\(\int (c+d x)^m \sin ^3(a+b x) \, dx\) [72]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 16, antiderivative size = 267 \[ \int (c+d x)^m \sin ^3(a+b x) \, dx=-\frac {3 e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i b (c+d x)}{d}\right )}{8 b}-\frac {3 e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i b (c+d x)}{d}\right )}{8 b}+\frac {3^{-1-m} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 i b (c+d x)}{d}\right )}{8 b}+\frac {3^{-1-m} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 i b (c+d x)}{d}\right )}{8 b} \]

[Out]

-3/8*exp(I*(a-b*c/d))*(d*x+c)^m*GAMMA(1+m,-I*b*(d*x+c)/d)/b/((-I*b*(d*x+c)/d)^m)-3/8*(d*x+c)^m*GAMMA(1+m,I*b*(
d*x+c)/d)/b/exp(I*(a-b*c/d))/((I*b*(d*x+c)/d)^m)+1/8*3^(-1-m)*exp(3*I*(a-b*c/d))*(d*x+c)^m*GAMMA(1+m,-3*I*b*(d
*x+c)/d)/b/((-I*b*(d*x+c)/d)^m)+1/8*3^(-1-m)*(d*x+c)^m*GAMMA(1+m,3*I*b*(d*x+c)/d)/b/exp(3*I*(a-b*c/d))/((I*b*(
d*x+c)/d)^m)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3393, 3389, 2212} \[ \int (c+d x)^m \sin ^3(a+b x) \, dx=-\frac {3 e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i b (c+d x)}{d}\right )}{8 b}+\frac {3^{-m-1} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {3 i b (c+d x)}{d}\right )}{8 b}-\frac {3 e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i b (c+d x)}{d}\right )}{8 b}+\frac {3^{-m-1} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {3 i b (c+d x)}{d}\right )}{8 b} \]

[In]

Int[(c + d*x)^m*Sin[a + b*x]^3,x]

[Out]

(-3*E^(I*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-I)*b*(c + d*x))/d])/(8*b*(((-I)*b*(c + d*x))/d)^m) - (3*(c
 + d*x)^m*Gamma[1 + m, (I*b*(c + d*x))/d])/(8*b*E^(I*(a - (b*c)/d))*((I*b*(c + d*x))/d)^m) + (3^(-1 - m)*E^((3
*I)*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-3*I)*b*(c + d*x))/d])/(8*b*(((-I)*b*(c + d*x))/d)^m) + (3^(-1 -
 m)*(c + d*x)^m*Gamma[1 + m, ((3*I)*b*(c + d*x))/d])/(8*b*E^((3*I)*(a - (b*c)/d))*((I*b*(c + d*x))/d)^m)

Rule 2212

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

Rule 3389

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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3}{4} (c+d x)^m \sin (a+b x)-\frac {1}{4} (c+d x)^m \sin (3 a+3 b x)\right ) \, dx \\ & = -\left (\frac {1}{4} \int (c+d x)^m \sin (3 a+3 b x) \, dx\right )+\frac {3}{4} \int (c+d x)^m \sin (a+b x) \, dx \\ & = -\left (\frac {1}{8} i \int e^{-i (3 a+3 b x)} (c+d x)^m \, dx\right )+\frac {1}{8} i \int e^{i (3 a+3 b x)} (c+d x)^m \, dx+\frac {3}{8} i \int e^{-i (a+b x)} (c+d x)^m \, dx-\frac {3}{8} i \int e^{i (a+b x)} (c+d x)^m \, dx \\ & = -\frac {3 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 )}{8 b}-\frac {3 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 )}{8 b}+\frac {3^{-1-m} e^{3 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 {3 i b (c+d x)}{d}\right )}{8 b}+\frac {3^{-1-m} e^{-3 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 {3 i b (c+d x)}{d}\right )}{8 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.94 \[ \int (c+d x)^m \sin ^3(a+b x) \, dx=\frac {3^{-1-m} e^{-\frac {3 i (b c+a d)}{d}} (c+d x)^m \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (-3^{2+m} e^{2 i \left (2 a+\frac {b c}{d}\right )} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {i b (c+d x)}{d}\right )-3^{2+m} e^{2 i a+\frac {4 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {i b (c+d x)}{d}\right )+e^{6 i a} \left (\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {3 i b (c+d x)}{d}\right )+e^{\frac {6 i b c}{d}} \left (-\frac {i b (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {3 i b (c+d x)}{d}\right )\right )}{8 b} \]

