3.1.66 \(\int \frac {1}{(a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}} \, dx\) [66]

Optimal. Leaf size=447 \[ \frac {b^{5/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{7/4} d e^{5/2}}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{7/4} d e^{5/2}}+\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}-\frac {a b^2 \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d e^2 \sqrt {e \sin (c+d x)}}-\frac {a b^2 \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right ) \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d e^2 \sqrt {e \sin (c+d x)}} \]

[Out]

b^(5/2)*arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/(-a^2+b^2)^(7/4)/d/e^(5/2)+b^(5/2)*arcta
nh(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))/(-a^2+b^2)^(7/4)/d/e^(5/2)+2/3*(b-a*cos(d*x+c))/(a^2
-b^2)/d/e/(e*sin(d*x+c))^(3/2)-2/3*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)*EllipticF(c
os(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/(a^2-b^2)/d/e^2/(e*sin(d*x+c))^(1/2)+a*b^2*(sin(1/2*c+1/4*P
i+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)/(a^2-b^2)/d/e^2/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(e*sin(d*x+c))^(1/2)+a*b^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)*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)/(a^2-b^2)/d/e^2/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.67, antiderivative size = 447, 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 = {2775, 2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \begin {gather*} \frac {b^{5/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac {2 a \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d e^2 \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}-\frac {a b^2 \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 )}{d e^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {e \sin (c+d x)}}-\frac {a b^2 \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 )}{d e^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {e \sin (c+d x)}}+\frac {2 (b-a \cos (c+d x))}{3 d e \left (a^2-b^2\right ) (e \sin (c+d x))^{3/2}}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{b^2-a^2}}\right )}{d e^{5/2} \left (b^2-a^2\right )^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(5/2)),x]

[Out]

(b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(7/4)*d*e^(5/2)) +
(b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(7/4)*d*e^(5/2))
+ (2*(b - a*Cos[c + d*x]))/(3*(a^2 - b^2)*d*e*(e*Sin[c + d*x])^(3/2)) + (2*a*EllipticF[(c - Pi/2 + d*x)/2, 2]
*Sqrt[Sin[c + d*x]])/(3*(a^2 - b^2)*d*e^2*Sqrt[e*Sin[c + d*x]]) - (a*b^2*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2
]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/((a^2 - b^2)*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*d*e^2*Sqrt[e*Sin[
c + d*x]]) - (a*b^2*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/((a^2
- b^2)*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*d*e^2*Sqrt[e*Sin[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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[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 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 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 2775

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*Cos
[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2781

