\(\int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{3/2}} \, dx\) [279]

Optimal result

Integrand size = 25, antiderivative size = 94 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{3/2}} \, dx=\frac {2 c (c \sin (a+b x))^{3/2}}{b d \sqrt {d \cos (a+b x)}}-\frac {3 c^2 \sqrt {d \cos (a+b x)} E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {c \sin (a+b x)}}{b d^2 \sqrt {\sin (2 a+2 b x)}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.15 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{3/2}} \, dx=\frac {2 \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {7}{4},\frac {11}{4},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{7/2}}{7 b c d \sqrt {d \cos (a+b x)}} \] Input:

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

Output:

(2*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[5/4, 7/4, 11/4, Sin[a + b*x]^2 
]*(c*Sin[a + b*x])^(7/2))/(7*b*c*d*Sqrt[d*Cos[a + b*x]])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3046, 3042, 3052, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3046

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{b d \sqrt {d \cos (a+b x)}}-\frac {3 c^2 \int \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}dx}{d^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{b d \sqrt {d \cos (a+b x)}}-\frac {3 c^2 \int \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}dx}{d^2}\)

\(\Big \downarrow \) 3052

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{b d \sqrt {d \cos (a+b x)}}-\frac {3 c^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)} \int \sqrt {\sin (2 a+2 b x)}dx}{d^2 \sqrt {\sin (2 a+2 b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{b d \sqrt {d \cos (a+b x)}}-\frac {3 c^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)} \int \sqrt {\sin (2 a+2 b x)}dx}{d^2 \sqrt {\sin (2 a+2 b x)}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{b d \sqrt {d \cos (a+b x)}}-\frac {3 c^2 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b d^2 \sqrt {\sin (2 a+2 b x)}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3046

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] + Simp[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] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
 

rule 3052

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

rule 3119

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(237\) vs. \(2(85)=170\).

Time = 5.38 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.53

method	result	size
default	\(\frac {\left (\left (-3 \cos \left (b x +a \right )-3\right ) \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}+\left (6 \cos \left (b x +a \right )+6\right ) \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}+2 \cos \left (b x +a \right )^{2}-6 \cos \left (b x +a \right )+4\right ) \sqrt {c \sin \left (b x +a \right )}\, c^{2} \csc \left (b x +a \right )}{2 b \sqrt {d \cos \left (b x +a \right )}\, d}\)	\(238\)

Input:

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

Output:

1/2/b*((-3*cos(b*x+a)-3)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a)-c 
sc(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)+csc(b*x+a)+1)^(1/2)+(6*cos(b*x+a)+6)*(2*cot(b*x+a)-2*csc(b*x+a 
)+2)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticE((-cot(b*x+a)+csc(b*x+a) 
+1)^(1/2),1/2*2^(1/2))*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)+2*cos(b*x+a)^2-6*c 
os(b*x+a)+4)*(c*sin(b*x+a))^(1/2)*c^2/(d*cos(b*x+a))^(1/2)/d*csc(b*x+a)
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(d)*sqrt(c)*int((sqrt(sin(a + b*x))*sqrt(cos(a + b*x))*sin(a + b*x)** 
2)/cos(a + b*x)**2,x)*c**2)/d**2