3.15.42 \(\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^2} \, dx\) [1442]
Optimal. Leaf size=331 \[ \frac {\sqrt {2} b g^2 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{a^2 \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \cos (e+f x)}}-\frac {\sqrt {2} b g^2 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{a^2 \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \cos (e+f x)}}+\frac {g \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{a d f (a+b \sin (e+f x))}+\frac {g^2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{2 a^2 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \]
[Out]
b*g^2*EllipticPi((d*sin(f*x+e))^(1/2)/d^(1/2)/(1+cos(f*x+e))^(1/2),-a/(b-(-a^2+b^2)^(1/2)),I)*2^(1/2)*cos(f*x+
e)^(1/2)/a^2/f/(-a^2+b^2)^(1/2)/d^(1/2)/(g*cos(f*x+e))^(1/2)-b*g^2*EllipticPi((d*sin(f*x+e))^(1/2)/d^(1/2)/(1+
cos(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)),I)*2^(1/2)*cos(f*x+e)^(1/2)/a^2/f/(-a^2+b^2)^(1/2)/d^(1/2)/(g*cos(f*
x+e))^(1/2)+g*(g*cos(f*x+e))^(1/2)*(d*sin(f*x+e))^(1/2)/a/d/f/(a+b*sin(f*x+e))-1/2*g^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))*sin(2*f*x+2*e)^(1/2)/a^2/f/(g*cos(f*x+e))^(1/2)/(d
*sin(f*x+e))^(1/2)
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Rubi [A]
time = 0.52, antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {2966, 2989,
2653, 2720, 2987, 2986, 1232} \begin {gather*} \frac {\sqrt {2} b g^2 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {b^2-a^2}};\left .\text {ArcSin}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a^2 \sqrt {d} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}-\frac {\sqrt {2} b g^2 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {b^2-a^2}};\left .\text {ArcSin}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right )\right |-1\right )}{a^2 \sqrt {d} f \sqrt {b^2-a^2} \sqrt {g \cos (e+f x)}}+\frac {g^2 \sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right )}{2 a^2 f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}+\frac {g \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{a d f (a+b \sin (e+f x))} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(g*Cos[e + f*x])^(3/2)/(Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^2),x]
[Out]
(Sqrt[2]*b*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[-(a/(b - Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]
*Sqrt[1 + Cos[e + f*x]])], -1])/(a^2*Sqrt[-a^2 + b^2]*Sqrt[d]*f*Sqrt[g*Cos[e + f*x]]) - (Sqrt[2]*b*g^2*Sqrt[Co
s[e + f*x]]*EllipticPi[-(a/(b + Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x]
])], -1])/(a^2*Sqrt[-a^2 + b^2]*Sqrt[d]*f*Sqrt[g*Cos[e + f*x]]) + (g*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Sin[e + f*x]]
)/(a*d*f*(a + b*Sin[e + f*x])) + (g^2*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]])/(2*a^2*f*Sqrt[g*Cos
[e + f*x]]*Sqrt[d*Sin[e + f*x]])
Rule 1232
Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[
a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 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]
Rule 2966
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_))/Sqrt[(d_.)*sin[(e_.) +
(f_.)*(x_)]], x_Symbol] :> Simp[(-g)*(g*Cos[e + f*x])^(p - 1)*Sqrt[d*Sin[e + f*x]]*((a + b*Sin[e + f*x])^(m +
1)/(a*d*f*(m + 1))), x] + Dist[g^2*((2*m + 3)/(2*a*(m + 1))), Int[(g*Cos[e + f*x])^(p - 2)*((a + b*Sin[e + f*x
])^(m + 1)/Sqrt[d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&
EqQ[m + p + 1/2, 0]
Rule 2986
Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]))
, x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Dist[2*Sqrt[2]*d*((b + q)/(f*q)), Subst[Int[1/((d*(b + q) + a*x^2
)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x] - Dist[2*Sqrt[2]*d*((b - q)/(f*q
)), Subst[Int[1/((d*(b - q) + a*x^2)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]],
x]] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 2987
Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(
x_)])), x_Symbol] :> Dist[Sqrt[Cos[e + f*x]]/Sqrt[g*Cos[e + f*x]], Int[Sqrt[d*Sin[e + f*x]]/(Sqrt[Cos[e + f*x]
]*(a + b*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
Rule 2989
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[1/a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] - Dist[b/(a*d), Int[(g*Cos[
e + f*x])^p*((d*Sin[e + f*x])^(n + 1)/(a + b*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2
- b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[-1, p, 1] && LtQ[n, 0]
Rubi steps
