3.237 \(\int \frac{1}{(a+b \sec (c+d x)) \sqrt{e \sin (c+d x)}} \, dx\)

Optimal. Leaf size=370 \[ \frac{2 \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{a d \sqrt{e \sin (c+d x)}}-\frac{b \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{\sqrt{a} d \sqrt{e} \left (a^2-b^2\right )^{3/4}}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{\sqrt{a} d \sqrt{e} \left (a^2-b^2\right )^{3/4}}+\frac{b^2 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a-\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a d \left (-a \sqrt{a^2-b^2}+a^2-b^2\right ) \sqrt{e \sin (c+d x)}}+\frac{b^2 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a+\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a d \left (a \sqrt{a^2-b^2}+a^2-b^2\right ) \sqrt{e \sin (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.780844, antiderivative size = 370, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {3872, 2867, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ -\frac{b \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{\sqrt{a} d \sqrt{e} \left (a^2-b^2\right )^{3/4}}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{\sqrt{a} d \sqrt{e} \left (a^2-b^2\right )^{3/4}}+\frac{b^2 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a-\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a d \left (-a \sqrt{a^2-b^2}+a^2-b^2\right ) \sqrt{e \sin (c+d x)}}+\frac{b^2 \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a+\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a d \left (a \sqrt{a^2-b^2}+a^2-b^2\right ) \sqrt{e \sin (c+d x)}}+\frac{2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a d \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

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

Rule 2702

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 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 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 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 212

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] &&
!GtQ[a/b, 0]

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 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]

Rubi steps

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

Mathematica [C]  time = 5.94347, size = 546, normalized size = 1.48 \[ \frac{2 \sqrt{\sin (c+d x)} \left (a \sqrt{\cos ^2(c+d x)}+b\right ) \left (\frac{b \left (-\log \left (-\sqrt{2} \sqrt{a} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}+a \sin (c+d x)\right )+\log \left (\sqrt{2} \sqrt{a} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}+a \sin (c+d x)\right )-2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{a} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )+2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )\right )}{4 \sqrt{2} \sqrt{a} \left (b^2-a^2\right )^{3/4}}-\frac{5 a \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \sqrt{\cos ^2(c+d x)} F_1\left (\frac{1}{4};-\frac{1}{2},1;\frac{5}{4};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right )}{\left (a^2 \sin ^2(c+d x)-a^2+b^2\right ) \left (2 \sin ^2(c+d x) \left (2 a^2 F_1\left (\frac{5}{4};-\frac{1}{2},2;\frac{9}{4};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right )+\left (b^2-a^2\right ) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right )\right )+5 \left (a^2-b^2\right ) F_1\left (\frac{1}{4};-\frac{1}{2},1;\frac{5}{4};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right )\right )}\right )}{d \sqrt{e \sin (c+d x)} (a \cos (c+d x)+b)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(b + a*Sqrt[Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]]*((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[Sin[c + d*x]] + a*Sin[c + d*x]] + Log[Sqrt[-a^2 + b^2] + Sqrt[2]*S
qrt[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[Cos[c + d*x]^2
]*Sqrt[Sin[c + d*x]])/((-a^2 + b^2 + a^2*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, 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))))/(d*(b + a*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])

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Maple [B]  time = 2.793, size = 937, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*b*e*(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/d*b*e*(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))-1/d/a*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/
2)*(a^2-b^2)/((a^2-b^2)^(1/2)-a)/(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))+1/d*a*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/((a^2-b^2)^(1/2)-a)/(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))-1/2/d/a*(-sin(d
*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/((a^2-b^2)^(1/2)-a)/(a+(a^2-b^2)^(1/2))/cos(d*x+c)/(e*s
in(d*x+c))^(1/2)*EllipticPi((-sin(d*x+c)+1)^(1/2),-a/((a^2-b^2)^(1/2)-a),1/2*2^(1/2))*b^2-1/2/d/a*(-sin(d*x+c)
+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/((a^2-b^2)^(1/2)-a)/(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))*b^2-1/2/d*(-sin(d*x+c)+1)^(1/2
)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(a^2-b^2)^(1/2)/((a^2-b^2)^(1/2)-a)/(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^2-b^2)^(1/2)-a),1/2*2^(1/2))*b^2+1/2/d*(-sin(d*x+c
)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(a^2-b^2)^(1/2)/((a^2-b^2)^(1/2)-a)/(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))*b^2

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{e \sin{\left (c + d x \right )}} \left (a + b \sec{\left (c + d x \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sqrt{e \sin \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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