3.3.39 \(\int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx\) [239]

Optimal. Leaf size=452 \[ -\frac {a^{3/2} b \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\left (a^2-b^2\right )^{7/4} d e^{5/2}}-\frac {a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt [4]{a^2-b^2} \sqrt {e}}\right )}{\left (a^2-b^2\right )^{7/4} d e^{5/2}}+\frac {2 (b-a \cos (c+d x))}{3 \left (a^2-b^2\right ) d e (e \sin (c+d x))^{3/2}}+\frac {2 a F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 \left (a^2-b^2\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right ) \left (a^2-b^2-a \sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b^2\right ) \left (a^2-b^2+a \sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}} \]

[Out]

-a^(3/2)*b*arctan(a^(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)-a^(3/2)*b*ar
ctanh(a^(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(
cos(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*
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*a/(a-(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^2-a*(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*a/(a+(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^2+a*(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(1/2)

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Rubi [A]
time = 0.76, antiderivative size = 452, 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 = {3957, 2945, 2946, 2721, 2720, 2781, 2886, 2884, 335, 218, 214, 211} \begin {gather*} \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 a}{a-\sqrt {a^2-b^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 \sqrt {a^2-b^2}+a^2-b^2\right ) \sqrt {e \sin (c+d x)}}+\frac {a b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^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 \sqrt {a^2-b^2}+a^2-b^2\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 {a^{3/2} b \text {ArcTan}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{5/2} \left (a^2-b^2\right )^{7/4}}-\frac {a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{5/2} \left (a^2-b^2\right )^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^(3/2)*b*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/((a^2 - b^2)^(7/4)*d*e^(5/2))
) - (a^(3/2)*b*ArcTanh[(Sqrt[a]*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*a)/(a - Sqrt[a^2 - b
^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/((a^2 - b^2)*(a^2 - b^2 - a*Sqrt[a^2 - b^2])*d*e^2*Sqrt[e*Sin
[c + d*x]]) + (a*b^2*EllipticPi[(2*a)/(a + Sqrt[a^2 - b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/((a^2
- b^2)*(a^2 - b^2 + a*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 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 2945

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

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \, dx &=-\int \frac {\cos (c+d x)}{(-b-a \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 {a b-\frac {1}{2} a^2 \cos (c+d x)}{(-b-a \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 {(a b) \int \frac {1}{(-b-a \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}-a \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}+a \sin (c+d x)\right )} \, dx}{2 \left (a^2-b^2\right )^{3/2} e^2}+\frac {\left (a^2 b\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (-a^2+b^2\right ) e^2+a^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 a^2 b\right ) \text {Subst}\left (\int \frac {1}{\left (-a^2+b^2\right ) e^2+a^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}-a \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}+a \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 a}{a-\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 (a-\sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \Pi \left (\frac {2 a}{a+\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 (a+\sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}-\frac {\left (a^2 b\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a^2-b^2} e-a x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d e^2}-\frac {\left (a^2 b\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a^2-b^2} e+a x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d e^2}\\ &=-\frac {a^{3/2} b \tan ^{-1}\left (\frac {\sqrt {a} \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 {a^{3/2} b \tanh ^{-1}\left (\frac {\sqrt {a} \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 a}{a-\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 (a-\sqrt {a^2-b^2}\right ) d e^2 \sqrt {e \sin (c+d x)}}+\frac {a b^2 \Pi \left (\frac {2 a}{a+\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 (a+\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 = 30.44, size = 1233, normalized size = 2.73 \begin {gather*} -\frac {a (b+a \cos (c+d x)) \sec (c+d x) \sin ^{\frac {5}{2}}(c+d x) \left (-\frac {2 a \cos ^2(c+d x) \left (b+a \sqrt {1-\sin ^2(c+d x)}\right ) \left (\frac {b \left (-2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )+2 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )+\log \left (\sqrt {-a^2+b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\sin (c+d x)}+a \sin (c+d x)\right )\right )}{4 \sqrt {2} \sqrt {a} \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 {a^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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )+2 \left (2 a^2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};\sin ^2(c+d x),\frac {a^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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (b^2+a^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(b+a \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {4 b \cos (c+d x) \left (b+a \sqrt {1-\sin ^2(c+d x)}\right ) \left (-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {a} \left (2 \text {ArcTan}\left (1-\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \text {ArcTan}\left (1+\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+\log \left (\sqrt {a^2-b^2}-(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )-\log \left (\sqrt {a^2-b^2}+(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+i a \sin (c+d x)\right )\right )}{\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 {a^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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )+2 \left (2 a^2 F_1\left (\frac {5}{4};\frac {1}{2},2;\frac {9}{4};\sin ^2(c+d x),\frac {a^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 {a^2 \sin ^2(c+d x)}{a^2-b^2}\right )\right ) \sin ^2(c+d x)\right ) \left (b^2+a^2 \left (-1+\sin ^2(c+d x)\right )\right )}\right )}{(b+a \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right )}{3 (a-b) (a+b) d (a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}}-\frac {2 (b-a \cos (c+d x)) (b+a \cos (c+d x)) \tan (c+d x)}{3 \left (-a^2+b^2\right ) d (a+b \sec (c+d x)) (e \sin (c+d x))^{5/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/3*(a*(b + a*Cos[c + d*x])*Sec[c + d*x]*Sin[c + d*x]^(5/2)*((-2*a*Cos[c + d*x]^2*(b + a*Sqrt[1 - Sin[c + d*x
]^2])*((b*(-2*ArcTan[1 - (Sqrt[2]*Sqrt[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt
[a]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[S
in[c + d*x]] + a*Sin[c + d*x]] + Log[Sqrt[-a^2 + b^2] + Sqrt[2]*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]]
+ a*Sin[c + d*x]]))/(4*Sqrt[2]*Sqrt[a]*(-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, (a^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)*App
ellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)] + 2*(2*a^2*AppellF1[5/4, -1/2, 2, 9
/4, Sin[c + d*x]^2, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)] + (-a^2 + b^2)*AppellF1[5/4, 1/2, 1, 9/4, Sin[c + d*x]^2
, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)])*Sin[c + d*x]^2)*(b^2 + a^2*(-1 + Sin[c + d*x]^2)))))/((b + a*Cos[c + d*x]
)*(1 - Sin[c + d*x]^2)) + (4*b*Cos[c + d*x]*(b + a*Sqrt[1 - Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[a]*(2*ArcTan[
1 - ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[a]*Sqrt[Sin[c + d*x]]
)/(a^2 - b^2)^(1/4)] + Log[Sqrt[a^2 - b^2] - (1 + I)*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*a*Sin[c
+ d*x]] - Log[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*a*Sin[c + d*x]]))/(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, (a^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, (a^2*
Sin[c + d*x]^2)/(a^2 - b^2)] + 2*(2*a^2*AppellF1[5/4, 1/2, 2, 9/4, Sin[c + d*x]^2, (a^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, (a^2*Sin[c + d*x]^2)/(a^2 - b^2)])*Sin[c + d*x
]^2)*(b^2 + a^2*(-1 + Sin[c + d*x]^2)))))/((b + a*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2])))/((a - b)*(a + b)*d
*(a + b*Sec[c + d*x])*(e*Sin[c + d*x])^(5/2)) - (2*(b - a*Cos[c + d*x])*(b + a*Cos[c + d*x])*Tan[c + d*x])/(3*
(-a^2 + b^2)*d*(a + b*Sec[c + d*x])*(e*Sin[c + d*x])^(5/2))

