3.5.78 \(\int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx\) [478]

3.5.78.1 Optimal result
3.5.78.2 Mathematica [C] (verified)
3.5.78.3 Rubi [A] (verified)
3.5.78.4 Maple [C] (warning: unable to verify)
3.5.78.5 Fricas [F]
3.5.78.6 Sympy [F(-1)]
3.5.78.7 Maxima [F]
3.5.78.8 Giac [F]
3.5.78.9 Mupad [F(-1)]

3.5.78.1 Optimal result

Integrand size = 25, antiderivative size = 135 \[ \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx=-\frac {a \sqrt {a \sin (e+f x)}}{6 b f \sqrt {b \sec (e+f x)}}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}+\frac {a^2 \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}}{12 b^2 f \sqrt {a \sin (e+f x)}} \]

output
1/3*(a*sin(f*x+e))^(5/2)/a/b/f/(b*sec(f*x+e))^(1/2)-1/6*a*(a*sin(f*x+e))^( 
1/2)/b/f/(b*sec(f*x+e))^(1/2)-1/12*a^2*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1 
/4*Pi+f*x)*EllipticF(cos(e+1/4*Pi+f*x),2^(1/2))*(b*sec(f*x+e))^(1/2)*sin(2 
*f*x+2*e)^(1/2)/b^2/f/(a*sin(f*x+e))^(1/2)
 
3.5.78.2 Mathematica [C] (verified)

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

Time = 0.48 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.64 \[ \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx=\frac {a \sqrt {a \sin (e+f x)} \left (-2 \cos (2 (e+f x))+\csc ^2(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\sec ^2(e+f x)\right ) \left (-\tan ^2(e+f x)\right )^{3/4}\right )}{12 b f \sqrt {b \sec (e+f x)}} \]

input
Integrate[(a*Sin[e + f*x])^(3/2)/(b*Sec[e + f*x])^(3/2),x]
 
output
(a*Sqrt[a*Sin[e + f*x]]*(-2*Cos[2*(e + f*x)] + Csc[e + f*x]^2*Hypergeometr 
ic2F1[1/2, 3/4, 3/2, Sec[e + f*x]^2]*(-Tan[e + f*x]^2)^(3/4)))/(12*b*f*Sqr 
t[b*Sec[e + f*x]])
 
3.5.78.3 Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3062, 3042, 3063, 3042, 3065, 3042, 3053, 3042, 3120}

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.

\(\displaystyle \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3062

\(\displaystyle \frac {\int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}dx}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}dx}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3063

\(\displaystyle \frac {\frac {1}{2} a^2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} a^2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3065

\(\displaystyle \frac {\frac {1}{2} a^2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} a^2 \sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}}dx-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3053

\(\displaystyle \frac {\frac {a^2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{2 \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 \sqrt {\sin (2 e+2 f x)} \sqrt {b \sec (e+f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{2 \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {a^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right ) \sqrt {b \sec (e+f x)}}{2 f \sqrt {a \sin (e+f x)}}-\frac {a b \sqrt {a \sin (e+f x)}}{f \sqrt {b \sec (e+f x)}}}{6 b^2}+\frac {(a \sin (e+f x))^{5/2}}{3 a b f \sqrt {b \sec (e+f x)}}\)

input
Int[(a*Sin[e + f*x])^(3/2)/(b*Sec[e + f*x])^(3/2),x]
 
output
(a*Sin[e + f*x])^(5/2)/(3*a*b*f*Sqrt[b*Sec[e + f*x]]) + (-((a*b*Sqrt[a*Sin 
[e + f*x]])/(f*Sqrt[b*Sec[e + f*x]])) + (a^2*EllipticF[e - Pi/4 + f*x, 2]* 
Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[2*e + 2*f*x]])/(2*f*Sqrt[a*Sin[e + f*x]]))/( 
6*b^2)
 

