3.73 \(\int \frac {\sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=438 \[ \frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} d \left (b^2-a^2\right )^{5/4}}-\frac {a \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} d \left (b^2-a^2\right )^{5/4}}-\frac {b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \left (a^2-b^2\right ) \sqrt {\sin (c+d x)}}+\frac {a^2 e \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{2 b d \left (a^2-b^2\right ) \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}+\frac {a^2 e \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{2 b d \left (a^2-b^2\right ) \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}} \]

[Out]

-b*(e*sin(d*x+c))^(3/2)/(a^2-b^2)/d/e/(a+b*cos(d*x+c))+1/2*a*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1
/4)/e^(1/2))*e^(1/2)/(-a^2+b^2)^(5/4)/d/b^(1/2)-1/2*a*arctanh(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^
(1/2))*e^(1/2)/(-a^2+b^2)^(5/4)/d/b^(1/2)-1/2*a^2*e*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d
*x)*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/b/(a^2-b^2)/d/(b-(
-a^2+b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-1/2*a^2*e*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*
EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/b/(a^2-b^2)/d/(b+(-a^2
+b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(
1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/(a^2-b^2)/d/sin(d*x+c)^(1/2)

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Rubi [A]  time = 0.91, antiderivative size = 438, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {2694, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} d \left (b^2-a^2\right )^{5/4}}-\frac {a \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} d \left (b^2-a^2\right )^{5/4}}-\frac {b (e \sin (c+d x))^{3/2}}{d e \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \left (a^2-b^2\right ) \sqrt {\sin (c+d x)}}+\frac {a^2 e \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{2 b d \left (a^2-b^2\right ) \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}+\frac {a^2 e \sqrt {\sin (c+d x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{2 b d \left (a^2-b^2\right ) \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[e]*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*Sqrt[b]*(-a^2 + b^2)^(5/4)*
d) - (a*Sqrt[e]*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(2*Sqrt[b]*(-a^2 + b^2)^
(5/4)*d) + (a^2*e*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*b*(a^
2 - b^2)*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (a^2*e*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c -
Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(2*b*(a^2 - b^2)*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (Elli
pticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)*d*Sqrt[Sin[c + d*x]]) - (b*(e*Sin[c + d*x])^(3
/2))/((a^2 - b^2)*d*e*(a + b*Cos[c + d*x]))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2639

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2694

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(f*g*(a^2 - b^2)*(m + 1)), x] + Dist[1/((a^2 - b^2)*(m +
1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + p + 2)*Sin[e + f*x]), x], x] /; F
reeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*p]

Rule 2701

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> With[{q = Rt[-a^2
+ b^2, 2]}, Dist[(a*g)/(2*b), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (-Dist[(a*g)/(2*b),
Int[1/(Sqrt[g*Cos[e + f*x]]*(q - b*Cos[e + f*x])), x], x] + Dist[(b*g)/f, Subst[Int[Sqrt[x]/(g^2*(a^2 - b^2)
+ b^2*x^2), x], x, g*Cos[e + f*x]], x])] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2867

