Integrand size = 20, antiderivative size = 85 \[ \int \cos ^3(a+b x) \sin ^q(2 a+2 b x) \, dx=-\frac {\cos ^3(a+b x) \cot (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1-q}{2},\frac {4+q}{2},\frac {6+q}{2},\cos ^2(a+b x)\right ) \sin ^2(a+b x)^{\frac {1-q}{2}} \sin ^q(2 a+2 b x)}{b (4+q)} \] Output:
-cos(b*x+a)^3*cot(b*x+a)*hypergeom([2+1/2*q, 1/2-1/2*q],[3+1/2*q],cos(b*x+ a)^2)*(sin(b*x+a)^2)^(1/2-1/2*q)*sin(2*b*x+2*a)^q/b/(4+q)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 10.83 (sec) , antiderivative size = 1801, normalized size of antiderivative = 21.19 \[ \int \cos ^3(a+b x) \sin ^q(2 a+2 b x) \, dx =\text {Too large to display} \] Input:
Integrate[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^q,x]
Output:
(2*(6*AppellF1[(1 + q)/2, -q, 2*(1 + q), (3 + q)/2, Tan[(a + b*x)/2]^2, -T an[(a + b*x)/2]^2] + 8*AppellF1[(1 + q)/2, -q, 2*(2 + q), (3 + q)/2, Tan[( a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - AppellF1[(1 + q)/2, -q, 1 + 2*q, (3 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 12*AppellF1[(1 + q)/2, -q, 3 + 2*q, (3 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Cos[a + b*x]^3*Sin[2*(a + b*x)]^q*Tan[(a + b*x)/2])/(b*((6*AppellF1[(1 + q)/2, -q, 2*(1 + q), (3 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 8*Appell F1[(1 + q)/2, -q, 2*(2 + q), (3 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x) /2]^2] - AppellF1[(1 + q)/2, -q, 1 + 2*q, (3 + q)/2, Tan[(a + b*x)/2]^2, - Tan[(a + b*x)/2]^2] - 12*AppellF1[(1 + q)/2, -q, 3 + 2*q, (3 + q)/2, Tan[( a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Sec[(a + b*x)/2]^2 + q*(6*AppellF1[(1 + q)/2, -q, 2*(1 + q), (3 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2 ] + 8*AppellF1[(1 + q)/2, -q, 2*(2 + q), (3 + q)/2, Tan[(a + b*x)/2]^2, -T an[(a + b*x)/2]^2] - AppellF1[(1 + q)/2, -q, 1 + 2*q, (3 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - 12*AppellF1[(1 + q)/2, -q, 3 + 2*q, (3 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Cos[2*(a + b*x)]*Sec[(a + b*x)/2]^2*Sec[a + b*x] + 4*q*(6*AppellF1[(1 + q)/2, -q, 2*(1 + q), (3 + q )/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] + 8*AppellF1[(1 + q)/2, -q, 2*(2 + q), (3 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - AppellF1[ (1 + q)/2, -q, 1 + 2*q, (3 + q)/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2...
Time = 0.35 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4797, 3042, 3056}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cos ^3(a+b x) \sin ^q(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (a+b x)^3 \sin (2 a+2 b x)^qdx\) |
\(\Big \downarrow \) 4797 |
\(\displaystyle \sin ^{-q}(a+b x) \sin ^q(2 a+2 b x) \cos ^{-q}(a+b x) \int \cos ^{q+3}(a+b x) \sin ^q(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin ^{-q}(a+b x) \sin ^q(2 a+2 b x) \cos ^{-q}(a+b x) \int \cos (a+b x)^{q+3} \sin (a+b x)^qdx\) |
\(\Big \downarrow \) 3056 |
\(\displaystyle -\frac {\cos ^3(a+b x) \cot (a+b x) \sin ^2(a+b x)^{\frac {1-q}{2}} \sin ^q(2 a+2 b x) \operatorname {Hypergeometric2F1}\left (\frac {1-q}{2},\frac {q+4}{2},\frac {q+6}{2},\cos ^2(a+b x)\right )}{b (q+4)}\) |
Input:
Int[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^q,x]
Output:
-((Cos[a + b*x]^3*Cot[a + b*x]*Hypergeometric2F1[(1 - q)/2, (4 + q)/2, (6 + q)/2, Cos[a + b*x]^2]*(Sin[a + b*x]^2)^((1 - q)/2)*Sin[2*a + 2*b*x]^q)/( b*(4 + q)))
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-b^(2*IntPart[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*F racPart[(n - 1)/2])*((a*Cos[e + f*x])^(m + 1)/(a*f*(m + 1)*(Sin[e + f*x]^2) ^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, C os[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] && SimplerQ[n, m]
Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^( p_), x_Symbol] :> Simp[(g*Sin[c + d*x])^p/((e*Cos[a + b*x])^p*Sin[a + b*x]^ p) Int[(e*Cos[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p]
\[\int \cos \left (b x +a \right )^{3} \sin \left (2 b x +2 a \right )^{q}d x\]
Input:
int(cos(b*x+a)^3*sin(2*b*x+2*a)^q,x)
Output:
int(cos(b*x+a)^3*sin(2*b*x+2*a)^q,x)
\[ \int \cos ^3(a+b x) \sin ^q(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{q} \cos \left (b x + a\right )^{3} \,d x } \] Input:
integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^q,x, algorithm="fricas")
Output:
integral(sin(2*b*x + 2*a)^q*cos(b*x + a)^3, x)
\[ \int \cos ^3(a+b x) \sin ^q(2 a+2 b x) \, dx=\int \sin ^{q}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )}\, dx \] Input:
integrate(cos(b*x+a)**3*sin(2*b*x+2*a)**q,x)
Output:
Integral(sin(2*a + 2*b*x)**q*cos(a + b*x)**3, x)
\[ \int \cos ^3(a+b x) \sin ^q(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{q} \cos \left (b x + a\right )^{3} \,d x } \] Input:
integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^q,x, algorithm="maxima")
Output:
integrate(sin(2*b*x + 2*a)^q*cos(b*x + a)^3, x)
\[ \int \cos ^3(a+b x) \sin ^q(2 a+2 b x) \, dx=\int { \sin \left (2 \, b x + 2 \, a\right )^{q} \cos \left (b x + a\right )^{3} \,d x } \] Input:
integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^q,x, algorithm="giac")
Output:
integrate(sin(2*b*x + 2*a)^q*cos(b*x + a)^3, x)
Timed out. \[ \int \cos ^3(a+b x) \sin ^q(2 a+2 b x) \, dx=\int {\cos \left (a+b\,x\right )}^3\,{\sin \left (2\,a+2\,b\,x\right )}^q \,d x \] Input:
int(cos(a + b*x)^3*sin(2*a + 2*b*x)^q,x)
Output:
int(cos(a + b*x)^3*sin(2*a + 2*b*x)^q, x)
\[ \int \cos ^3(a+b x) \sin ^q(2 a+2 b x) \, dx=\int \sin \left (2 b x +2 a \right )^{q} \cos \left (b x +a \right )^{3}d x \] Input:
int(cos(b*x+a)^3*sin(2*b*x+2*a)^q,x)
Output:
int(sin(2*a + 2*b*x)**q*cos(a + b*x)**3,x)