3.1.84 \(\int \frac {\sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^3} \, dx\) [84]

Optimal. Leaf size=529 \[ -\frac {\left (3 a^2+2 b^2\right ) \sqrt {e} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 \sqrt {b} \left (-a^2+b^2\right )^{9/4} d}+\frac {\left (3 a^2+2 b^2\right ) \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 \sqrt {b} \left (-a^2+b^2\right )^{9/4} d}+\frac {a \left (3 a^2+2 b^2\right ) e \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{8 b \left (a^2-b^2\right )^2 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \left (3 a^2+2 b^2\right ) e \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{8 b \left (a^2-b^2\right )^2 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt {\sin (c+d x)}}-\frac {b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))} \]

[Out]

-1/2*b*(e*sin(d*x+c))^(3/2)/(a^2-b^2)/d/e/(a+b*cos(d*x+c))^2-5/4*a*b*(e*sin(d*x+c))^(3/2)/(a^2-b^2)^2/d/e/(a+b
*cos(d*x+c))-1/8*(3*a^2+2*b^2)*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
)^(9/4)/d/b^(1/2)+1/8*(3*a^2+2*b^2)*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)^(9/4)/d/b^(1/2)-1/8*a*(3*a^2+2*b^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)*El
lipticPi(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)^2/d/(b-(-a^2
+b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-1/8*a*(3*a^2+2*b^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)^2/
d/(b+(-a^2+b^2)^(1/2))/(e*sin(d*x+c))^(1/2)-5/4*a*(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)^2/d/sin(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.76, antiderivative size = 529, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2773, 2943, 2946, 2721, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \begin {gather*} -\frac {\sqrt {e} \left (3 a^2+2 b^2\right ) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 \sqrt {b} d \left (b^2-a^2\right )^{9/4}}+\frac {\sqrt {e} \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{8 \sqrt {b} d \left (b^2-a^2\right )^{9/4}}-\frac {5 a b (e \sin (c+d x))^{3/2}}{4 d e \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac {b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac {5 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{4 d \left (a^2-b^2\right )^2 \sqrt {\sin (c+d x)}}+\frac {a e \left (3 a^2+2 b^2\right ) \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 )}{8 b d \left (a^2-b^2\right )^2 \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}+\frac {a e \left (3 a^2+2 b^2\right ) \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 )}{8 b d \left (a^2-b^2\right )^2 \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[e*Sin[c + d*x]]/(a + b*Cos[c + d*x])^3,x]

[Out]

-1/8*((3*a^2 + 2*b^2)*Sqrt[e]*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(Sqrt[b]*(-
a^2 + b^2)^(9/4)*d) + ((3*a^2 + 2*b^2)*Sqrt[e]*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt
[e])])/(8*Sqrt[b]*(-a^2 + b^2)^(9/4)*d) + (a*(3*a^2 + 2*b^2)*e*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - P
i/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b*(a^2 - b^2)^2*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (a*(3
*a^2 + 2*b^2)*e*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b*(a^2
- b^2)^2*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (5*a*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c +
d*x]])/(4*(a^2 - b^2)^2*d*Sqrt[Sin[c + d*x]]) - (b*(e*Sin[c + d*x])^(3/2))/(2*(a^2 - b^2)*d*e*(a + b*Cos[c +
d*x])^2) - (5*a*b*(e*Sin[c + d*x])^(3/2))/(4*(a^2 - b^2)^2*d*e*(a + b*Cos[c + d*x]))

Rule 211

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

Rule 214

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

Rule 304

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 335

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 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 2773

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] /;
FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*p]

Rule 2780

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 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 2886

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/
(c + d))*Sin[e + f*x]]), 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 2943

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*(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)*S
imp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p},
x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 2946

