\(\int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx\) [79]

Optimal result

Integrand size = 24, antiderivative size = 162 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {(c+d x)^{1+m}}{8 d (1+m)}+\frac {i 2^{-2 (3+m)} e^{4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {4 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-2 (3+m)} e^{-4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 i b (c+d x)}{d}\right )}{b} \]

[Out]

1/8*(d*x+c)^(1+m)/d/(1+m)+I*exp(4*I*(a-b*c/d))*(d*x+c)^m*GAMMA(1+m,-4*I*b*(d*x+c)/d)/(2^(6+2*m))/b/((-I*b*(d*x
+c)/d)^m)-I*(d*x+c)^m*GAMMA(1+m,4*I*b*(d*x+c)/d)/(2^(6+2*m))/b/exp(4*I*(a-b*c/d))/((I*b*(d*x+c)/d)^m)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4491, 3388, 2212} \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {i 2^{-2 (m+3)} e^{4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {4 i b (c+d x)}{d}\right )}{b}-\frac {i 2^{-2 (m+3)} e^{-4 i \left (a-\frac {b c}{d}\right )} (c+d x)^m \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {4 i b (c+d x)}{d}\right )}{b}+\frac {(c+d x)^{m+1}}{8 d (m+1)} \]

[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 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 4491

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*} \text {integral}& = \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] (verified)

Time = 3.78 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.96 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=-\frac {(c+d x)^m \left (-8 b (c+d x)-i 4^{-m} d e^{4 i \left (a-\frac {b c}{d}\right )} (1+m) \left (-\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {4 i b (c+d x)}{d}\right )+i 4^{-m} d e^{-4 i \left (a-\frac {b c}{d}\right )} (1+m) \left (\frac {i b (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 i b (c+d x)}{d}\right )\right )}{64 b d (1+m)} \]

[In]

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

[Out]

-1/64*((c + d*x)^m*(-8*b*(c + d*x) - (I*d*E^((4*I)*(a - (b*c)/d))*(1 + m)*Gamma[1 + m, ((-4*I)*b*(c + d*x))/d]
)/(4^m*(((-I)*b*(c + d*x))/d)^m) + (I*d*(1 + m)*Gamma[1 + m, ((4*I)*b*(c + d*x))/d])/(4^m*E^((4*I)*(a - (b*c)/
d))*((I*b*(c + d*x))/d)^m)))/(b*d*(1 + m))

Maple [F]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.84 \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {{\left (i \, d m + i \, d\right )} e^{\left (-\frac {d m \log \left (-\frac {4 i \, b}{d}\right ) + 4 i \, b c - 4 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {4 \, {\left (i \, b d x + i \, b c\right )}}{d}\right ) + {\left (-i \, d m - i \, d\right )} e^{\left (-\frac {d m \log \left (\frac {4 i \, b}{d}\right ) - 4 i \, b c + 4 i \, a d}{d}\right )} \Gamma \left (m + 1, -\frac {4 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) + 8 \, {\left (b d x + b c\right )} {\left (d x + c\right )}^{m}}{64 \, {\left (b d m + b d\right )}} \]

[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 + 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 - I*b*c)/d) + 8*(b*d*x + b*
c)*(d*x + c)^m)/(b*d*m + b*d)

Sympy [F(-2)]

Exception generated. \[ \int (c+d x)^m \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[In]

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

[Out]

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