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

Optimal. Leaf size=162 \[ \frac{i 2^{-2 (m+3)} e^{4 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{4 i b (c+d x)}{d}\right )}{b}-\frac{i 2^{-2 (m+3)} e^{-4 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{4 i b (c+d x)}{d}\right )}{b}+\frac{(c+d x)^{m+1}}{8 d (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.2048, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4406, 3307, 2181} \[ \frac{i 2^{-2 (m+3)} e^{4 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (-\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{4 i b (c+d x)}{d}\right )}{b}-\frac{i 2^{-2 (m+3)} e^{-4 i \left (a-\frac{b c}{d}\right )} (c+d x)^m \left (\frac{i b (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{4 i b (c+d x)}{d}\right )}{b}+\frac{(c+d x)^{m+1}}{8 d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

\begin{align*} \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx &=\int \left (\frac{1}{8} (c+d x)^m-\frac{1}{8} (c+d x)^m \cos (4 a+4 b x)\right ) \, dx\\ &=\frac{(c+d x)^{1+m}}{8 d (1+m)}-\frac{1}{8} \int (c+d x)^m \cos (4 a+4 b x) \, dx\\ &=\frac{(c+d x)^{1+m}}{8 d (1+m)}-\frac{1}{16} \int e^{-i (4 a+4 b x)} (c+d x)^m \, dx-\frac{1}{16} \int e^{i (4 a+4 b x)} (c+d x)^m \, dx\\ &=\frac{(c+d x)^{1+m}}{8 d (1+m)}+\frac{i 4^{-3-m} e^{4 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{4 i b (c+d x)}{d}\right )}{b}-\frac{i 4^{-3-m} e^{-4 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{4 i b (c+d x)}{d}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 1.10385, size = 213, normalized size = 1.31 \[ \frac{4^{-m-3} (c+d x)^m \left (\frac{b^2 (c+d x)^2}{d^2}\right )^{-m} \left (-i d (m+1) \left (-\frac{i b (c+d x)}{d}\right )^m \left (\cos \left (4 a-\frac{4 b c}{d}\right )-i \sin \left (4 a-\frac{4 b c}{d}\right )\right ) \text{Gamma}\left (m+1,\frac{4 i b (c+d x)}{d}\right )+i d (m+1) \left (\frac{i b (c+d x)}{d}\right )^m \left (\cos \left (4 a-\frac{4 b c}{d}\right )+i \sin \left (4 a-\frac{4 b c}{d}\right )\right ) \text{Gamma}\left (m+1,-\frac{4 i b (c+d x)}{d}\right )+b 2^{2 m+3} (c+d x) \left (\frac{b^2 (c+d x)^2}{d^2}\right )^m\right )}{b d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.209, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{m} \left ( \cos \left ( bx+a \right ) \right ) ^{2} \left ( \sin \left ( bx+a \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (d m + d\right )} \int{\left (d x + c\right )}^{m} \cos \left (4 \, b x + 4 \, a\right )\,{d x} - e^{\left (m \log \left (d x + c\right ) + \log \left (d x + c\right )\right )}}{8 \,{\left (d m + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((d*m + d)*integrate((d*x + c)^m*cos(4*b*x + 4*a), x) - e^(m*log(d*x + c) + log(d*x + c)))/(d*m + d)

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Fricas [A]  time = 0.526546, size = 342, normalized size = 2.11 \begin{align*} \frac{{\left (-i \, d m - i \, d\right )} e^{\left (-\frac{d m \log \left (\frac{4 i \, b}{d}\right ) - 4 i \, b c + 4 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac{4 i \, b d x + 4 i \, b c}{d}\right ) +{\left (i \, d m + i \, d\right )} e^{\left (-\frac{d m \log \left (-\frac{4 i \, b}{d}\right ) + 4 i \, b c - 4 i \, a d}{d}\right )} \Gamma \left (m + 1, \frac{-4 i \, b d x - 4 i \, b c}{d}\right ) + 8 \,{\left (b d x + b c\right )}{\left (d x + c\right )}^{m}}{64 \,{\left (b d m + b d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*((-I*d*m - I*d)*e^(-(d*m*log(4*I*b/d) - 4*I*b*c + 4*I*a*d)/d)*gamma(m + 1, (4*I*b*d*x + 4*I*b*c)/d) + (I*
d*m + I*d)*e^(-(d*m*log(-4*I*b/d) + 4*I*b*c - 4*I*a*d)/d)*gamma(m + 1, (-4*I*b*d*x - 4*I*b*c)/d) + 8*(b*d*x +
b*c)*(d*x + c)^m)/(b*d*m + b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{m} \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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