3.238 \(\int \frac {1}{(a+b \sec (c+d x)) (e \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=430 \[ \frac {\sqrt {a} b \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{3/2} \left (a^2-b^2\right )^{5/4}}-\frac {\sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{3/2} \left (a^2-b^2\right )^{5/4}}-\frac {2 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \left (a^2-b^2\right ) \sqrt {\sin (c+d x)}}+\frac {2 (b-a \cos (c+d x))}{d e \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}-\frac {b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e \left (a^2-b^2\right ) \left (a-\sqrt {a^2-b^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e \left (a^2-b^2\right ) \left (\sqrt {a^2-b^2}+a\right ) \sqrt {e \sin (c+d x)}} \]

[Out]

b*arctan(a^(1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))*a^(1/2)/(a^2-b^2)^(5/4)/d/e^(3/2)-b*arctanh(a^(
1/2)*(e*sin(d*x+c))^(1/2)/(a^2-b^2)^(1/4)/e^(1/2))*a^(1/2)/(a^2-b^2)^(5/4)/d/e^(3/2)+2*(b-a*cos(d*x+c))/(a^2-b
^2)/d/e/(e*sin(d*x+c))^(1/2)+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/(a-(a^2-b^2)^(1/2))/(e*s
in(d*x+c))^(1/2)+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/(a+(a^2-b^2)^(1/2))/(e*sin(d*x+c))^(
1/2)+2*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)/d/e^2/sin(d*x+c)^(1/2)

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Rubi [A]  time = 1.04, antiderivative size = 430, 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 = {3872, 2866, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac {\sqrt {a} b \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{3/2} \left (a^2-b^2\right )^{5/4}}-\frac {\sqrt {a} b \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {e \sin (c+d x)}}{\sqrt {e} \sqrt [4]{a^2-b^2}}\right )}{d e^{3/2} \left (a^2-b^2\right )^{5/4}}-\frac {2 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d e^2 \left (a^2-b^2\right ) \sqrt {\sin (c+d x)}}+\frac {2 (b-a \cos (c+d x))}{d e \left (a^2-b^2\right ) \sqrt {e \sin (c+d x)}}-\frac {b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a-\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e \left (a^2-b^2\right ) \left (a-\sqrt {a^2-b^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {b^2 \sqrt {\sin (c+d x)} \Pi \left (\frac {2 a}{a+\sqrt {a^2-b^2}};\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d e \left (a^2-b^2\right ) \left (\sqrt {a^2-b^2}+a\right ) \sqrt {e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Sec[c + d*x])*(e*Sin[c + d*x])^(3/2)),x]

[Out]

(Sqrt[a]*b*ArcTan[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/((a^2 - b^2)^(5/4)*d*e^(3/2)) -
(Sqrt[a]*b*ArcTanh[(Sqrt[a]*Sqrt[e*Sin[c + d*x]])/((a^2 - b^2)^(1/4)*Sqrt[e])])/((a^2 - b^2)^(5/4)*d*e^(3/2))
+ (2*(b - a*Cos[c + d*x]))/((a^2 - b^2)*d*e*Sqrt[e*Sin[c + d*x]]) - (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 - Sqrt[a^2 - b^2])*d*e*Sqrt[e*Sin[c + d*x]]) -
(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 + Sqrt
[a^2 - b^2])*d*e*Sqrt[e*Sin[c + d*x]]) - (2*a*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/((a^2 - b
^2)*d*e^2*Sqrt[Sin[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 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 2866