\begin {align*} \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^2} \, dx &=\frac {g \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{a d f (a+b \sin (e+f x))}+\frac {g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx}{2 a}\\ &=\frac {g \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{a d f (a+b \sin (e+f x))}+\frac {g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \, dx}{2 a^2}-\frac {\left (b g^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{2 a^2 d}\\ &=\frac {g \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{a d f (a+b \sin (e+f x))}-\frac {\left (b g^2 \sqrt {\cos (e+f x)}\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))} \, dx}{2 a^2 d \sqrt {g \cos (e+f x)}}+\frac {\left (g^2 \sqrt {\sin (2 e+2 f x)}\right ) \int \frac {1}{\sqrt {\sin (2 e+2 f x)}} \, dx}{2 a^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ &=\frac {g \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{a d f (a+b \sin (e+f x))}+\frac {g^2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{2 a^2 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}-\frac {\left (\sqrt {2} b \left (1-\frac {b}{\sqrt {-a^2+b^2}}\right ) g^2 \sqrt {\cos (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\left (b-\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {1+\cos (e+f x)}}\right )}{a^2 f \sqrt {g \cos (e+f x)}}-\frac {\left (\sqrt {2} b \left (1+\frac {b}{\sqrt {-a^2+b^2}}\right ) g^2 \sqrt {\cos (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\left (b+\sqrt {-a^2+b^2}\right ) d+a x^2\right ) \sqrt {1-\frac {x^4}{d^2}}} \, dx,x,\frac {\sqrt {d \sin (e+f x)}}{\sqrt {1+\cos (e+f x)}}\right )}{a^2 f \sqrt {g \cos (e+f x)}}\\ &=\frac {\sqrt {2} b g^2 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b-\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{a^2 \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \cos (e+f x)}}-\frac {\sqrt {2} b g^2 \sqrt {\cos (e+f x)} \Pi \left (-\frac {a}{b+\sqrt {-a^2+b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right )\right |-1\right )}{a^2 \sqrt {-a^2+b^2} \sqrt {d} f \sqrt {g \cos (e+f x)}}+\frac {g \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}{a d f (a+b \sin (e+f x))}+\frac {g^2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{2 a^2 f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in
optimal.
time = 25.32, size = 717, normalized size = 2.17 \begin {gather*} -\frac {(g \cos (e+f x))^{3/2} \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {3}{4},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \sqrt {\cos (e+f x)}}{\left (1-\cos ^2(e+f x)\right )^{3/4} \left (5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {3}{4},1;\frac {5}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+\left (-4 b^2 F_1\left (\frac {5}{4};\frac {3}{4},2;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+3 \left (a^2-b^2\right ) F_1\left (\frac {5}{4};\frac {7}{4},1;\frac {9}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )\right ) \cos ^2(e+f x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(e+f x)\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) b \left (2 \tan ^{-1}\left (1-\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )-2 \tan ^{-1}\left (1+\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )+\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}-\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )-\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}+\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )\right )}{\sqrt {a} \left (-a^2+b^2\right )^{3/4}}\right ) \sin (e+f x)}{a f \cos ^{\frac {3}{2}}(e+f x) \sqrt [4]{1-\cos ^2(e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}+\frac {(g \cos (e+f x))^{3/2} \tan (e+f x)}{a f \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(g*Cos[e + f*x])^(3/2)/(Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^2),x]
[Out]
-(((g*Cos[e + f*x])^(3/2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 3/4, 1, 5/4, Cos[e
+ f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/((1 - Cos[e + f*x]^2)^(3/4)*(5*(a^2 - b^2)*Ap
pellF1[1/4, 3/4, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (-4*b^2*AppellF1[5/4, 3/4, 2, 9/
4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + 3*(a^2 - b^2)*AppellF1[5/4, 7/4, 1, 9/4, Cos[e + f*x]^
2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*b*(2*
ArcTan[1 - ((1 + I)*Sqrt[a]*Sqrt[Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*(-1 + Cos[e + f*x]^2)^(1/4))] - 2*ArcTan[1
+ ((1 + I)*Sqrt[a]*Sqrt[Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*(-1 + Cos[e + f*x]^2)^(1/4))] + Log[Sqrt[-a^2 + b^
2] + (I*a*Cos[e + f*x])/Sqrt[-1 + Cos[e + f*x]^2] - ((1 + I)*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]])/(-
1 + Cos[e + f*x]^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] + (I*a*Cos[e + f*x])/Sqrt[-1 + Cos[e + f*x]^2] + ((1 + I)*Sq
rt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]])/(-1 + Cos[e + f*x]^2)^(1/4)]))/(Sqrt[a]*(-a^2 + b^2)^(3/4)))*Sin[
e + f*x])/(a*f*Cos[e + f*x]^(3/2)*(1 - Cos[e + f*x]^2)^(1/4)*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x]))) + ((g
*Cos[e + f*x])^(3/2)*Tan[e + f*x])/(a*f*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x]))
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(3489\) vs.
