\(\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx\) [131]
Optimal result
Integrand size = 42, antiderivative size = 121 \[
\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac {2 g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}
\]
[Out]
2*(g*cos(f*x+e))^(5/2)/f/g/(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2)-2*g*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(
1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(g*cos(f*x+e))^(1/2)/c/f/(a+a*sin(f*x+e)
)^(1/2)/(c-c*sin(f*x+e))^(1/2)
Rubi [A] (verified)
Time = 0.40 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of
steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2931, 2921, 2721, 2719}
\[
\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {2 g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}
\]
[In]
Int[(g*Cos[e + f*x])^(3/2)/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)),x]
[Out]
(2*(g*Cos[e + f*x])^(5/2))/(f*g*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)) - (2*g*Sqrt[Cos[e + f*x]]
*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(c*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])
Rule 2719
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]
Rule 2721
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]
Rule 2921
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)]]), x_Symbol] :> Dist[g*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), In
t[(g*Cos[e + f*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2
, 0]
Rule 2931
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^
n/(a*f*g*(2*m + p + 1))), x] + Dist[(m + n + p + 1)/(a*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f
*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] && E
qQ[a^2 - b^2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && !LtQ[m, n, -1] && IntegersQ[2*m, 2*n, 2*p]
Rubi steps \begin{align*}
\text {integral}& = \frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx}{c} \\ & = \frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac {(g \cos (e+f x)) \int \sqrt {g \cos (e+f x)} \, dx}{c \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac {\left (g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \, dx}{c \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac {2 g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\
\end{align*}
Mathematica [A] (verified)
Time = 3.83 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.22
\[
\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {2 (g \cos (e+f x))^{3/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \left (-\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\sqrt {\cos (e+f x)} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )}{c f \cos ^{\frac {3}{2}}(e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}}
\]
[In]
Integrate[(g*Cos[e + f*x])^(3/2)/(Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)),x]
[Out]
(2*(g*Cos[e + f*x])^(3/2)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(EllipticE[(e + f*x)/2, 2]*(-Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2]) + Sqrt[Cos[e + f*x]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])))/(c*f*Cos[e + f*x]^(3/2)*Sqrt[
a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]])
Maple [C] (verified)
Result contains complex when optimal does not.
Time = 0.83 (sec) , antiderivative size = 799, normalized size of antiderivative = 6.60
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| method | result | size |
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| default |
\(\text {Expression too large to display}\) |
\(799\) |
| risch |
\(\text {Expression too large to display}\) |
\(1145\) |
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[In]
int((g*cos(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/2*I/f*(4*cos(f*x+e)^2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot
(f*x+e)),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)-4*cos(f*x+e)^2*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*
x+e)))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+8*EllipticE(I*(csc(f*
x+e)-cot(f*x+e)),I)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^
(1/2)*cos(f*x+e)-8*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))
^(1/2)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^
(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)-4*(1/(1+cos(f*x+e)))^(1/2)*(
cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+I
*ln(2*(2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)-cos(f*x+e)+1)/
(1+cos(f*x+e)))*cos(f*x+e)-I*ln((2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+2*(-cos(f*x+e)/(1+cos(f*x+e
))^2)^(1/2)-cos(f*x+e)+1)/(1+cos(f*x+e)))*cos(f*x+e)+4*I*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*I*(
-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*sin(f*x+e)+4*I*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2))*cos(f*x+e)*(g*cos(f*x
+e))^(1/2)*g/(a*(1+sin(f*x+e)))^(1/2)/(-c*(sin(f*x+e)-1))^(1/2)/(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)/(1+cos(f*
x+e))^3/c
Fricas [C] (verification not implemented)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.31
\[
\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=-\frac {2 \, \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} g - \sqrt {a c g} {\left (i \, \sqrt {2} g \sin \left (f x + e\right ) - i \, \sqrt {2} g\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - \sqrt {a c g} {\left (-i \, \sqrt {2} g \sin \left (f x + e\right ) + i \, \sqrt {2} g\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{a c^{2} f \sin \left (f x + e\right ) - a c^{2} f}
\]
[In]
integrate((g*cos(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
-(2*sqrt(g*cos(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*g - sqrt(a*c*g)*(I*sqrt(2)*g*sin(f
*x + e) - I*sqrt(2)*g)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) - sqr
t(a*c*g)*(-I*sqrt(2)*g*sin(f*x + e) + I*sqrt(2)*g)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x +
e) - I*sin(f*x + e))))/(a*c^2*f*sin(f*x + e) - a*c^2*f)
Sympy [F(-1)]
Timed out. \[
\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out}
\]
[In]
integrate((g*cos(f*x+e))**(3/2)/(c-c*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e))**(1/2),x)
[Out]
Timed out
Maxima [F]
\[
\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}}}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x }
\]
[In]
integrate((g*cos(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate((g*cos(f*x + e))^(3/2)/(sqrt(a*sin(f*x + e) + a)*(-c*sin(f*x + e) + c)^(3/2)), x)
Giac [F(-2)]
Exception generated. \[
\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError}
\]
[In]
integrate((g*cos(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%{[%%%{%%{[-2692572175765579235328,0]:[1,0,%%%{-1,[1]%%
%}]%%},[2]%
Mupad [F(-1)]
Timed out. \[
\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}}{\sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x
\]
[In]
int((g*cos(e + f*x))^(3/2)/((a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(3/2)),x)
[Out]
int((g*cos(e + f*x))^(3/2)/((a + a*sin(e + f*x))^(1/2)*(c - c*sin(e + f*x))^(3/2)), x)