3.73 \(\int (c+d x)^m \sin ^2(a+b x) \, dx\)

Optimal. Leaf size=162 \[ \frac {i 2^{-m-3} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-m-3} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {(c+d x)^{m+1}}{2 d (m+1)} \]

[Out]

1/2*(d*x+c)^(1+m)/d/(1+m)+I*2^(-3-m)*exp(2*I*(a-b*c/d))*(d*x+c)^m*GAMMA(1+m,-2*I*b*(d*x+c)/d)/b/((-I*b*(d*x+c)
/d)^m)-I*2^(-3-m)*(d*x+c)^m*GAMMA(1+m,2*I*b*(d*x+c)/d)/b/exp(2*I*(a-b*c/d))/((I*b*(d*x+c)/d)^m)

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Rubi [A]  time = 0.22, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3312, 3307, 2181} \[ \frac {i 2^{-m-3} e^{2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {2 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-m-3} e^{-2 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {2 i b (c+d x)}{d}\right )}{b}+\frac {(c+d x)^{m+1}}{2 d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^m*Sin[a + b*x]^2,x]

[Out]

(c + d*x)^(1 + m)/(2*d*(1 + m)) + (I*2^(-3 - m)*E^((2*I)*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-2*I)*b*(c
+ d*x))/d])/(b*(((-I)*b*(c + d*x))/d)^m) - (I*2^(-3 - m)*(c + d*x)^m*Gamma[1 + m, ((2*I)*b*(c + d*x))/d])/(b*E
^((2*I)*(a - (b*c)/d))*((I*b*(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 3307

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

Rule 3312

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*} \int (c+d x)^m \sin ^2(a+b x) \, dx &=\int \left (\frac {1}{2} (c+d x)^m-\frac {1}{2} (c+d x)^m \cos (2 a+2 b x)\right ) \, dx\\ &=\frac {(c+d x)^{1+m}}{2 d (1+m)}-\frac {1}{2} \int (c+d x)^m \cos (2 a+2 b x) \, dx\\ &=\frac {(c+d x)^{1+m}}{2 d (1+m)}-\frac {1}{4} \int e^{-i (2 a+2 b x)} (c+d x)^m \, dx-\frac {1}{4} \int e^{i (2 a+2 b x)} (c+d x)^m \, dx\\ &=\frac {(c+d x)^{1+m}}{2 d (1+m)}+\frac {i 2^{-3-m} e^{2 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 {2 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-3-m} e^{-2 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 {2 i b (c+d x)}{d}\right )}{b}\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 211, normalized size = 1.30 \[ \frac {2^{-m-3} (c+d x)^m \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \left (-i d (m+1) \left (-\frac {i b (c+d x)}{d}\right )^m \left (\cos \left (2 a-\frac {2 b c}{d}\right )-i \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \Gamma \left (m+1,\frac {2 i b (c+d x)}{d}\right )+i d (m+1) \left (\frac {i b (c+d x)}{d}\right )^m \left (\cos \left (2 a-\frac {2 b c}{d}\right )+i \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \Gamma \left (m+1,-\frac {2 i b (c+d x)}{d}\right )+b 2^{m+2} (c+d x) \left (\frac {b^2 (c+d x)^2}{d^2}\right )^m\right )}{b d (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^m*Sin[a + b*x]^2,x]

[Out]

(2^(-3 - m)*(c + d*x)^m*(2^(2 + m)*b*(c + d*x)*((b^2*(c + d*x)^2)/d^2)^m - I*d*(1 + m)*(((-I)*b*(c + d*x))/d)^
m*Gamma[1 + m, ((2*I)*b*(c + d*x))/d]*(Cos[2*a - (2*b*c)/d] - I*Sin[2*a - (2*b*c)/d]) + I*d*(1 + m)*((I*b*(c +
d*x))/d)^m*Gamma[1 + m, ((-2*I)*b*(c + d*x))/d]*(Cos[2*a - (2*b*c)/d] + I*Sin[2*a - (2*b*c)/d])))/(b*d*(1 + m
)*((b^2*(c + d*x)^2)/d^2)^m)

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fricas [A]  time = 0.81, size = 134, normalized size = 0.83 \[ \frac {{\left (-i \, d m - i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, b}{d}\right ) - 2 i \, b c + 2 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {2 i \, b d x + 2 i \, b c}{d}\right ) + {\left (i \, d m + i \, d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 i \, b}{d}\right ) + 2 i \, b c - 2 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {-2 i \, b d x - 2 i \, b c}{d}\right ) + 4 \, {\left (b d x + b c\right )} {\left (d x + c\right )}^{m}}{8 \, {\left (b d m + b d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*sin(b*x+a)^2,x, algorithm="fricas")

[Out]

1/8*((-I*d*m - I*d)*e^(-(d*m*log(2*I*b/d) - 2*I*b*c + 2*I*a*d)/d)*gamma(m + 1, (2*I*b*d*x + 2*I*b*c)/d) + (I*d
*m + I*d)*e^(-(d*m*log(-2*I*b/d) + 2*I*b*c - 2*I*a*d)/d)*gamma(m + 1, (-2*I*b*d*x - 2*I*b*c)/d) + 4*(b*d*x + b
*c)*(d*x + c)^m)/(b*d*m + b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*sin(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^m*sin(b*x + a)^2, x)

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{m} \left (\sin ^{2}\left (b x +a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^m*sin(b*x+a)^2,x)

[Out]

int((d*x+c)^m*sin(b*x+a)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (d m + d\right )} \int {\left (d x + c\right )}^{m} \cos \left (2 \, b x + 2 \, a\right )\,{d x} - e^{\left (m \log \left (d x + c\right ) + \log \left (d x + c\right )\right )}}{2 \, {\left (d m + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*sin(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^2*(c + d*x)^m,x)

[Out]

int(sin(a + b*x)^2*(c + d*x)^m, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{m} \sin ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**m*sin(b*x+a)**2,x)

[Out]

Integral((c + d*x)**m*sin(a + b*x)**2, x)

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