3.148 \(\int (c+d x)^m (a+a \sin (e+f x)) \, dx\)

Optimal. Leaf size=148 \[ -\frac{a e^{i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i f (c+d x)}{d}\right )}{2 f}-\frac{a e^{-i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{i f (c+d x)}{d}\right )}{2 f}+\frac{a (c+d x)^{m+1}}{d (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.144218, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3317, 3308, 2181} \[ -\frac{a e^{i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i f (c+d x)}{d}\right )}{2 f}-\frac{a e^{-i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{i f (c+d x)}{d}\right )}{2 f}+\frac{a (c+d x)^{m+1}}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(c + d*x)^(1 + m))/(d*(1 + m)) - (a*E^(I*(e - (c*f)/d))*(c + d*x)^m*Gamma[1 + m, ((-I)*f*(c + d*x))/d])/(2*
f*(((-I)*f*(c + d*x))/d)^m) - (a*(c + d*x)^m*Gamma[1 + m, (I*f*(c + d*x))/d])/(2*E^(I*(e - (c*f)/d))*f*((I*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+a \sin (e+f x)) \, dx &=\int \left (a (c+d x)^m+a (c+d x)^m \sin (e+f x)\right ) \, dx\\ &=\frac{a (c+d x)^{1+m}}{d (1+m)}+a \int (c+d x)^m \sin (e+f x) \, dx\\ &=\frac{a (c+d x)^{1+m}}{d (1+m)}+\frac{1}{2} (i a) \int e^{-i (e+f x)} (c+d x)^m \, dx-\frac{1}{2} (i a) \int e^{i (e+f x)} (c+d x)^m \, dx\\ &=\frac{a (c+d x)^{1+m}}{d (1+m)}-\frac{a e^{i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (-\frac{i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac{i f (c+d x)}{d}\right )}{2 f}-\frac{a e^{-i \left (e-\frac{c f}{d}\right )} (c+d x)^m \left (\frac{i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac{i f (c+d x)}{d}\right )}{2 f}\\ \end{align*}

Mathematica [A]  time = 2.71381, size = 199, normalized size = 1.34 \[ -\frac{a (c+d x)^m (\sin (e+f x)+1) \left (d (m+1) \left (-\frac{i f (c+d x)}{d}\right )^{-m} \left (\cos \left (e-\frac{c f}{d}\right )+i \sin \left (e-\frac{c f}{d}\right )\right ) \text{Gamma}\left (m+1,-\frac{i f (c+d x)}{d}\right )+d (m+1) \left (\frac{i f (c+d x)}{d}\right )^{-m} \left (\cos \left (e-\frac{c f}{d}\right )-i \sin \left (e-\frac{c f}{d}\right )\right ) \text{Gamma}\left (m+1,\frac{i f (c+d x)}{d}\right )-2 c f-2 d (e+f x)+2 d e\right )}{2 d f (m+1) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.093, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{m} \left ( a+a\sin \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+a*sin(f*x+e)),x)

[Out]

int((d*x+c)^m*(a+a*sin(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+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.89878, size = 319, normalized size = 2.16 \begin{align*} -\frac{{\left (a d m + a d\right )} e^{\left (-\frac{d m \log \left (\frac{i \, f}{d}\right ) + i \, d e - i \, c f}{d}\right )} \Gamma \left (m + 1, \frac{i \, d f x + i \, c f}{d}\right ) +{\left (a d m + a d\right )} e^{\left (-\frac{d m \log \left (-\frac{i \, f}{d}\right ) - i \, d e + i \, c f}{d}\right )} \Gamma \left (m + 1, \frac{-i \, d f x - i \, 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+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

-1/2*((a*d*m + a*d)*e^(-(d*m*log(I*f/d) + I*d*e - I*c*f)/d)*gamma(m + 1, (I*d*f*x + I*c*f)/d) + (a*d*m + a*d)*
e^(-(d*m*log(-I*f/d) - I*d*e + I*c*f)/d)*gamma(m + 1, (-I*d*f*x - I*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]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \left (c + d x\right )^{m} \sin{\left (e + f x \right )}\, dx + \int \left (c + d x\right )^{m}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral((c + d*x)**m*sin(e + f*x), x) + Integral((c + d*x)**m, x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \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+a*sin(f*x+e)),x, algorithm="giac")

[Out]

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