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

Optimal. Leaf size=135 \[ \frac{i a e^{e-\frac{c f}{d}} (c+d x)^m \left (-\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{f (c+d x)}{d}\right )}{2 f}+\frac{i a e^{\frac{c f}{d}-e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{f (c+d x)}{d}\right )}{2 f}+\frac{a (c+d x)^{m+1}}{d (m+1)} \]

[Out]

(a*(c + d*x)^(1 + m))/(d*(1 + m)) + ((I/2)*a*E^(e - (c*f)/d)*(c + d*x)^m*Gamma[1 + m, -((f*(c + d*x))/d)])/(f*
(-((f*(c + d*x))/d))^m) + ((I/2)*a*E^(-e + (c*f)/d)*(c + d*x)^m*Gamma[1 + m, (f*(c + d*x))/d])/(f*((f*(c + d*x
))/d)^m)

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Rubi [A]  time = 0.153541, antiderivative size = 135, 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, 3308, 2181} \[ \frac{i a e^{e-\frac{c f}{d}} (c+d x)^m \left (-\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{f (c+d x)}{d}\right )}{2 f}+\frac{i a e^{\frac{c f}{d}-e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{f (c+d x)}{d}\right )}{2 f}+\frac{a (c+d x)^{m+1}}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(c + d*x)^(1 + m))/(d*(1 + m)) + ((I/2)*a*E^(e - (c*f)/d)*(c + d*x)^m*Gamma[1 + m, -((f*(c + d*x))/d)])/(f*
(-((f*(c + d*x))/d))^m) + ((I/2)*a*E^(-e + (c*f)/d)*(c + d*x)^m*Gamma[1 + m, (f*(c + d*x))/d])/(f*((f*(c + d*x
))/d)^m)

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 3308

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

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

Mathematica [A]  time = 0.55988, size = 207, normalized size = 1.53 \[ -\frac{a e^{-\frac{c f}{d}-e} (c+d x)^m (\sinh (e+f x)-i) \left (-\frac{f^2 (c+d x)^2}{d^2}\right )^{-m} \left (d e^{2 e} (m+1) \left (f \left (\frac{c}{d}+x\right )\right )^m \text{Gamma}\left (m+1,-\frac{f (c+d x)}{d}\right )+d (m+1) e^{\frac{2 c f}{d}} \left (-\frac{f (c+d x)}{d}\right )^m \text{Gamma}\left (m+1,\frac{f (c+d x)}{d}\right )-2 i f (c+d x) e^{\frac{c f}{d}+e} \left (-\frac{f^2 (c+d x)^2}{d^2}\right )^m\right )}{2 d f (m+1) \left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*E^(-e - (c*f)/d)*(c + d*x)^m*((-2*I)*E^(e + (c*f)/d)*f*(c + d*x)*(-((f^2*(c + d*x)^2)/d^2))^m + d*E^(2*e)*
(1 + m)*(f*(c/d + x))^m*Gamma[1 + m, -((f*(c + d*x))/d)] + d*E^((2*c*f)/d)*(1 + m)*(-((f*(c + d*x))/d))^m*Gamm
a[1 + m, (f*(c + d*x))/d])*(-I + Sinh[e + f*x]))/(2*d*f*(1 + m)*(-((f^2*(c + d*x)^2)/d^2))^m*(Cosh[(e + f*x)/2
] + I*Sinh[(e + f*x)/2])^2)

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{m} \left ( a+ia\sinh \left ( fx+e \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.54164, size = 301, normalized size = 2.23 \begin{align*} \frac{{\left (i \, a d m + i \, a d\right )} e^{\left (-\frac{d m \log \left (\frac{f}{d}\right ) + d e - c f}{d}\right )} \Gamma \left (m + 1, \frac{d f x + c f}{d}\right ) +{\left (i \, a d m + i \, a d\right )} e^{\left (-\frac{d m \log \left (-\frac{f}{d}\right ) - d e + c f}{d}\right )} \Gamma \left (m + 1, -\frac{d f x + c f}{d}\right ) + 2 \,{\left (a d f x + a c f\right )}{\left (d x + c\right )}^{m}}{2 \,{\left (d f m + d f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \sinh \left (f x + e\right ) + a\right )}{\left (d x + c\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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