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]))/((a_) + (b_.)*sin[(e_.) + (
f_.)*(x_)]), x_Symbol] :> Dist[d/b, Int[(g*Cos[e + f*x])^p, x], x] + Dist[(b*c - a*d)/b, Int[(g*Cos[e + f*x])^
p/(a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^2} \, dx &=-\frac {b (e \sin (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}+\frac {\int \frac {\left (-a-\frac {1}{2} b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{-a^2+b^2}\\ &=-\frac {b (e \sin (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}+\frac {\int \sqrt {e \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )}+\frac {a \int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {b (e \sin (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}-\frac {\left (a^2 e\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{4 b \left (a^2-b^2\right )}+\frac {\left (a^2 e\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{4 b \left (a^2-b^2\right )}-\frac {(a b e) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \sin (c+d x)\right )}{2 \left (a^2-b^2\right ) d}+\frac {\sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)} \, dx}{2 \left (a^2-b^2\right ) \sqrt {\sin (c+d x)}}\\ &=\frac {E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) d \sqrt {\sin (c+d x)}}-\frac {b (e \sin (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}-\frac {(a b e) \operatorname {Subst}\left (\int \frac {x^2}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (a^2-b^2\right ) d}-\frac {\left (a^2 e \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{4 b \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {\left (a^2 e \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{4 b \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}\\ &=\frac {a^2 e \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{2 b \left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a^2 e \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{2 b \left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) d \sqrt {\sin (c+d x)}}-\frac {b (e \sin (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}+\frac {(a e) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{2 \left (a^2-b^2\right ) d}-\frac {(a e) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{2 \left (a^2-b^2\right ) d}\\ &=\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 \sqrt {b} \left (-a^2+b^2\right )^{5/4} d}-\frac {a \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{2 \sqrt {b} \left (-a^2+b^2\right )^{5/4} d}+\frac {a^2 e \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{2 b \left (a^2-b^2\right ) \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a^2 e \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{2 b \left (a^2-b^2\right ) \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{\left (a^2-b^2\right ) d \sqrt {\sin (c+d x)}}-\frac {b (e \sin (c+d x))^{3/2}}{\left (a^2-b^2\right ) d e (a+b \cos (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 13.94, size = 786, normalized size = 1.79 \[ \frac {b \sin (c+d x) \sqrt {e \sin (c+d x)}}{d \left (b^2-a^2\right ) (a+b \cos (c+d x))}+\frac {\sqrt {e \sin (c+d x)} \left (\frac {\cos ^2(c+d x) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (8 b^{5/2} \sin ^{\frac {3}{2}}(c+d x) F_1\left (\frac {3}{4};-\frac {1}{2},1;\frac {7}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{b^2-a^2}\right )+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (-\log \left (-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+\sqrt {a^2-b^2}+b \sin (c+d x)\right )+\log \left (\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+\sqrt {a^2-b^2}+b \sin (c+d x)\right )+2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )\right )\right )}{12 \sqrt {b} \left (b^2-a^2\right ) \left (1-\sin ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac {4 a \cos (c+d x) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (\frac {a \sin ^{\frac {3}{2}}(c+d x) F_1\left (\frac {3}{4};\frac {1}{2},1;\frac {7}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{b^2-a^2}\right )}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (-\log \left (-(1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}+i b \sin (c+d x)\right )+\log \left ((1+i) \sqrt {b} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}+i b \sin (c+d x)\right )+2 \tan ^{-1}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )\right )}{\sqrt {b} \sqrt [4]{b^2-a^2}}\right )}{\sqrt {1-\sin ^2(c+d x)} (a+b \cos (c+d x))}\right )}{2 d (a-b) (a+b) \sqrt {\sin (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(b*Sin[c + d*x]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)*d*(a + b*Cos[c + d*x])) + (Sqrt[e*Sin[c + d*x]]*((Cos[c +
d*x]^2*(3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] -
2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(
a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4
)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x]]) + 8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Sin[c + d*x]^2, (b^2*Sin[c + d
*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(3/2))*(a + b*Sqrt[1 - Sin[c + d*x]^2]))/(12*Sqrt[b]*(-a^2 + b^2)*(a + b*Cos
[c + d*x])*(1 - Sin[c + d*x]^2)) + (4*a*Cos[c + d*x]*(((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Sin[c +
d*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt
[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]] + Log[Sqrt[-a^2 + b^2
] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)) +
(a*AppellF1[3/4, 1/2, 1, 7/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(3/2))/(3*(a^2
- b^2)))*(a + b*Sqrt[1 - Sin[c + d*x]^2]))/((a + b*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2])))/(2*(a - b)*(a + b
)*d*Sqrt[Sin[c + d*x]])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(1/2)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \sin \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(1/2)/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 1.21, size = 1384, normalized size = 3.16 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(d*x+c))^(1/2)/(a+b*cos(d*x+c))^2,x)

[Out]

-1/d*e^3*a*b*(e*sin(d*x+c))^(3/2)/(a^2*e^2-b^2*e^2)/(-b^2*cos(d*x+c)^2*e^2+a^2*e^2)-1/8/d*e^3*a/b/(a^2*e^2-b^2
*e^2)/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*ln((e*sin(d*x+c)-(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2
)+(e^2*(a^2-b^2)/b^2)^(1/2))/(e*sin(d*x+c)+(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^
2)/b^2)^(1/2)))-1/4/d*e^3*a/b/(a^2*e^2-b^2*e^2)/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2*(a^2-b^2
)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)+1)-1/4/d*e^3*a/b/(a^2*e^2-b^2*e^2)/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*arctan(
2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)-1)+1/2/d*e/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/b^2*(-sin(d*
x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*EllipticPi((-sin(d*x+c)+1)^(1/2),
1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))+1/2/d*e/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/b^2*(-sin(d*x+c)+1)^(1/2)*(2*sin
(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*EllipticPi((-sin(d*x+c)+1)^(1/2),1/(1+(-a^2+b^2)^(1/2
)/b),1/2*2^(1/2))+1/d*e*sin(d*x+c)^2*cos(d*x+c)/(e*sin(d*x+c))^(1/2)*b^2/(a^2-b^2)/(-cos(d*x+c)^2*b^2+a^2)-1/d
*e/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ell
ipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+1/2/d*e/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)*(-sin(d*x+c)+1)^(1
/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-3/4/d*e/cos(d*x+c)/(e
*sin(d*x+c))^(1/2)*a^2/(a^2-b^2)/b^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(1-(-a^2+b^
2)^(1/2)/b)*EllipticPi((-sin(d*x+c)+1)^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))+1/2/d*e/cos(d*x+c)/(e*sin(d
*x+c))^(1/2)/(a^2-b^2)*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*El
lipticPi((-sin(d*x+c)+1)^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-3/4/d*e/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*a
^2/(a^2-b^2)/b^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*Elliptic
Pi((-sin(d*x+c)+1)^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))+1/2/d*e/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^
2)*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*EllipticPi((-sin(d*x+c
)+1)^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \sin \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(1/2)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {e\,\sin \left (c+d\,x\right )}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(c + d*x))^(1/2)/(a + b*cos(c + d*x))^2,x)

[Out]

int((e*sin(c + d*x))^(1/2)/(a + b*cos(c + d*x))^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))**(1/2)/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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