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[
-a^2 + b^2, 2]}, Dist[-a/(2*q), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Dist[b*(g/f), Sub
st[Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e
+ f*x]]*(q - b*Cos[e + f*x])), 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 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 {1}{(a+b \cos (c+d x)) (e \sin (c+d x))^{5/2}} \, dx &=\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}-\frac {2 \int \frac {-\frac {a^2}{2}+\frac {3 b^2}{2}-\frac {1}{2} a b \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right ) e^2}\\ &=\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right ) e^2}-\frac {b^2 \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx}{\left (a^2-b^2\right ) e^2}\\ &=\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}-\frac {\left (a b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 \left (-a^2+b^2\right )^{3/2} e^2}-\frac {\left (a b^2\right ) \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 \left (-a^2+b^2\right )^{3/2} e^2}+\frac {b^3 \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \sin (c+d x)\right )}{\left (a^2-b^2\right ) d e}+\frac {\left (a \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right ) e^2 \sqrt {e \sin (c+d x)}}\\ &=\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {\left (2 b^3\right ) \text {Subst}\left (\int \frac {1}{\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 e}-\frac {\left (a b^2 \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}{2 \left (-a^2+b^2\right )^{3/2} e^2 \sqrt {e \sin (c+d x)}}-\frac {\left (a b^2 \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}{2 \left (-a^2+b^2\right )^{3/2} e^2 \sqrt {e \sin (c+d x)}}\\ &=\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \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)}}{\left (-a^2+b^2\right )^{3/2} \left (b-\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}-\frac {a b^2 \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)}}{\left (-a^2+b^2\right )^{3/2} \left (b+\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {b^3 \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e-b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (-a^2+b^2\right )^{3/2} d e^2}+\frac {b^3 \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} e+b x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (-a^2+b^2\right )^{3/2} d e^2}\\ &=\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{7/4} d e^{5/2}}+\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{7/4} d e^{5/2}}+\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \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)}}{\left (-a^2+b^2\right )^{3/2} \left (b-\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}-\frac {a b^2 \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)}}{\left (-a^2+b^2\right )^{3/2} \left (b+\sqrt {-a^2+b^2}\right ) d e^2 \sqrt {e \sin (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 = 31.77, size = 1192, normalized size = 2.67 \begin {gather*} -\frac {2 (-b+a \cos (c+d x)) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (e \sin (c+d x))^{5/2}}+\frac {\sin ^{\frac {5}{2}}(c+d x) \left (\frac {2 a b \cos ^2(c+d x) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (\frac {a \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 )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}+\frac {5 b \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\sin (c+d x)} \sqrt {1-\sin ^2(c+d x)}}{\left (-5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )+2 \left (2 b^2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )+\left (a^2-b^2\right ) F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {2 \left (a^2-3 b^2\right ) \cos (c+d x) \left (a+b \sqrt {1-\sin ^2(c+d x)}\right ) \left (-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {b} \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 )}{\left (-a^2+b^2\right )^{3/4}}+\frac {5 a \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right ) \sqrt {\sin (c+d x)}}{\sqrt {1-\sin ^2(c+d x)} \left (5 \left (a^2-b^2\right ) F_1\left (\frac {1}{4};\frac {1}{2},1;\frac {5}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )-2 \left (2 b^2 F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )+\left (-a^2+b^2\right ) F_1\left (\frac {5}{4};\frac {3}{2},1;\frac {9}{4};\sin ^2(c+d x),\frac {b^2 \sin ^2(c+d x)}{-a^2+b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (a^2+b^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(a+b \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{3 (a-b) (a+b) d (e \sin (c+d x))^{5/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(5/2)),x]

[Out]

(-2*(-b + a*Cos[c + d*x])*Sin[c + d*x])/(3*(a^2 - b^2)*d*(e*Sin[c + d*x])^(5/2)) + (Sin[c + d*x]^(5/2)*((2*a*b
*Cos[c + d*x]^2*(a + b*Sqrt[1 - Sin[c + d*x]^2])*((a*(-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] - S
qrt[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]]))/(4*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)) + (5*b*(a^2 - b
^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Sin[c + d*x]]*Sqrt[1 -
Sin[c + d*x]^2])/((-5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^
2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + (a^2 - b^2)*Ap
pellF1[5/4, 1/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)])*Sin[c + d*x]^2)*(a^2 + b^2*(-1 +
Sin[c + d*x]^2)))))/((a + b*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (2*(a^2 - 3*b^2)*Cos[c + d*x]*(a + b*Sqrt[1
- Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[b]*(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)*Sq
rt[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]]))/(-a^2 + b^2)^(3/4) + (5*a*(a^2 - b^2)*AppellF1[1/4, 1/2
, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Sin[c + d*x]])/(Sqrt[1 - Sin[c + d*x]^2]*(5*
(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[
5/4, 1/2, 2, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4,
Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)])*Sin[c + d*x]^2)*(a^2 + b^2*(-1 + Sin[c + d*x]^2)))))/((a
+ b*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2])))/(3*(a - b)*(a + b)*d*(e*Sin[c + d*x])^(5/2))

