\(\int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx\) [269]

Optimal result

Integrand size = 25, antiderivative size = 98 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2 c \sqrt {c \sin (a+b x)}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)}}{3 b d^2 \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}} \]

[Out]

2/3*c*(c*sin(b*x+a))^(1/2)/b/d/(d*cos(b*x+a))^(3/2)+1/3*c^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*Elli
pticF(cos(a+1/4*Pi+b*x),2^(1/2))*sin(2*b*x+2*a)^(1/2)/b/d^2/(d*cos(b*x+a))^(1/2)/(c*sin(b*x+a))^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2646, 2653, 2720} \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2 c \sqrt {c \sin (a+b x)}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \sqrt {\sin (2 a+2 b x)} \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b d^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}} \]

[In]

Int[(c*Sin[a + b*x])^(3/2)/(d*Cos[a + b*x])^(5/2),x]

[Out]

(2*c*Sqrt[c*Sin[a + b*x]])/(3*b*d*(d*Cos[a + b*x])^(3/2)) - (c^2*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2
*b*x]])/(3*b*d^2*Sqrt[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b*x]])

Rule 2646

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(a*Sin[e
+ f*x])^(m - 1)*((b*Cos[e + f*x])^(n + 1)/(b*f*(n + 1))), x] + Dist[a^2*((m - 1)/(b^2*(n + 1))), Int[(a*Sin[e
+ f*x])^(m - 2)*(b*Cos[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Int
egersQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2653

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {2 c \sqrt {c \sin (a+b x)}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \int \frac {1}{\sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}} \, dx}{3 d^2} \\ & = \frac {2 c \sqrt {c \sin (a+b x)}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {\left (c^2 \sqrt {\sin (2 a+2 b x)}\right ) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx}{3 d^2 \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}} \\ & = \frac {2 c \sqrt {c \sin (a+b x)}}{3 b d (d \cos (a+b x))^{3/2}}-\frac {c^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)}}{3 b d^2 \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.68 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {2 \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {7}{4},\frac {9}{4},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{5/2}}{5 b c d (d \cos (a+b x))^{3/2}} \]

[In]

Integrate[(c*Sin[a + b*x])^(3/2)/(d*Cos[a + b*x])^(5/2),x]

[Out]

(2*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[5/4, 7/4, 9/4, Sin[a + b*x]^2]*(c*Sin[a + b*x])^(5/2))/(5*b*c*d*(d
*Cos[a + b*x])^(3/2))

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.17

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {c \sin \left (b x +a \right )}\, c \left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \cot \left (b x +a \right )+\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )+1}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, F\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \csc \left (b x +a \right )-\sqrt {2}\, \sec \left (b x +a \right )\right )}{3 b \sqrt {d \cos \left (b x +a \right )}\, d^{2}}\)	\(213\)

[In]

int((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/b*2^(1/2)*(c*sin(b*x+a))^(1/2)*c/(d*cos(b*x+a))^(1/2)/d^2*((-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-c
sc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticF((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*cot(b
*x+a)+(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a)+1)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticF
((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*csc(b*x+a)-2^(1/2)*sec(b*x+a))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.10 \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\frac {\sqrt {i \, c d} c \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + \sqrt {-i \, c d} c \cos \left (b x + a\right )^{2} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) + 2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} c}{3 \, b d^{3} \cos \left (b x + a\right )^{2}} \]

[In]

integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

1/3*(sqrt(I*c*d)*c*cos(b*x + a)^2*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1) + sqrt(-I*c*d)*c*cos(b
*x + a)^2*elliptic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) + 2*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*
c)/(b*d^3*cos(b*x + a)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((c*sin(b*x+a))**(3/2)/(d*cos(b*x+a))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate((c*sin(b*x + a))^(3/2)/(d*cos(b*x + a))^(5/2), x)

Giac [F]

\[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {3}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((c*sin(b*x+a))^(3/2)/(d*cos(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate((c*sin(b*x + a))^(3/2)/(d*cos(b*x + a))^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{3/2}}{(d \cos (a+b x))^{5/2}} \, dx=\int \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^{3/2}}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{5/2}} \,d x \]

[In]

int((c*sin(a + b*x))^(3/2)/(d*cos(a + b*x))^(5/2),x)

[Out]

int((c*sin(a + b*x))^(3/2)/(d*cos(a + b*x))^(5/2), x)