3.147 \(\int (c+d x)^m (a+a \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=299 \[ -\frac {a^2 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i f (c+d x)}{d}\right )}{f}+\frac {i a^2 2^{-m-3} e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {a^2 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i f (c+d x)}{d}\right )}{f}-\frac {i a^2 2^{-m-3} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {3 a^2 (c+d x)^{m+1}}{2 d (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.37, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3318, 3312, 3307, 2181, 3308} \[ -\frac {a^2 e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {i f (c+d x)}{d}\right )}{f}+\frac {i a^2 2^{-m-3} e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {a^2 e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {i f (c+d x)}{d}\right )}{f}-\frac {i a^2 2^{-m-3} e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {3 a^2 (c+d x)^{m+1}}{2 d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^m*(a + a*Sin[e + f*x])^2,x]

[Out]

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

Rule 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rubi steps

\begin {align*} \int (c+d x)^m (a+a \sin (e+f x))^2 \, dx &=\left (4 a^2\right ) \int (c+d x)^m \sin ^4\left (\frac {1}{2} \left (e+\frac {\pi }{2}\right )+\frac {f x}{2}\right ) \, dx\\ &=\left (4 a^2\right ) \int \left (\frac {3}{8} (c+d x)^m-\frac {1}{8} (c+d x)^m \cos (2 e+2 f x)+\frac {1}{2} (c+d x)^m \sin (e+f x)\right ) \, dx\\ &=\frac {3 a^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {1}{2} a^2 \int (c+d x)^m \cos (2 e+2 f x) \, dx+\left (2 a^2\right ) \int (c+d x)^m \sin (e+f x) \, dx\\ &=\frac {3 a^2 (c+d x)^{1+m}}{2 d (1+m)}+\left (i a^2\right ) \int e^{-i (e+f x)} (c+d x)^m \, dx-\left (i a^2\right ) \int e^{i (e+f x)} (c+d x)^m \, dx-\frac {1}{4} a^2 \int e^{-i (2 e+2 f x)} (c+d x)^m \, dx-\frac {1}{4} a^2 \int e^{i (2 e+2 f x)} (c+d x)^m \, dx\\ &=\frac {3 a^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {a^2 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 )}{f}-\frac {a^2 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 )}{f}+\frac {i 2^{-3-m} a^2 e^{2 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 {2 i f (c+d x)}{d}\right )}{f}-\frac {i 2^{-3-m} a^2 e^{-2 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 {2 i f (c+d x)}{d}\right )}{f}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 260, normalized size = 0.87 \[ \frac {1}{8} a^2 (c+d x)^m \left (-\frac {8 e^{i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i f (c+d x)}{d}\right )}{f}+\frac {i 2^{-m} e^{2 i \left (e-\frac {c f}{d}\right )} \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {8 e^{-i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i f (c+d x)}{d}\right )}{f}-\frac {i 2^{-m} e^{-2 i \left (e-\frac {c f}{d}\right )} \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 i f (c+d x)}{d}\right )}{f}+\frac {12 (c+d x)}{d (m+1)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(c + d*x)^m*((12*(c + d*x))/(d*(1 + m)) - (8*E^(I*(e - (c*f)/d))*Gamma[1 + m, ((-I)*f*(c + d*x))/d])/(f*(
((-I)*f*(c + d*x))/d)^m) - (8*Gamma[1 + m, (I*f*(c + d*x))/d])/(E^(I*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m) +
(I*E^((2*I)*(e - (c*f)/d))*Gamma[1 + m, ((-2*I)*f*(c + d*x))/d])/(2^m*f*(((-I)*f*(c + d*x))/d)^m) - (I*Gamma[
1 + m, ((2*I)*f*(c + d*x))/d])/(2^m*E^((2*I)*(e - (c*f)/d))*f*((I*f*(c + d*x))/d)^m)))/8

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fricas [A]  time = 0.83, size = 266, normalized size = 0.89 \[ \frac {{\left (-i \, a^{2} d m - i \, a^{2} d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) + 2 i \, d e - 2 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {2 i \, d f x + 2 i \, c f}{d}\right ) - 8 \, {\left (a^{2} d m + a^{2} d\right )} e^{\left (-\frac {d m \log \left (\frac {i \, f}{d}\right ) + i \, d e - i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {i \, d f x + i \, c f}{d}\right ) - 8 \, {\left (a^{2} d m + a^{2} d\right )} e^{\left (-\frac {d m \log \left (-\frac {i \, f}{d}\right ) - i \, d e + i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-i \, d f x - i \, c f}{d}\right ) + {\left (i \, a^{2} d m + i \, a^{2} d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 i \, f}{d}\right ) - 2 i \, d e + 2 i \, c f}{d}\right )} \Gamma \left (m + 1, \frac {-2 i \, d f x - 2 i \, c f}{d}\right ) + 12 \, {\left (a^{2} d f x + a^{2} c f\right )} {\left (d x + c\right )}^{m}}{8 \, {\left (d f m + d f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((-I*a^2*d*m - I*a^2*d)*e^(-(d*m*log(2*I*f/d) + 2*I*d*e - 2*I*c*f)/d)*gamma(m + 1, (2*I*d*f*x + 2*I*c*f)/d
) - 8*(a^2*d*m + a^2*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) - 8*(a^2*d*m
+ a^2*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) + (I*a^2*d*m + I*a^2*d)*
e^(-(d*m*log(-2*I*f/d) - 2*I*d*e + 2*I*c*f)/d)*gamma(m + 1, (-2*I*d*f*x - 2*I*c*f)/d) + 12*(a^2*d*f*x + a^2*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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (d x + c\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{m} \left (a +a \sin \left (f x +e \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (d x + c\right )}^{m + 1} a^{2}}{d {\left (m + 1\right )}} + \frac {a^{2} e^{\left (m \log \left (d x + c\right ) + \log \left (d x + c\right )\right )} - {\left (a^{2} d m + a^{2} d\right )} \int {\left (d x + c\right )}^{m} \cos \left (2 \, f x + 2 \, e\right )\,{d x} + 4 \, {\left (a^{2} d m + a^{2} d\right )} \int {\left (d x + c\right )}^{m} \sin \left (f x + e\right )\,{d x}}{2 \, {\left (d m + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x + c)^(m + 1)*a^2/(d*(m + 1)) + 1/2*(a^2*e^(m*log(d*x + c) + log(d*x + c)) - (a^2*d*m + a^2*d)*integrate((
d*x + c)^m*cos(2*f*x + 2*e), x) + 4*(a^2*d*m + a^2*d)*integrate((d*x + c)^m*sin(f*x + e), x))/(d*m + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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