3.5.78.3.1 Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3053
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ 
)]]), x_Symbol] :> Simp[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 3062
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(a*Sin[e + f*x])^(m + 1)*((b*Sec[e + f*x])^(n + 1)/(a 
*b*f*(m - n))), x] - Simp[(n + 1)/(b^2*(m - n))   Int[(a*Sin[e + f*x])^m*(b 
*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] & 
& NeQ[m - n, 0] && IntegersQ[2*m, 2*n]
 

rule 3063
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(-a)*b*(a*Sin[e + f*x])^(m - 1)*((b*Sec[e + f*x])^(n 
- 1)/(f*(m - n))), x] + Simp[a^2*((m - 1)/(m - n))   Int[(a*Sin[e + f*x])^( 
m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] 
 && NeQ[m - n, 0] && IntegersQ[2*m, 2*n]
 

rule 3065
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(b*Cos[e + f*x])^n*(b*Sec[e + f*x])^n   Int[(a*Sin[e 
+ f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && Int 
egerQ[m - 1/2] && IntegerQ[n - 1/2]
 

rule 3120
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]
 
3.5.78.4 Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.97 (sec) , antiderivative size = 1746, normalized size of antiderivative = 12.93

method result size
default \(\text {Expression too large to display}\) \(1746\)

input
int((a*sin(f*x+e))^(3/2)/(b*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 
output
-1/48/f*2^(1/2)*(-6*I*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x 
+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/2)*EllipticPi((-cot(f*x+e)+csc(f*x 
+e)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(f*x+e)-6*I*(-cot(f*x+e)+csc(f*x+e) 
+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/2)*El 
lipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))+6*I*(-cot 
(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-cs 
c(f*x+e))^(1/2)*EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2+1/2*I,1/2* 
2^(1/2))-6*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2 
)*(cot(f*x+e)-csc(f*x+e))^(1/2)*EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2 
),1/2+1/2*I,1/2*2^(1/2))*cos(f*x+e)+8*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(co 
t(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/2)*EllipticF((-cot 
(f*x+e)+csc(f*x+e)+1)^(1/2),1/2*2^(1/2))*cos(f*x+e)+6*I*(-cot(f*x+e)+csc(f 
*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/ 
2)*EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos( 
f*x+e)-6*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)* 
(cot(f*x+e)-csc(f*x+e))^(1/2)*EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2), 
1/2-1/2*I,1/2*2^(1/2))*cos(f*x+e)+8*2^(1/2)*cos(f*x+e)^3*sin(f*x+e)-6*(-co 
t(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-c 
sc(f*x+e))^(1/2)*EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2+1/2*I,1/2 
*2^(1/2))+8*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^...
 
3.5.78.5 Fricas [F]

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

input
integrate((a*sin(f*x+e))^(3/2)/(b*sec(f*x+e))^(3/2),x, algorithm="fricas")
 
output
integral(sqrt(b*sec(f*x + e))*sqrt(a*sin(f*x + e))*a*sin(f*x + e)/(b^2*sec 
(f*x + e)^2), x)
 
3.5.78.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(a \sin (e+f x))^{3/2}}{(b \sec (e+f x))^{3/2}} \, dx=\text {Timed out} \]

input
integrate((a*sin(f*x+e))**(3/2)/(b*sec(f*x+e))**(3/2),x)
 
output
Timed out
 
3.5.78.7 Maxima [F]

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

input
integrate((a*sin(f*x+e))^(3/2)/(b*sec(f*x+e))^(3/2),x, algorithm="maxima")
 
output
integrate((a*sin(f*x + e))^(3/2)/(b*sec(f*x + e))^(3/2), x)
 
3.5.78.8 Giac [F]

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

input
integrate((a*sin(f*x+e))^(3/2)/(b*sec(f*x+e))^(3/2),x, algorithm="giac")
 
output
integrate((a*sin(f*x + e))^(3/2)/(b*sec(f*x + e))^(3/2), x)
 
3.5.78.9 Mupad [F(-1)]

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

input
int((a*sin(e + f*x))^(3/2)/(b/cos(e + f*x))^(3/2),x)
 
output
int((a*sin(e + f*x))^(3/2)/(b/cos(e + f*x))^(3/2), x)