Integrand size = 23, antiderivative size = 76 \[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=-\frac {c (d \cos (a+b x))^{1+n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(a+b x)\right ) (c \sin (a+b x))^{3/2}}{b d (1+n) \sin ^2(a+b x)^{3/4}} \] Output:
-c*(d*cos(b*x+a))^(1+n)*hypergeom([-3/4, 1/2+1/2*n],[3/2+1/2*n],cos(b*x+a) ^2)*(c*sin(b*x+a))^(3/2)/b/d/(1+n)/(sin(b*x+a)^2)^(3/4)
Leaf count is larger than twice the leaf count of optimal. \(158\) vs. \(2(76)=152\).
Time = 0.96 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.08 \[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\frac {(d \cos (a+b x))^n \cot (a+b x) \left (-\left ((3+n) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(a+b x)\right )\right )-(3+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(a+b x)\right )+(1+n) \cos ^2(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3+n}{2},\frac {5+n}{2},\cos ^2(a+b x)\right )\right ) (c \sin (a+b x))^{5/2}}{2 b (1+n) (3+n) \sin ^2(a+b x)^{3/4}} \] Input:
Integrate[(d*Cos[a + b*x])^n*(c*Sin[a + b*x])^(5/2),x]
Output:
((d*Cos[a + b*x])^n*Cot[a + b*x]*(-((3 + n)*Hypergeometric2F1[-3/4, (1 + n )/2, (3 + n)/2, Cos[a + b*x]^2]) - (3 + n)*Hypergeometric2F1[1/4, (1 + n)/ 2, (3 + n)/2, Cos[a + b*x]^2] + (1 + n)*Cos[a + b*x]^2*Hypergeometric2F1[1 /4, (3 + n)/2, (5 + n)/2, Cos[a + b*x]^2])*(c*Sin[a + b*x])^(5/2))/(2*b*(1 + n)*(3 + n)*(Sin[a + b*x]^2)^(3/4))
Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {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.
\(\displaystyle \int (c \sin (a+b x))^{5/2} (d \cos (a+b x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c \sin (a+b x))^{5/2} (d \cos (a+b x))^ndx\) |
\(\Big \downarrow \) 3056 |
\(\displaystyle -\frac {c (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{n+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {n+1}{2},\frac {n+3}{2},\cos ^2(a+b x)\right )}{b d (n+1) \sin ^2(a+b x)^{3/4}}\) |
Input:
Int[(d*Cos[a + b*x])^n*(c*Sin[a + b*x])^(5/2),x]
Output:
-((c*(d*Cos[a + b*x])^(1 + n)*Hypergeometric2F1[-3/4, (1 + n)/2, (3 + n)/2 , Cos[a + b*x]^2]*(c*Sin[a + b*x])^(3/2))/(b*d*(1 + n)*(Sin[a + b*x]^2)^(3 /4)))
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]
\[\int \left (d \cos \left (b x +a \right )\right )^{n} \left (c \sin \left (b x +a \right )\right )^{\frac {5}{2}}d x\]
Input:
int((d*cos(b*x+a))^n*(c*sin(b*x+a))^(5/2),x)
Output:
int((d*cos(b*x+a))^n*(c*sin(b*x+a))^(5/2),x)
\[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \left (d \cos \left (b x + a\right )\right )^{n} \,d x } \] Input:
integrate((d*cos(b*x+a))^n*(c*sin(b*x+a))^(5/2),x, algorithm="fricas")
Output:
integral(-(c^2*cos(b*x + a)^2 - c^2)*sqrt(c*sin(b*x + a))*(d*cos(b*x + a)) ^n, x)
Timed out. \[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate((d*cos(b*x+a))**n*(c*sin(b*x+a))**(5/2),x)
Output:
Timed out
\[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \left (d \cos \left (b x + a\right )\right )^{n} \,d x } \] Input:
integrate((d*cos(b*x+a))^n*(c*sin(b*x+a))^(5/2),x, algorithm="maxima")
Output:
integrate((c*sin(b*x + a))^(5/2)*(d*cos(b*x + a))^n, x)
\[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \left (d \cos \left (b x + a\right )\right )^{n} \,d x } \] Input:
integrate((d*cos(b*x+a))^n*(c*sin(b*x+a))^(5/2),x, algorithm="giac")
Output:
integrate((c*sin(b*x + a))^(5/2)*(d*cos(b*x + a))^n, x)
Timed out. \[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\int {\left (d\,\cos \left (a+b\,x\right )\right )}^n\,{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2} \,d x \] Input:
int((d*cos(a + b*x))^n*(c*sin(a + b*x))^(5/2),x)
Output:
int((d*cos(a + b*x))^n*(c*sin(a + b*x))^(5/2), x)
\[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=d^{n} \sqrt {c}\, \left (\int \sqrt {\sin \left (b x +a \right )}\, \cos \left (b x +a \right )^{n} \sin \left (b x +a \right )^{2}d x \right ) c^{2} \] Input:
int((d*cos(b*x+a))^n*(c*sin(b*x+a))^(5/2),x)
Output:
d**n*sqrt(c)*int(sqrt(sin(a + b*x))*cos(a + b*x)**n*sin(a + b*x)**2,x)*c** 2