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))^3} \, dx &=-\frac {b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {\int \frac {\left (-2 a+\frac {1}{2} b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{(a+b \cos (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}+\frac {\int \frac {\left (\frac {1}{2} \left (4 a^2+b^2\right )+\frac {5}{4} a b \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac {b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}+\frac {(5 a) \int \sqrt {e \sin (c+d x)} \, dx}{8 \left (a^2-b^2\right )^2}+\frac {\left (3 a^2+2 b^2\right ) \int \frac {\sqrt {e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{8 \left (a^2-b^2\right )^2}\\ &=-\frac {b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}-\frac {\left (a \left (3 a^2+2 b^2\right ) e\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{16 b \left (a^2-b^2\right )^2}+\frac {\left (a \left (3 a^2+2 b^2\right ) e\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{16 b \left (a^2-b^2\right )^2}-\frac {\left (b \left (3 a^2+2 b^2\right ) e\right ) \text {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 )}{8 \left (a^2-b^2\right )^2 d}+\frac {\left (5 a \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{8 \left (a^2-b^2\right )^2 \sqrt {\sin (c+d x)}}\\ &=\frac {5 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt {\sin (c+d x)}}-\frac {b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}-\frac {\left (b \left (3 a^2+2 b^2\right ) e\right ) \text {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 )}{4 \left (a^2-b^2\right )^2 d}-\frac {\left (a \left (3 a^2+2 b^2\right ) 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}{16 b \left (a^2-b^2\right )^2 \sqrt {e \sin (c+d x)}}+\frac {\left (a \left (3 a^2+2 b^2\right ) 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}{16 b \left (a^2-b^2\right )^2 \sqrt {e \sin (c+d x)}}\\ &=\frac {a \left (3 a^2+2 b^2\right ) 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)}}{8 b \left (a^2-b^2\right )^2 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \left (3 a^2+2 b^2\right ) 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)}}{8 b \left (a^2-b^2\right )^2 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt {\sin (c+d x)}}-\frac {b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}+\frac {\left (\left (3 a^2+2 b^2\right ) e\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{8 \left (a^2-b^2\right )^2 d}-\frac {\left (\left (3 a^2+2 b^2\right ) e\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac {\left (3 a^2+2 b^2\right ) \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 \sqrt {b} \left (-a^2+b^2\right )^{9/4} d}+\frac {\left (3 a^2+2 b^2\right ) \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{8 \sqrt {b} \left (-a^2+b^2\right )^{9/4} d}+\frac {a \left (3 a^2+2 b^2\right ) 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)}}{8 b \left (a^2-b^2\right )^2 \left (b-\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \left (3 a^2+2 b^2\right ) 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)}}{8 b \left (a^2-b^2\right )^2 \left (b+\sqrt {-a^2+b^2}\right ) d \sqrt {e \sin (c+d x)}}+\frac {5 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt {\sin (c+d x)}}-\frac {b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac {5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d 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 = 21.77, size = 748, normalized size = 1.41 \begin {gather*} \frac {\sqrt {e \sin (c+d x)} \left (-\frac {2 b \left (7 a^2-2 b^2+5 a b \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (a+b \cos (c+d x))^2}+\frac {\cos (c+d x) \left (a+b \sqrt {\cos ^2(c+d x)}\right ) \left (\frac {5 a \sec (c+d x) \left (3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+b \sin (c+d x)\right )\right )+8 b^{5/2} 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)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)\right )}{\sqrt {b} \left (-a^2+b^2\right )}+\frac {48 \left (4 a^2+b^2\right ) \left (\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \text {ArcTan}\left (1-\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \text {ArcTan}\left (1+\frac {(1+i) \sqrt {b} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+i b \sin (c+d x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}+\frac {a 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)}{-a^2+b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}\right )}{\sqrt {\cos ^2(c+d x)}}\right )}{12 (a-b)^2 (a+b)^2 (a+b \cos (c+d x)) \sqrt {\sin (c+d x)}}\right )}{8 d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[e*Sin[c + d*x]]/(a + b*Cos[c + d*x])^3,x]

[Out]

(Sqrt[e*Sin[c + d*x]]*((-2*b*(7*a^2 - 2*b^2 + 5*a*b*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(a + b*Cos[c +
d*x])^2) + (Cos[c + d*x]*(a + b*Sqrt[Cos[c + d*x]^2])*((5*a*Sec[c + d*x]*(3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(2*Arc
Tan[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)))/(S
qrt[b]*(-a^2 + b^2)) + (48*(4*a^2 + b^2)*(((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))))/Sq
rt[Cos[c + d*x]^2]))/(12*(a - b)^2*(a + b)^2*(a + b*Cos[c + d*x])*Sqrt[Sin[c + d*x]])))/(8*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2889\) vs. \(2(553)=1106\).
time = 0.55, size = 2890, normalized size = 5.46

method	result	size
default	\(\text {Expression too large to display}\)	\(2890\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

