\(\int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx\) [366]

Optimal result

Integrand size = 23, antiderivative size = 76 \[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=-\frac {c (d \cos (a+b x))^{1+n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(a+b x)\right ) (c \sin (a+b x))^{3/2}}{b d (1+n) \sin ^2(a+b x)^{3/4}} \]

[Out]

-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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2656} \[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=-\frac {c (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{n+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {n+1}{2},\frac {n+3}{2},\cos ^2(a+b x)\right )}{b d (n+1) \sin ^2(a+b x)^{3/4}} \]

[In]

Int[(d*Cos[a + b*x])^n*(c*Sin[a + b*x])^(5/2),x]

[Out]

-((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)))

Rule 2656

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^(2*IntPar
t[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*FracPart[(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, Cos[e + f*x]^2], x] /; FreeQ[{a
, b, e, f, m, n}, x] && SimplerQ[n, m]

Rubi steps \begin{align*} \text {integral}& = -\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}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(158\) vs. \(2(76)=152\).

Time = 0.60 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.08 \[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\frac {(d \cos (a+b x))^n \cot (a+b x) \left (-\left ((3+n) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(a+b x)\right )\right )-(3+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(a+b x)\right )+(1+n) \cos ^2(a+b x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3+n}{2},\frac {5+n}{2},\cos ^2(a+b x)\right )\right ) (c \sin (a+b x))^{5/2}}{2 b (1+n) (3+n) \sin ^2(a+b x)^{3/4}} \]

[In]

Integrate[(d*Cos[a + b*x])^n*(c*Sin[a + b*x])^(5/2),x]

[Out]

((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))

Maple [F]

\[\int \left (d \cos \left (b x +a \right )\right )^{n} \left (c \sin \left (b x +a \right )\right )^{\frac {5}{2}}d x\]

[In]

int((d*cos(b*x+a))^n*(c*sin(b*x+a))^(5/2),x)

[Out]

int((d*cos(b*x+a))^n*(c*sin(b*x+a))^(5/2),x)

Fricas [F]

\[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \left (d \cos \left (b x + a\right )\right )^{n} \,d x } \]

[In]

integrate((d*cos(b*x+a))^n*(c*sin(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

integral(-(c^2*cos(b*x + a)^2 - c^2)*sqrt(c*sin(b*x + a))*(d*cos(b*x + a))^n, x)

Sympy [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((d*cos(b*x+a))**n*(c*sin(b*x+a))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \left (d \cos \left (b x + a\right )\right )^{n} \,d x } \]

[In]

integrate((d*cos(b*x+a))^n*(c*sin(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a))^(5/2)*(d*cos(b*x + a))^n, x)

Giac [F]

\[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \left (d \cos \left (b x + a\right )\right )^{n} \,d x } \]

[In]

integrate((d*cos(b*x+a))^n*(c*sin(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a))^(5/2)*(d*cos(b*x + a))^n, x)

Mupad [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^n (c \sin (a+b x))^{5/2} \, dx=\int {\left (d\,\cos \left (a+b\,x\right )\right )}^n\,{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2} \,d x \]

[In]

int((d*cos(a + b*x))^n*(c*sin(a + b*x))^(5/2),x)

[Out]

int((d*cos(a + b*x))^n*(c*sin(a + b*x))^(5/2), x)