\(2(308)=616\).
time = 0.63, size = 3490, normalized size = 10.54
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e))^2/(d*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
[Out]
-1/2/f*(2*2^(1/2)*cos(f*x+e)^2*(-a^2+b^2)^(1/2)*a^2-((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x
+e))/sin(f*x+e))^(1/2)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/
sin(f*x+e))^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*b^2-((-1+cos(f*x+e)+sin(f*x+e))/sin(
f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(
-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*a*b^2-((-1+cos(f*x+e)+sin(f
*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*Ell
ipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-a^2+b^2)^(1/2)
*b^2+((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*(-(-1+cos(f*x+e)-sin(f*x
+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2
*2^(1/2))*a*b^2-2*(-a^2+b^2)^(1/2)*2^(1/2)*cos(f*x+e)*a^2-sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^
(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*
x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a*b-sin(f*x+e)*(-(-
1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(
f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))
*(-a^2+b^2)^(1/2)*a*b+((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*(-(-1+c
os(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(-b+(-a^2+
b^2)^(1/2)+a),1/2*2^(1/2))*b^3-((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2
)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/
(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*b^3+2*sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f
*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/
sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a*b-2*(-a^2+b^2)^(1/2)*sin(f*x+e)*EllipticF((-(-1+cos(f*x+e)-s
in(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f
*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*a^2-sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*
x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+
cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*a^2*b+sin(f*x+e)*(-(-1+cos(f*x
+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(
1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*a^2*b+2*
(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/
sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*cos(f
*x+e)^2*a*b+sin(f*x+e)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2
*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+co
s(f*x+e))/sin(f*x+e))^(1/2)*a*b^2-sin(f*x+e)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+si
n(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x
+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*a*b^2-2*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+
cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x
+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*a*b-2*cos(f*x+e)^2*(-a^2+b^2)^(1/2)*2^(1/2)*a*b+2*cos(f*x
+e)*(-a^2+b^2)^(1/2)*2^(1/2)*a*b+EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1
/2)-a),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/
2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*cos(f*x+e)^2*b^3-(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos
(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e
))/sin(f*x+e))^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))*cos(f*x+e)^2*b^3+2*(-(-1+cos(f*x+e)-sin(f*x+e))/si
n(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(
-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-a^2+b^2)^(1/2)*b^2+EllipticPi((-(-1+cos(f*x+e)-sin(
f*x+e))/sin(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2
)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*(-a^2+b^2)^(1/2)*cos(f*x+e)
^2*b^2-EllipticPi((-(-1+cos(f*x+e)-sin(f*x+e))/...
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e))^2/(d*sin(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate((g*cos(f*x + e))^(3/2)/((b*sin(f*x + e) + a)^2*sqrt(d*sin(f*x + e))), x)
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e))^2/(d*sin(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (g \cos {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\sqrt {d \sin {\left (e + f x \right )}} \left (a + b \sin {\left (e + f x \right )}\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*cos(f*x+e))**(3/2)/(a+b*sin(f*x+e))**2/(d*sin(f*x+e))**(1/2),x)
[Out]
Integral((g*cos(e + f*x))**(3/2)/(sqrt(d*sin(e + f*x))*(a + b*sin(e + f*x))**2), x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e))^2/(d*sin(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate((g*cos(f*x + e))^(3/2)/((b*sin(f*x + e) + a)^2*sqrt(d*sin(f*x + e))), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}}{\sqrt {d\,\sin \left (e+f\,x\right )}\,{\left (a+b\,\sin \left (e+f\,x\right )\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((g*cos(e + f*x))^(3/2)/((d*sin(e + f*x))^(1/2)*(a + b*sin(e + f*x))^2),x)
[Out]
int((g*cos(e + f*x))^(3/2)/((d*sin(e + f*x))^(1/2)*(a + b*sin(e + f*x))^2), x)
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