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

Optimal. Leaf size=275 \[ -\frac {i e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i b (c+d x)}{d}\right )}{8 b}+\frac {i 3^{-m-1} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {3 i b (c+d x)}{d}\right )}{8 b}+\frac {i e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {i b (c+d x)}{d}\right )}{8 b}-\frac {i 3^{-m-1} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {3 i b (c+d x)}{d}\right )}{8 b} \]

[Out]

-1/8*I*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)+1/8*I*(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*I*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*I*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)

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Rubi [A]  time = 0.33, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4406, 3307, 2181} \[ -\frac {i e^{i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {i b (c+d x)}{d}\right )}{8 b}+\frac {i 3^{-m-1} e^{3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {3 i b (c+d x)}{d}\right )}{8 b}+\frac {i e^{-i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {i b (c+d x)}{d}\right )}{8 b}-\frac {i 3^{-m-1} e^{-3 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {3 i b (c+d x)}{d}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/8)*E^(I*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-I)*b*(c + d*x))/d])/(b*(((-I)*b*(c + d*x))/d)^m) + ((I
/8)*(c + d*x)^m*Gamma[1 + m, (I*b*(c + d*x))/d])/(b*E^(I*(a - (b*c)/d))*((I*b*(c + d*x))/d)^m) + ((I/8)*3^(-1
- m)*E^((3*I)*(a - (b*c)/d))*(c + d*x)^m*Gamma[1 + m, ((-3*I)*b*(c + d*x))/d])/(b*(((-I)*b*(c + d*x))/d)^m) -
((I/8)*3^(-1 - m)*(c + d*x)^m*Gamma[1 + m, ((3*I)*b*(c + d*x))/d])/(b*E^((3*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 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.69, size = 237, normalized size = 0.86 \[ -\frac {i e^{-\frac {3 i (a d+b c)}{d}} (c+d x)^m \left (\left (\frac {i b (c+d x)}{d}\right )^{-m} \left (3^{-m} \left (e^{\frac {6 i b c}{d}} \Gamma \left (m+1,\frac {3 i b (c+d x)}{d}\right )-e^{6 i a} \left (\frac {i b (c+d x)}{d}\right )^{2 m} \left (\frac {b^2 (c+d x)^2}{d^2}\right )^{-m} \Gamma \left (m+1,-\frac {3 i b (c+d x)}{d}\right )\right )-3 e^{2 i a+\frac {4 i b c}{d}} \Gamma \left (m+1,\frac {i b (c+d x)}{d}\right )\right )+3 e^{2 i \left (2 a+\frac {b c}{d}\right )} \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {i b (c+d x)}{d}\right )\right )}{24 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/24*I)*(c + d*x)^m*((3*E^((2*I)*(2*a + (b*c)/d))*Gamma[1 + m, ((-I)*b*(c + d*x))/d])/(((-I)*b*(c + d*x))/d
)^m + (-3*E^((2*I)*a + ((4*I)*b*c)/d)*Gamma[1 + m, (I*b*(c + d*x))/d] + (-((E^((6*I)*a)*((I*b*(c + d*x))/d)^(2
*m)*Gamma[1 + m, ((-3*I)*b*(c + d*x))/d])/((b^2*(c + d*x)^2)/d^2)^m) + E^(((6*I)*b*c)/d)*Gamma[1 + m, ((3*I)*b
*(c + d*x))/d])/3^m)/((I*b*(c + d*x))/d)^m))/(b*E^(((3*I)*(b*c + a*d))/d))

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fricas [A]  time = 0.61, size = 186, normalized size = 0.68 \[ \frac {-i \, e^{\left (-\frac {d m \log \left (\frac {3 i \, b}{d}\right ) - 3 i \, b c + 3 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {3 i \, b d x + 3 i \, b c}{d}\right ) + 3 i \, e^{\left (-\frac {d m \log \left (\frac {i \, b}{d}\right ) - i \, b c + i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {i \, b d x + i \, b c}{d}\right ) - 3 i \, e^{\left (-\frac {d m \log \left (-\frac {i \, b}{d}\right ) + i \, b c - i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {-i \, b d x - i \, b c}{d}\right ) + i \, e^{\left (-\frac {d m \log \left (-\frac {3 i \, b}{d}\right ) + 3 i \, b c - 3 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac {-3 i \, b d x - 3 i \, b c}{d}\right )}{24 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(-I*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 + 3*I*b*c)/d) + 3*I*e^(-(d*m*lo
g(I*b/d) - I*b*c + I*a*d)/d)*gamma(m + 1, (I*b*d*x + I*b*c)/d) - 3*I*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) + I*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
- 3*I*b*c)/d))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{m} \cos \left (b x +a \right ) \left (\sin ^{2}\left (b x +a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(a + b*x)*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 \[ \int \left (c + d x\right )^{m} \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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