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 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]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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))^{3/2}} \, dx &=-\int \frac {\cos (c+d x)}{(-b-a \cos (c+d x)) (e \sin (c+d x))^{3/2}} \, dx\\ &=\frac {2 (b-a \cos (c+d x))}{\left (a^2-b^2\right ) d e \sqrt {e \sin (c+d x)}}+\frac {2 \int \frac {\left (a b+\frac {1}{2} a^2 \cos (c+d x)\right ) \sqrt {e \sin (c+d x)}}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right ) e^2}\\ &=\frac {2 (b-a \cos (c+d x))}{\left (a^2-b^2\right ) d e \sqrt {e \sin (c+d x)}}-\frac {a \int \sqrt {e \sin (c+d x)} \, dx}{\left (a^2-b^2\right ) e^2}+\frac {(a b) \int \frac {\sqrt {e \sin (c+d x)}}{-b-a \cos (c+d x)} \, dx}{\left (a^2-b^2\right ) e^2}\\ &=\frac {2 (b-a \cos (c+d x))}{\left (a^2-b^2\right ) d e \sqrt {e \sin (c+d x)}}+\frac {b^2 \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 ) e}-\frac {b^2 \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 ) e}+\frac {\left (a^2 b\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x}}{\left (-a^2+b^2\right ) e^2+a^2 x^2} \, dx,x,e \sin (c+d x)\right )}{\left (a^2-b^2\right ) d e}-\frac {\left (a \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{\left (a^2-b^2\right ) e^2 \sqrt {\sin (c+d x)}}\\ &=\frac {2 (b-a \cos (c+d x))}{\left (a^2-b^2\right ) d e \sqrt {e \sin (c+d x)}}-\frac {2 a 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 e^2 \sqrt {\sin (c+d x)}}+\frac {\left (2 a^2 b\right ) \operatorname {Subst}\left (\int \frac {x^2}{\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 (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 ) e \sqrt {e \sin (c+d x)}}-\frac {\left (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 ) e \sqrt {e \sin (c+d x)}}\\ &=\frac {2 (b-a \cos (c+d x))}{\left (a^2-b^2\right ) d e \sqrt {e \sin (c+d x)}}-\frac {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-\sqrt {a^2-b^2}\right ) d e \sqrt {e \sin (c+d x)}}-\frac {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+\sqrt {a^2-b^2}\right ) d e \sqrt {e \sin (c+d x)}}-\frac {2 a 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 e^2 \sqrt {\sin (c+d x)}}-\frac {(a b) \operatorname {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 ) d e}+\frac {(a b) \operatorname {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 ) d e}\\ &=\frac {\sqrt {a} 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 )^{5/4} d e^{3/2}}-\frac {\sqrt {a} 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 )^{5/4} d e^{3/2}}+\frac {2 (b-a \cos (c+d x))}{\left (a^2-b^2\right ) d e \sqrt {e \sin (c+d x)}}-\frac {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-\sqrt {a^2-b^2}\right ) d e \sqrt {e \sin (c+d x)}}-\frac {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+\sqrt {a^2-b^2}\right ) d e \sqrt {e \sin (c+d x)}}-\frac {2 a 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 e^2 \sqrt {\sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 14.48, size = 834, normalized size = 1.94 \[ -\frac {a (b+a \cos (c+d x)) \sec (c+d x) \left (\frac {\left (8 F_1\left (\frac {3}{4};-\frac {1}{2},1;\frac {7}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac {3}{2}}(c+d x) a^{5/2}+3 \sqrt {2} b \left (b^2-a^2\right )^{3/4} \left (2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )-\log \left (a \sin (c+d x)-\sqrt {2} \sqrt {a} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}\right )+\log \left (a \sin (c+d x)+\sqrt {2} \sqrt {a} \sqrt [4]{b^2-a^2} \sqrt {\sin (c+d x)}+\sqrt {b^2-a^2}\right )\right )\right ) \left (\sqrt {1-\sin ^2(c+d x)} a+b\right ) \cos ^2(c+d x)}{12 \sqrt {a} \left (a^2-b^2\right ) (b+a \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac {4 b \left (\frac {b F_1\left (\frac {3}{4};\frac {1}{2},1;\frac {7}{4};\sin ^2(c+d x),\frac {a^2 \sin ^2(c+d x)}{a^2-b^2}\right ) \sin ^{\frac {3}{2}}(c+d x)}{3 \left (b^2-a^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \tan ^{-1}\left (1-\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (i a \sin (c+d x)-(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+\sqrt {a^2-b^2}\right )+\log \left (i a \sin (c+d x)+(1+i) \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\sin (c+d x)}+\sqrt {a^2-b^2}\right )\right )}{\sqrt {a} \sqrt [4]{a^2-b^2}}\right ) \left (\sqrt {1-\sin ^2(c+d x)} a+b\right ) \cos (c+d x)}{(b+a \cos (c+d x)) \sqrt {1-\sin ^2(c+d x)}}\right ) \sin ^{\frac {3}{2}}(c+d x)}{(a-b) (a+b) d (a+b \sec (c+d x)) (e \sin (c+d x))^{3/2}}-\frac {2 (b-a \cos (c+d x)) (b+a \cos (c+d x)) \tan (c+d x)}{\left (b^2-a^2\right ) d (a+b \sec (c+d x)) (e \sin (c+d x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*Sec[c + d*x])*(e*Sin[c + d*x])^(3/2)),x]