[In]

Integrate[(c + d*x)^m*Sin[a + b*x]^3,x]

[Out]

(3^(-1 - m)*(c + d*x)^m*(-(3^(2 + m)*E^((2*I)*(2*a + (b*c)/d))*((I*b*(c + d*x))/d)^m*Gamma[1 + m, ((-I)*b*(c +
 d*x))/d]) - 3^(2 + m)*E^((2*I)*a + ((4*I)*b*c)/d)*(((-I)*b*(c + d*x))/d)^m*Gamma[1 + m, (I*b*(c + d*x))/d] +
E^((6*I)*a)*((I*b*(c + d*x))/d)^m*Gamma[1 + m, ((-3*I)*b*(c + d*x))/d] + E^(((6*I)*b*c)/d)*(((-I)*b*(c + d*x))
/d)^m*Gamma[1 + m, ((3*I)*b*(c + d*x))/d]))/(8*b*E^(((3*I)*(b*c + a*d))/d)*((b^2*(c + d*x)^2)/d^2)^m)

Maple [F]

\[\int \left (d x +c \right )^{m} \left (\sin ^{3}\left (b x +a \right )\right )d x\]

[In]

int((d*x+c)^m*sin(b*x+a)^3,x)

[Out]

int((d*x+c)^m*sin(b*x+a)^3,x)

Fricas [A] (verification not implemented)

none

Time = 0.11 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.70 \[ \int (c+d x)^m \sin ^3(a+b x) \, dx=-\frac {9 \, e^{\left (-\frac {d m \log \left (\frac {i \, b}{d}\right ) - i \, b c + i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {i \, b d x + i \, b c}{d}\right ) - e^{\left (-\frac {d m \log \left (-\frac {3 i \, b}{d}\right ) + 3 i \, b c - 3 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + 9 \, e^{\left (-\frac {d m \log \left (-\frac {i \, b}{d}\right ) + i \, b c - i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {-i \, b d x - i \, b c}{d}\right ) - e^{\left (-\frac {d m \log \left (\frac {3 i \, b}{d}\right ) - 3 i \, b c + 3 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )}{24 \, b} \]

[In]

integrate((d*x+c)^m*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/24*(9*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(-3*I*b/d) +
3*I*b*c - 3*I*a*d)/d)*gamma(m + 1, -3*(I*b*d*x + I*b*c)/d) + 9*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(3*I*b/d) - 3*I*b*c + 3*I*a*d)/d)*gamma(m + 1, -3*(-I*b*d*x - I*b*c
)/d))/b

Sympy [F]

\[ \int (c+d x)^m \sin ^3(a+b x) \, dx=\int \left (c + d x\right )^{m} \sin ^{3}{\left (a + b x \right )}\, dx \]

[In]

integrate((d*x+c)**m*sin(b*x+a)**3,x)

[Out]

Integral((c + d*x)**m*sin(a + b*x)**3, x)

Maxima [F]

\[ \int (c+d x)^m \sin ^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \sin \left (b x + a\right )^{3} \,d x } \]

[In]

integrate((d*x+c)^m*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

integrate((d*x + c)^m*sin(b*x + a)^3, x)

Giac [F]

\[ \int (c+d x)^m \sin ^3(a+b x) \, dx=\int { {\left (d x + c\right )}^{m} \sin \left (b x + a\right )^{3} \,d x } \]

[In]

integrate((d*x+c)^m*sin(b*x+a)^3,x, algorithm="giac")

[Out]

integrate((d*x + c)^m*sin(b*x + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^m \sin ^3(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^m \,d x \]

[In]

int(sin(a + b*x)^3*(c + d*x)^m,x)

[Out]

int(sin(a + b*x)^3*(c + d*x)^m, x)