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Maple [A]
time = 0.26, size = 820, normalized size = 1.83

method	result	size
default	\(\frac {\frac {b^{3} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {e \sin \left (d x +c \right )+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}{e \sin \left (d x +c \right )-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}\right )}{4 e \left (a -b \right ) \left (a +b \right ) \left (a^{2} e^{2}-b^{2} e^{2}\right )}+\frac {b^{3} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}+1\right )}{2 e \left (a -b \right ) \left (a +b \right ) \left (a^{2} e^{2}-b^{2} e^{2}\right )}+\frac {b^{3} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}-1\right )}{2 e \left (a -b \right ) \left (a +b \right ) \left (a^{2} e^{2}-b^{2} e^{2}\right )}+\frac {2 b}{3 e \left (a^{2}-b^{2}\right ) \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {\sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) e \sin \left (d x +c \right )}\, a \left (\frac {\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 \left (a^{2}-b^{2}\right ) \sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) e \sin \left (d x +c \right )}\, \left (\cos ^{2}\left (d x +c \right )-1\right )}-\frac {b^{2} \left (-\frac {\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {1}{1-\frac {\sqrt {-a^{2}+b^{2}}}{b}}, \frac {\sqrt {2}}{2}\right )}{2 b \sqrt {-a^{2}+b^{2}}\, \sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) e \sin \left (d x +c \right )}\, \left (1-\frac {\sqrt {-a^{2}+b^{2}}}{b}\right )}+\frac {\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticPi \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {1}{1+\frac {\sqrt {-a^{2}+b^{2}}}{b}}, \frac {\sqrt {2}}{2}\right )}{2 b \sqrt {-a^{2}+b^{2}}\, \sqrt {\left (\cos ^{2}\left (d x +c \right )\right ) e \sin \left (d x +c \right )}\, \left (1+\frac {\sqrt {-a^{2}+b^{2}}}{b}\right )}\right )}{\left (a -b \right ) \left (a +b \right )}\right )}{e^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(820\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(d*x+c))/(e*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

(1/4*b^3/e/(a-b)/(a+b)*(e^2*(a^2-b^2)/b^2)^(1/4)/(a^2*e^2-b^2*e^2)*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)))+1/2*b^3/e/(a-b)/(a+b)*(e^2*(a^2-b^2)/b^2)^(1/4)/(a^2*e^2-b^
2*e^2)*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*b^3/e/(a-b)/(a+b)*(e^2*(a^
2-b^2)/b^2)^(1/4)/(a^2*e^2-b^2*e^2)*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)+2
/3*b/e/(a^2-b^2)/(e*sin(d*x+c))^(3/2)+(cos(d*x+c)^2*e*sin(d*x+c))^(1/2)*a/e^2*(1/3/(a^2-b^2)/(cos(d*x+c)^2*e*s
in(d*x+c))^(1/2)/(cos(d*x+c)^2-1)*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticF((-s
in(d*x+c)+1)^(1/2),1/2*2^(1/2))+2*cos(d*x+c)^2*sin(d*x+c))-1/(a-b)/(a+b)*b^2*(-1/2/b/(-a^2+b^2)^(1/2)*(-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))+1/2/b/(-a^2+b^2)^(1/2)*(-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)*Ellipt
icPi((-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(1/(a+b*cos(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

e^(-5/2)*integrate(1/((b*cos(d*x + c) + a)*sin(d*x + c)^(5/2)), 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(1/(a+b*cos(d*x+c))/(e*sin(d*x+c))^(5/2),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 {1}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {5}{2}} \left (a + b \cos {\left (c + d x \right )}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c))/(e*sin(d*x+c))**(5/2),x)

[Out]

Integral(1/((e*sin(c + d*x))**(5/2)*(a + b*cos(c + d*x))), 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(1/(a+b*cos(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(e^(-5/2)/((b*cos(d*x + c) + a)*sin(d*x + c)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*sin(c + d*x))^(5/2)*(a + b*cos(c + d*x))),x)

[Out]

int(1/((e*sin(c + d*x))^(5/2)*(a + b*cos(c + d*x))), x)

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