[Out]

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

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fricas [F]  time = 175.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {e \sin \left (d x + c\right )}}{a e^{2} \cos \left (d x + c\right )^{2} - a e^{2} + {\left (b e^{2} \cos \left (d x + c\right )^{2} - b e^{2}\right )} \sec \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 7.71, size = 1083, normalized size = 2.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/e*b/(a-b)/(a+b)/(e^2*(a^2-b^2)/a^2)^(1/4)*arctan((e*sin(d*x+c))^(1/2)/(e^2*(a^2-b^2)/a^2)^(1/4))-1/2/d/e*b
/(a-b)/(a+b)/(e^2*(a^2-b^2)/a^2)^(1/4)*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)))+2/d/e*b/(a^2-b^2)/(e*sin(d*x+c))^(1/2)+1/2/d*b^2/e/(a^2-b^2)^(1/2)/(a+(a^2-b^2)
^(1/2))/(-a+(a^2-b^2)^(1/2))/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*EllipticPi((-sin(d*x+c)+1)^(1/2),a/(a+(a^2-b^2)^(
1/2)),1/2*2^(1/2))*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)-1/2/d*b^2/e/(a^2-b^2)/(a+(a^2
-b^2)^(1/2))/(-a+(a^2-b^2)^(1/2))/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)
*sin(d*x+c)^(1/2)*a*EllipticPi((-sin(d*x+c)+1)^(1/2),a/(a+(a^2-b^2)^(1/2)),1/2*2^(1/2))-2/d*b^2/e/(a^2-b^2)/(a
+(a^2-b^2)^(1/2))/(-a+(a^2-b^2)^(1/2))/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(
1/2))*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*a+1/d*b^2/e/(a^2-b^2)/(a+(a^2-b^2)^(1/2))/
(-a+(a^2-b^2)^(1/2))/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*(-sin(d*x+c)
+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*a-1/2/d*b^2/e/(a^2-b^2)^(1/2)/(a+(a^2-b^2)^(1/2))/(-a+(a^2-b
^2)^(1/2))/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellip
ticPi((-sin(d*x+c)+1)^(1/2),-a/(-a+(a^2-b^2)^(1/2)),1/2*2^(1/2))-1/2/d*b^2/e/(a^2-b^2)/(a+(a^2-b^2)^(1/2))/(-a
+(a^2-b^2)^(1/2))/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2
)*a*EllipticPi((-sin(d*x+c)+1)^(1/2),-a/(-a+(a^2-b^2)^(1/2)),1/2*2^(1/2))+2/d*b^2/e/(a^2-b^2)/(a+(a^2-b^2)^(1/
2))/(-a+(a^2-b^2)^(1/2))*cos(d*x+c)/(e*sin(d*x+c))^(1/2)*a

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (c+d\,x\right )}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}} \left (a + b \sec {\left (c + d x \right )}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((e*sin(c + d*x))**(3/2)*(a + b*sec(c + d*x))), x)

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