ( fx+e \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)),x)

[Out]

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

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Maxima [A]  time = 1.04286, size = 89, normalized size = 1.78 \begin{align*} \frac{1}{2} \, a d x^{2} + a c x + \frac{1}{2} i \, a 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{i \, a 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.63207, size = 192, normalized size = 3.84 \begin{align*} \frac{{\left (i \, a d f x + i \, a c f + i \, a d +{\left (i \, a d f x + i \, a c f - i \, a d\right )} e^{\left (2 \, f x + 2 \, e\right )} +{\left (a d f^{2} x^{2} + 2 \, a c f^{2} x\right )} e^{\left (f x + e\right )}\right )} e^{\left (-f x - e\right )}}{2 \, 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.943545, size = 177, normalized size = 3.54 \begin{align*} a c x + \frac{a d x^{2}}{2} + \begin{cases} \frac{\left (\left (2 i a c f^{5} e^{e} + 2 i a d f^{5} x e^{e} + 2 i a d f^{4} e^{e}\right ) e^{- f x} + \left (2 i a c f^{5} e^{3 e} + 2 i a d f^{5} x e^{3 e} - 2 i a d f^{4} e^{3 e}\right ) e^{f x}\right ) e^{- 2 e}}{4 f^{6}} & \text{for}\: 4 f^{6} e^{2 e} \neq 0 \\\frac{x^{2} \left (i a d e^{2 e} - i a d\right ) e^{- e}}{4} + \frac{x \left (i a c e^{2 e} - i a c\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)*(a+I*a*sinh(f*x+e)),x)

[Out]

a*c*x + a*d*x**2/2 + Piecewise((((2*I*a*c*f**5*exp(e) + 2*I*a*d*f**5*x*exp(e) + 2*I*a*d*f**4*exp(e))*exp(-f*x)
+ (2*I*a*c*f**5*exp(3*e) + 2*I*a*d*f**5*x*exp(3*e) - 2*I*a*d*f**4*exp(3*e))*exp(f*x))*exp(-2*e)/(4*f**6), Ne(
4*f**6*exp(2*e), 0)), (x**2*(I*a*d*exp(2*e) - I*a*d)*exp(-e)/4 + x*(I*a*c*exp(2*e) - I*a*c)*exp(-e)/2, True))

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Giac [A]  time = 1.29728, size = 96, normalized size = 1.92 \begin{align*} \frac{1}{2} \, a d x^{2} + a c x - \frac{{\left (-i \, a d f x - i \, a c f + i \, a d\right )} e^{\left (f x + e\right )}}{2 \, f^{2}} - \frac{{\left (-i \, a d f x - i \, a c f - i \, a d\right )} e^{\left (-f x - e\right )}}{2 \, 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)),x, algorithm="giac")

[Out]

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