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

Optimal. Leaf size=356 \[ \frac{b \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{3/2} d \sqrt [4]{a^2-b^2}}-\frac{b \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{3/2} d \sqrt [4]{a^2-b^2}}-\frac{b^2 e \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a-\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^2 d \left (a-\sqrt{a^2-b^2}\right ) \sqrt{e \sin (c+d x)}}-\frac{b^2 e \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a+\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^2 d \left (\sqrt{a^2-b^2}+a\right ) \sqrt{e \sin (c+d x)}}+\frac{2 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{a d \sqrt{\sin (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.76152, antiderivative size = 356, 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, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac{b \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{3/2} d \sqrt [4]{a^2-b^2}}-\frac{b \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{a^2-b^2}}\right )}{a^{3/2} d \sqrt [4]{a^2-b^2}}-\frac{b^2 e \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a-\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^2 d \left (a-\sqrt{a^2-b^2}\right ) \sqrt{e \sin (c+d x)}}-\frac{b^2 e \sqrt{\sin (c+d x)} \Pi \left (\frac{2 a}{a+\sqrt{a^2-b^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{a^2 d \left (\sqrt{a^2-b^2}+a\right ) \sqrt{e \sin (c+d x)}}+\frac{2 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{a d \sqrt{\sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [C]  time = 20.0788, size = 351, normalized size = 0.99 \[ \frac{\sqrt{e \sin (c+d x)} \left (a \sqrt{\cos ^2(c+d x)}+b\right ) \left (8 a^{5/2} \sin ^{\frac{3}{2}}(c+d x) F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{a^2 \sin ^2(c+d x)}{a^2-b^2}\right )+3 \sqrt{2} b \left (b^2-a^2\right )^{3/4} \left (-\log \left (-\sqrt{2} \sqrt{a} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}+a \sin (c+d x)\right )+\log \left (\sqrt{2} \sqrt{a} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}+a \sin (c+d x)\right )+2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{a} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )\right )\right )}{12 a^{3/2} d \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} (a \cos (c+d x)+b)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Sqrt[Cos[c + d*x]^2])*Sqrt[e*Sin[c + d*x]]*(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[S
qrt[-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)*AppellF
1[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)))/(12*a^(3/2)*(a^2 -
b^2)*d*(b + a*Cos[c + d*x])*Sqrt[Sin[c + d*x]])

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Maple [B]  time = 2.525, size = 919, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(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{\sqrt{e \sin \left (d x + c\right )}}{b \sec \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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