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Maple [A]
time = 0.39, size = 657, normalized size = 1.45

method	result	size
default	\(\frac {\frac {b \,a^{2} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}} \ln \left (\frac {\sqrt {e \sin \left (d x +c \right )}+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}{\sqrt {e \sin \left (d x +c \right )}-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{2 e \left (a +b \right ) \left (a -b \right ) \left (-a^{2} e^{2}+b^{2} e^{2}\right )}+\frac {b \,a^{2} \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{a^{2}}\right )^{\frac {1}{4}}}\right )}{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}}}{a}}, \frac {\sqrt {2}}{2}\right )}{2 \sqrt {a^{2}-b^{2}}\, a \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}}}{a}\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}}}{a}}, \frac {\sqrt {2}}{2}\right )}{2 \sqrt {a^{2}-b^{2}}\, a \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}}}{a}\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}\)	\(657\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/2/e*b/(a+b)/(a-b)*a^2*(e^2*(a^2-b^2)/a^2)^(1/4)/(-a^2*e^2+b^2*e^2)*ln(((e*sin(d*x+c))^(1/2)+(e^2*(a^2-b^2)/
a^2)^(1/4))/((e*sin(d*x+c))^(1/2)-(e^2*(a^2-b^2)/a^2)^(1/4)))+1/e*b/(a+b)/(a-b)*a^2*(e^2*(a^2-b^2)/a^2)^(1/4)/
(-a^2*e^2+b^2*e^2)*arctan((e*sin(d*x+c))^(1/2)/(e^2*(a^2-b^2)/a^2)^(1/4))+2/3/e*b/(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*sin(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((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+2*c
os(d*x+c)^2*sin(d*x+c))-1/(a-b)/(a+b)*b^2*(-1/2/(a^2-b^2)^(1/2)/a*(-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)/a)*EllipticPi((-sin(d*x+c)+1)^(1/2),1/(
1-(a^2-b^2)^(1/2)/a),1/2*2^(1/2))+1/2/(a^2-b^2)^(1/2)/a*(-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)/a)*EllipticPi((-sin(d*x+c)+1)^(1/2),1/(1+(a^2-b^2
)^(1/2)/a),1/2*2^(1/2))))/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

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Maxima [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*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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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*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5010 deep

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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*sec(d*x+c))/(e*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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