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

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

[Out]

a*(d*x+c)^(1+m)/d/(1+m)-1/2*b*exp(I*(e-c*f/d))*(d*x+c)^m*GAMMA(1+m,-I*f*(d*x+c)/d)/f/((-I*f*(d*x+c)/d)^m)-1/2*
b*(d*x+c)^m*GAMMA(1+m,I*f*(d*x+c)/d)/exp(I*(e-c*f/d))/f/((I*f*(d*x+c)/d)^m)

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Rubi [A]  time = 0.15, antiderivative size = 148, normalized size of antiderivative = 1.00, 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} \[ -\frac {b e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {i f (c+d x)}{d}\right )}{2 f}-\frac {b e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {i f (c+d x)}{d}\right )}{2 f}+\frac {a (c+d x)^{m+1}}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(c + d*x)^(1 + m))/(d*(1 + m)) - (b*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) - (b*(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 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]

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 138, normalized size = 0.93 \[ \frac {1}{2} (c+d x)^m \left (\frac {2 a (c+d x)}{d (m+1)}-\frac {b e^{i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i f (c+d x)}{d}\right )}{f}-\frac {b e^{-i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i f (c+d x)}{d}\right )}{f}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 136, normalized size = 0.92 \[ -\frac {{\left (b d m + b d\right )} e^{\left (-\frac {d m \log \left (\frac {i \, f}{d}\right ) + i \, d e - i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {i \, d f x + i \, c f}{d}\right ) + {\left (b d m + b d\right )} e^{\left (-\frac {d m \log \left (-\frac {i \, f}{d}\right ) - i \, d e + i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-i \, d f x - i \, c f}{d}\right ) - 2 \, {\left (a d f x + a c f\right )} {\left (d x + c\right )}^{m}}{2 \, {\left (d f m + d f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((b*d*m + b*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) + (b*d*m + b*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right ) + a\right )} {\left (d x + c\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{m} \left (a +b \sin \left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int {\left (d x + c\right )}^{m} \sin \left (f x + e\right )\,{d x} + \frac {{\left (d x + c\right )}^{m + 1} a}{d {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate((d*x + c)^m*sin(f*x + e), x) + (d*x + c)^(m + 1)*a/(d*(m + 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+b\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (e + f x \right )}\right ) \left (c + d x\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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