(-3/4*e*b^3/(-b^2*cos(d*x+c)^2*e^2+a^2*e^2)^2/(a^4-2*a^2*b^2+b^4)*(e*sin(d*x+c))^(7/2)*a^2-1/2*e*b^5/(-b^2*cos
(d*x+c)^2*e^2+a^2*e^2)^2/(a^4-2*a^2*b^2+b^4)*(e*sin(d*x+c))^(7/2)-7/4*e^3*b/(-b^2*cos(d*x+c)^2*e^2+a^2*e^2)^2/
(a^2-b^2)*(e*sin(d*x+c))^(3/2)*a^2-1/2*e^3*b^3/(-b^2*cos(d*x+c)^2*e^2+a^2*e^2)^2/(a^2-b^2)*(e*sin(d*x+c))^(3/2
)-3/32*e/b/(a^4-2*a^2*b^2+b^4)/(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)))*a^2-1/16*e*b/(a^4-2*a^2*b^2+b^4)/(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)))-3/16*e/b/(a^4-2*a^2*b^2+b^
4)/(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)*a^2-1/8*
e*b/(a^4-2*a^2*b^2+b^4)/(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)-3/16*e/b/(a^4-2*a^2*b^2+b^4)/(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)*a^2-1/8*e*b/(a^4-2*a^2*b^2+b^4)/(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)-(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*e*a*(3/2*b^2/e/a^2/(a^2-
b^2)*sin(d*x+c)*(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-cos(d*x+c)^2*b^2+a^2)-3/2/a^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)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),
1/2*2^(1/2))+3/4/a^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)/(cos(d*x+c)^2*e*s
in(d*x+c))^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-9/8/(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)/(cos(d*x+c)^2*e*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))+3/4/a^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)/(cos(d*x+c)^2*e*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))-9/8/(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)/(cos(d*x+c)^2*e*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))+3/4/a^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)/(cos
(d*x+c)^2*e*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))-4*a^2*(1/4*b^2/e/a^2/(a^2-b^2)*sin(d*x+c)*(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-cos(d*x+c)^2*b^2+a
^2)^2+1/16*b^2*(11*a^2-6*b^2)/a^4/(a^2-b^2)^2/e*sin(d*x+c)*(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(-cos(d*x+c)^2*b^
2+a^2)-11/16/a^2/(a^2-b^2)^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin
(d*x+c))^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+3/8/a^4/(a^2-b^2)^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d
*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))
*b^2+11/32/a^2/(a^2-b^2)^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d
*x+c))^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-3/16/a^4/(a^2-b^2)^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*
x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*
b^2-21/64/(a^2-b^2)^2/b^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*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))+7/16
/a^2/(a^2-b^2)^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*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))-3/16/a^4/(a^2
-b^2)^2*b^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*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))-21/64/(a^2-b^2)^2/
b^2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*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))+7/16/a^2/(a^2-b^2)^2*(-sin
(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*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))-3/16/a^4/(a^2-b^2)^2*b^2*(-sin(d*x+
c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*E
llipticPi((-sin(d*x+c)+1)^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))))/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

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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((e*sin(d*x+c))^(1/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

e^(1/2)*integrate(sqrt(sin(d*x + c))/(b*cos(d*x + c) + a)^3, 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((e*sin(d*x+c))^(1/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [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((e*sin(d*x+c))**(1/2)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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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((e*sin(d*x+c))^(1/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(e^(1/2)*sqrt(sin(d*x + c))/(b*cos(d*x + c) + a)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {e\,\sin \left (c+d\,x\right )}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3} \,d x \end {gather*}

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))^3,x)

[Out]

int((e*sin(c + d*x))^(1/2)/(a + b*cos(c + d*x))^3, x)

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