3.2.75 \(\int (c+d x)^m (a+b \sin (e+f x))^2 \, dx\) [175]

Optimal. Leaf size=318 \[ \frac {a^2 (c+d x)^{1+m}}{d (1+m)}+\frac {b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {a b e^{i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right )}{f}-\frac {a b e^{-i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right )}{f}+\frac {i 2^{-3-m} b^2 e^{2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )}{f}-\frac {i 2^{-3-m} b^2 e^{-2 i \left (e-\frac {c f}{d}\right )} (c+d x)^m \left (\frac {i f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )}{f} \]

[Out]

a^2*(d*x+c)^(1+m)/d/(1+m)+1/2*b^2*(d*x+c)^(1+m)/d/(1+m)-a*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)-a*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)+I*2^(-
3-m)*b^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)*b^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 = 0.26, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3398, 3389, 2212, 3393, 3388} \begin {gather*} -\frac {a 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 )}{f}-\frac {a 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 )}{f}+\frac {i b^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 {i b^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 (c+d x)^{m+1}}{d (m+1)}+\frac {b^2 (c+d x)^{m+1}}{2 d (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(c + d*x)^(1 + m))/(d*(1 + m)) + (b^2*(c + d*x)^(1 + m))/(2*d*(1 + m)) - (a*b*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*b*(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)*b^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)*b^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 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 3388

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 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])
)

Rule 3398

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))^2 \, dx &=\int \left (a^2 (c+d x)^m+2 a b (c+d x)^m \sin (e+f x)+b^2 (c+d x)^m \sin ^2(e+f x)\right ) \, dx\\ &=\frac {a^2 (c+d x)^{1+m}}{d (1+m)}+(2 a b) \int (c+d x)^m \sin (e+f x) \, dx+b^2 \int (c+d x)^m \sin ^2(e+f x) \, dx\\ &=\frac {a^2 (c+d x)^{1+m}}{d (1+m)}+(i a b) \int e^{-i (e+f x)} (c+d x)^m \, dx-(i a b) \int e^{i (e+f x)} (c+d x)^m \, dx+b^2 \int \left (\frac {1}{2} (c+d x)^m-\frac {1}{2} (c+d x)^m \cos (2 e+2 f x)\right ) \, dx\\ &=\frac {a^2 (c+d x)^{1+m}}{d (1+m)}+\frac {b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {a 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 )}{f}-\frac {a 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 )}{f}-\frac {1}{2} b^2 \int (c+d x)^m \cos (2 e+2 f x) \, dx\\ &=\frac {a^2 (c+d x)^{1+m}}{d (1+m)}+\frac {b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {a 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 )}{f}-\frac {a 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 )}{f}-\frac {1}{4} b^2 \int e^{-i (2 e+2 f x)} (c+d x)^m \, dx-\frac {1}{4} b^2 \int e^{i (2 e+2 f x)} (c+d x)^m \, dx\\ &=\frac {a^2 (c+d x)^{1+m}}{d (1+m)}+\frac {b^2 (c+d x)^{1+m}}{2 d (1+m)}-\frac {a 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 )}{f}-\frac {a 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 )}{f}+\frac {i 2^{-3-m} b^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} b^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 [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(707\) vs. \(2(318)=636\).
time = 9.48, size = 707, normalized size = 2.22 \begin {gather*} \frac {2^{-3-m} (c+d x)^m \left (\frac {f^2 (c+d x)^2}{d^2}\right )^{-m} \left (2^{3+m} a^2 c f \left (\frac {f^2 (c+d x)^2}{d^2}\right )^m+2^{2+m} b^2 c f \left (\frac {f^2 (c+d x)^2}{d^2}\right )^m+2^{3+m} a^2 d f x \left (\frac {f^2 (c+d x)^2}{d^2}\right )^m+2^{2+m} b^2 d f x \left (\frac {f^2 (c+d x)^2}{d^2}\right )^m+i b^2 d \left (\frac {i f (c+d x)}{d}\right )^m \cos \left (2 e-\frac {2 c f}{d}\right ) \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )+i b^2 d m \left (\frac {i f (c+d x)}{d}\right )^m \cos \left (2 e-\frac {2 c f}{d}\right ) \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right )-i b^2 d \left (-\frac {i f (c+d x)}{d}\right )^m \cos \left (2 e-\frac {2 c f}{d}\right ) \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )-i b^2 d m \left (-\frac {i f (c+d x)}{d}\right )^m \cos \left (2 e-\frac {2 c f}{d}\right ) \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right )-b^2 d \left (\frac {i f (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )-b^2 d m \left (\frac {i f (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {2 i f (c+d x)}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )-b^2 d \left (-\frac {i f (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )-b^2 d m \left (-\frac {i f (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {2 i f (c+d x)}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )-2^{3+m} a b d (1+m) \left (-\frac {i f (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {i f (c+d x)}{d}\right ) \left (\cos \left (e-\frac {c f}{d}\right )-i \sin \left (e-\frac {c f}{d}\right )\right )-2^{3+m} a b d (1+m) \left (\frac {i f (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {i f (c+d x)}{d}\right ) \left (\cos \left (e-\frac {c f}{d}\right )+i \sin \left (e-\frac {c f}{d}\right )\right )\right )}{d f (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \left (d x +c \right )^{m} \left (a +b \sin \left (f x +e \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.10, size = 282, normalized size = 0.89 \begin {gather*} -\frac {8 \, {\left (a b d m + a b d\right )} e^{\left (-\frac {d m \log \left (\frac {i \, f}{d}\right ) - i \, c f + i \, d e}{d}\right )} \Gamma \left (m + 1, \frac {i \, d f x + i \, c f}{d}\right ) - {\left (i \, b^{2} d m + i \, b^{2} d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 i \, f}{d}\right ) + 2 i \, c f - 2 i \, d e}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) + 8 \, {\left (a b d m + a b d\right )} e^{\left (-\frac {d m \log \left (-\frac {i \, f}{d}\right ) + i \, c f - i \, d e}{d}\right )} \Gamma \left (m + 1, \frac {-i \, d f x - i \, c f}{d}\right ) - {\left (-i \, b^{2} d m - i \, b^{2} d\right )} e^{\left (-\frac {d m \log \left (\frac {2 i \, f}{d}\right ) - 2 i \, c f + 2 i \, d e}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (-i \, d f x - i \, c f\right )}}{d}\right ) - 4 \, {\left ({\left (2 \, a^{2} + b^{2}\right )} d f x + {\left (2 \, a^{2} + b^{2}\right )} c f\right )} {\left (d x + c\right )}^{m}}{8 \, {\left (d f m + d f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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