3.74 \(\int (c+d x)^m \sin (a+b x) \, dx\)

Optimal. Leaf size=127 \[ -\frac{e^{i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i b (c+d x)}{d}\right )}{2 b}-\frac{e^{-i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{i b (c+d x)}{d}\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0884659, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3308, 2181} \[ -\frac{e^{i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i b (c+d x)}{d}\right )}{2 b}-\frac{e^{-i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{i b (c+d x)}{d}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^m*Sin[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.0493769, size = 121, normalized size = 0.95 \[ \frac{e^{-\frac{i (a d+b c)}{d}} (c+d x)^m \left (-e^{2 i a} \left (-\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{i b (c+d x)}{d}\right )-e^{\frac{2 i b c}{d}} \left (\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{i b (c+d x)}{d}\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^m*Sin[a + b*x],x]

[Out]

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

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{m}\sin \left ( bx+a \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^m*sin(b*x+a),x)

[Out]

int((d*x+c)^m*sin(b*x+a),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{m} \sin \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*sin(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.82587, size = 219, normalized size = 1.72 \begin{align*} -\frac{e^{\left (-\frac{d m \log \left (\frac{i \, b}{d}\right ) - i \, b c + i \, a d}{d}\right )} \Gamma \left (m + 1, \frac{i \, b d x + i \, b c}{d}\right ) + e^{\left (-\frac{d m \log \left (-\frac{i \, b}{d}\right ) + i \, b c - i \, a d}{d}\right )} \Gamma \left (m + 1, \frac{-i \, b d x - i \, b c}{d}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*sin(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{m} \sin{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*sin(b*x+a),x, algorithm="giac")

[Out]

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