\(\int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{7/2}} \, dx\) [280]

Optimal result

Integrand size = 25, antiderivative size = 133 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2 c (c \sin (a+b x))^{3/2}}{5 b d (d \cos (a+b x))^{5/2}}-\frac {6 c (c \sin (a+b x))^{3/2}}{5 b d^3 \sqrt {d \cos (a+b x)}}+\frac {6 c^2 \sqrt {d \cos (a+b x)} E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b d^4 \sqrt {\sin (2 a+2 b x)}} \] Output:

2/5*c*(c*sin(b*x+a))^(3/2)/b/d/(d*cos(b*x+a))^(5/2)-6/5*c*(c*sin(b*x+a))^( 
3/2)/b/d^3/(d*cos(b*x+a))^(1/2)-6/5*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^4/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 = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.53 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{7/2}} \, dx=\frac {2 \cos ^2(a+b x)^{5/4} \cot (a+b x) \operatorname {Hypergeometric2F1}\left (\frac {7}{4},\frac {9}{4},\frac {11}{4},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{9/2}}{7 b c^2 (d \cos (a+b x))^{7/2}} \] Input:

Integrate[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(7/2),x]
 

Output:

(2*(Cos[a + b*x]^2)^(5/4)*Cot[a + b*x]*Hypergeometric2F1[7/4, 9/4, 11/4, S 
in[a + b*x]^2]*(c*Sin[a + b*x])^(9/2))/(7*b*c^2*(d*Cos[a + b*x])^(7/2))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3046, 3042, 3051, 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.

\(\displaystyle \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{7/2}}dx\)

\(\Big \downarrow \) 3046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3051

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 c (c \sin (a+b x))^{3/2}}{5 b d (d \cos (a+b x))^{5/2}}-\frac {3 c^2 \left (\frac {2 (c \sin (a+b x))^{3/2}}{b c d \sqrt {d \cos (a+b x)}}-\frac {2 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b d^2 \sqrt {\sin (2 a+2 b x)}}\right )}{5 d^2}\)

Input:

Int[(c*Sin[a + b*x])^(5/2)/(d*Cos[a + b*x])^(7/2),x]
 

Output:

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

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 3051

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

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. \(258\) vs. \(2(114)=228\).

Time = 4.83 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.95

method	result	size
default	\(\frac {\sqrt {c \sin \left (b x +a \right )}\, c^{2} \left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (3 \cot \left (b x +a \right )+3 \csc \left (b x +a \right )\right )+\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-6 \cot \left (b x +a \right )-6 \csc \left (b x +a \right )\right )+6 \cot \left (b x +a \right )-8 \csc \left (b x +a \right )+2 \csc \left (b x +a \right ) \sec \left (b x +a \right )^{2}\right )}{5 b \sqrt {d \cos \left (b x +a \right )}\, d^{3}}\)	\(259\)

Input:

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/5/b*(c*sin(b*x+a))^(1/2)*c^2/(d*cos(b*x+a))^(1/2)/d^3*((-cot(b*x+a)+csc( 
b*x+a)+1)^(1/2)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(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))*(3*cot(b*x 
+a)+3*csc(b*x+a))+(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(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))*(-6*cot(b*x+a)-6*csc(b*x+a))+6*cot(b*x+a)-8*csc(b 
*x+a)+2*csc(b*x+a)*sec(b*x+a)^2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.59 \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{7/2}} \, dx=\frac {3 i \, \sqrt {i \, c d} c^{2} \cos \left (b x + a\right )^{3} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) - 3 i \, \sqrt {-i \, c d} c^{2} \cos \left (b x + a\right )^{3} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - 3 i \, \sqrt {i \, c d} c^{2} \cos \left (b x + a\right )^{3} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) + 3 i \, \sqrt {-i \, c d} c^{2} \cos \left (b x + a\right )^{3} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) - 2 \, {\left (3 \, c^{2} \cos \left (b x + a\right )^{2} - c^{2}\right )} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )}{5 \, b d^{4} \cos \left (b x + a\right )^{3}} \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(7/2),x, algorithm="fricas")
 

Output:

1/5*(3*I*sqrt(I*c*d)*c^2*cos(b*x + a)^3*elliptic_e(arcsin(cos(b*x + a) + I 
*sin(b*x + a)), -1) - 3*I*sqrt(-I*c*d)*c^2*cos(b*x + a)^3*elliptic_e(arcsi 
n(cos(b*x + a) - I*sin(b*x + a)), -1) - 3*I*sqrt(I*c*d)*c^2*cos(b*x + a)^3 
*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1) + 3*I*sqrt(-I*c*d)* 
c^2*cos(b*x + a)^3*elliptic_f(arcsin(cos(b*x + a) - I*sin(b*x + a)), -1) - 
 2*(3*c^2*cos(b*x + a)^2 - c^2)*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))* 
sin(b*x + a))/(b*d^4*cos(b*x + a)^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((c*sin(b*x+a))**(5/2)/(d*cos(b*x+a))**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{7/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(7/2),x, algorithm="maxima")
 

Output:

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(7/2), x)
 

Giac [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{7/2}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\left (d \cos \left (b x + a\right )\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(7/2),x, algorithm="giac")
 

Output:

integrate((c*sin(b*x + a))^(5/2)/(d*cos(b*x + a))^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{7/2}} \, dx=\int \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2}}{{\left (d\,\cos \left (a+b\,x\right )\right )}^{7/2}} \,d x \] Input:

int((c*sin(a + b*x))^(5/2)/(d*cos(a + b*x))^(7/2),x)
 

Output:

int((c*sin(a + b*x))^(5/2)/(d*cos(a + b*x))^(7/2), x)
 

Reduce [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{(d \cos (a+b x))^{7/2}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \left (\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}\, \sin \left (b x +a \right )^{2}}{\cos \left (b x +a \right )^{4}}d x \right ) c^{2}}{d^{4}} \] Input:

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(7/2),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)**4,x)*c**2)/d**4