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

Optimal. Leaf size=307 \[ \frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{d \sqrt{e} \left (b^2-a^2\right )^{3/4}}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{d \sqrt{e} \left (b^2-a^2\right )^{3/4}}+\frac{a \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{a \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \sin (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.588308, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2702, 2807, 2805, 329, 212, 208, 205} \[ \frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{d \sqrt{e} \left (b^2-a^2\right )^{3/4}}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{d \sqrt{e} \left (b^2-a^2\right )^{3/4}}+\frac{a \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{d \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{a \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{d \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(3/4)*d*Sqrt[e]) +
(Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/((-a^2 + b^2)^(3/4)*d*Sqrt[e])
+ (a*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/((a^2 - b*(b - Sqrt[
-a^2 + b^2]))*d*Sqrt[e*Sin[c + d*x]]) + (a*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqr
t[Sin[c + d*x]])/((a^2 - b*(b + Sqrt[-a^2 + b^2]))*d*Sqrt[e*Sin[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 \cos (c+d x)) \sqrt{e \sin (c+d x)}} \, dx &=-\frac{a \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{2 \sqrt{-a^2+b^2}}-\frac{a \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{2 \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+b^2 x^2\right )} \, dx,x,e \sin (c+d x)\right )}{d}\\ &=-\frac{(2 b e) \operatorname{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 )}{d}-\frac{\left (a \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 \sqrt{-a^2+b^2} \sqrt{e \sin (c+d x)}}-\frac{\left (a \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 \sqrt{-a^2+b^2} \sqrt{e \sin (c+d x)}}\\ &=\frac{a \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 \left (b-\sqrt{-a^2+b^2}\right )\right ) d \sqrt{e \sin (c+d x)}}+\frac{a \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 \left (b+\sqrt{-a^2+b^2}\right )\right ) d \sqrt{e \sin (c+d x)}}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e-b 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+b x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{\sqrt{-a^2+b^2} d}\\ &=\frac{\sqrt{b} \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 )^{3/4} d \sqrt{e}}+\frac{\sqrt{b} \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 )^{3/4} d \sqrt{e}}+\frac{a \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 \left (b-\sqrt{-a^2+b^2}\right )\right ) d \sqrt{e \sin (c+d x)}}+\frac{a \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 \left (b+\sqrt{-a^2+b^2}\right )\right ) d \sqrt{e \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.02917, size = 265, normalized size = 0.86 \[ \frac{10 (a+b) \sqrt{e \sin (c+d x)} F_1\left (\frac{1}{4};-\frac{1}{2},1;\frac{5}{4};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (c+d x)\right )}{a+b}\right )}{d e (a+b \cos (c+d x)) \left (2 \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left ((a+b) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (c+d x)\right )}{a+b}\right )-2 (a-b) F_1\left (\frac{5}{4};-\frac{1}{2},2;\frac{9}{4};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (c+d x)\right )}{a+b}\right )\right )+5 (a+b) F_1\left (\frac{1}{4};-\frac{1}{2},1;\frac{5}{4};-\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\frac{(a-b) \tan ^2\left (\frac{1}{2} (c+d x)\right )}{a+b}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(10*(a + b)*AppellF1[1/4, -1/2, 1, 5/4, -Tan[(c + d*x)/2]^2, -(((a - b)*Tan[(c + d*x)/2]^2)/(a + b))]*Sqrt[e*S
in[c + d*x]])/(d*e*(a + b*Cos[c + d*x])*(5*(a + b)*AppellF1[1/4, -1/2, 1, 5/4, -Tan[(c + d*x)/2]^2, -(((a - b)
*Tan[(c + d*x)/2]^2)/(a + b))] + 2*(-2*(a - b)*AppellF1[5/4, -1/2, 2, 9/4, -Tan[(c + d*x)/2]^2, -(((a - b)*Tan
[(c + d*x)/2]^2)/(a + b))] + (a + b)*AppellF1[5/4, 1/2, 1, 9/4, -Tan[(c + d*x)/2]^2, -(((a - b)*Tan[(c + d*x)/
2]^2)/(a + b))])*Tan[(c + d*x)/2]^2))

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Maple [B]  time = 5.006, size = 855, normalized size = 2.8 \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*cos(d*x+c))/(e*sin(d*x+c))^(1/2),x)

[Out]

-1/4/d*b*e*(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*s
in(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/d*b*e*(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/d*b*e*(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/d*a*(1-sin(d*x+c))^(1/2)*(2+2*sin
(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(-b+(-a^2+b^2)^(1/2))/(b+(-a^2+b^2)^(1/2))/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*Ell
ipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))+1/2/d*a*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c
))^(1/2)*sin(d*x+c)^(1/2)/(-b+(-a^2+b^2)^(1/2))/(b+(-a^2+b^2)^(1/2))/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*EllipticP
i((1-sin(d*x+c))^(1/2),1/(b+(-a^2+b^2)^(1/2))*b,1/2*2^(1/2))+1/2/d*a*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/
2)*sin(d*x+c)^(1/2)/(-a^2+b^2)^(1/2)/(-b+(-a^2+b^2)^(1/2))/(b+(-a^2+b^2)^(1/2))/cos(d*x+c)/(e*sin(d*x+c))^(1/2
)*EllipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^2+b^2)^(1/2)),1/2*2^(1/2))*b-1/2/d*a*(1-sin(d*x+c))^(1/2)*(2+2*si
n(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(-a^2+b^2)^(1/2)/(-b+(-a^2+b^2)^(1/2))/(b+(-a^2+b^2)^(1/2))/cos(d*x+c)/(e*sin
(d*x+c))^(1/2)*EllipticPi((1-sin(d*x+c))^(1/2),1/(b+(-a^2+b^2)^(1/2))*b,1/2*2^(1/2))*b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \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*cos(d*x+c))/(e*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*cos(d*x + c) + a)*sqrt(e*sin(d*x + c))), 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(1/(a+b*cos(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 \cos{\left (c + d x \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \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*cos(d*x+c))/(e*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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