3.21 \(\int \frac{\sqrt{e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=204 \[ \frac{2 \sqrt{2} \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)} \Pi \left (\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{\cos (c+d x)+1}}\right )\right |-1\right )}{d \sqrt{b-a} \sqrt{a+b} \sqrt{\sin (c+d x)}}-\frac{2 \sqrt{2} \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)} \Pi \left (-\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{\cos (c+d x)+1}}\right )\right |-1\right )}{d \sqrt{b-a} \sqrt{a+b} \sqrt{\sin (c+d x)}} \]

[Out]

(-2*Sqrt[2]*Sqrt[Cos[c + d*x]]*EllipticPi[-(Sqrt[-a + b]/Sqrt[a + b]), ArcSin[Sqrt[Sin[c + d*x]]/Sqrt[1 + Cos[
c + d*x]]], -1]*Sqrt[e*Tan[c + d*x]])/(Sqrt[-a + b]*Sqrt[a + b]*d*Sqrt[Sin[c + d*x]]) + (2*Sqrt[2]*Sqrt[Cos[c
+ d*x]]*EllipticPi[Sqrt[-a + b]/Sqrt[a + b], ArcSin[Sqrt[Sin[c + d*x]]/Sqrt[1 + Cos[c + d*x]]], -1]*Sqrt[e*Tan
[c + d*x]])/(Sqrt[-a + b]*Sqrt[a + b]*d*Sqrt[Sin[c + d*x]])

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Rubi [A]  time = 0.583039, antiderivative size = 204, 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 = {2733, 2730, 2906, 2905, 490, 1213, 537} \[ \frac{2 \sqrt{2} \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)} \Pi \left (\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{\cos (c+d x)+1}}\right )\right |-1\right )}{d \sqrt{b-a} \sqrt{a+b} \sqrt{\sin (c+d x)}}-\frac{2 \sqrt{2} \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)} \Pi \left (-\frac{\sqrt{b-a}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{\cos (c+d x)+1}}\right )\right |-1\right )}{d \sqrt{b-a} \sqrt{a+b} \sqrt{\sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*Sqrt[Cos[c + d*x]]*EllipticPi[-(Sqrt[-a + b]/Sqrt[a + b]), ArcSin[Sqrt[Sin[c + d*x]]/Sqrt[1 + Cos[
c + d*x]]], -1]*Sqrt[e*Tan[c + d*x]])/(Sqrt[-a + b]*Sqrt[a + b]*d*Sqrt[Sin[c + d*x]]) + (2*Sqrt[2]*Sqrt[Cos[c
+ d*x]]*EllipticPi[Sqrt[-a + b]/Sqrt[a + b], ArcSin[Sqrt[Sin[c + d*x]]/Sqrt[1 + Cos[c + d*x]]], -1]*Sqrt[e*Tan
[c + d*x]])/(Sqrt[-a + b]*Sqrt[a + b]*d*Sqrt[Sin[c + d*x]])

Rule 2733

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[g^(2*
IntPart[p])*(g*Cot[e + f*x])^FracPart[p]*(g*Tan[e + f*x])^FracPart[p], Int[(a + b*Sin[e + f*x])^m/(g*Tan[e + f
*x])^p, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]

Rule 2730

Int[1/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(g_)*tan[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[
e + f*x]]/(Sqrt[Cos[e + f*x]]*Sqrt[g*Tan[e + f*x]]), Int[Sqrt[Cos[e + f*x]]/(Sqrt[Sin[e + f*x]]*(a + b*Sin[e +
f*x])), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2906

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[(d_)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x
_)])), x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]], Int[Sqrt[g*Cos[e + f*x]]/(Sqrt[Sin[e + f*x]]
*(a + b*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2905

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]))
, x_Symbol] :> Dist[(-4*Sqrt[2]*g)/f, Subst[Int[x^2/(((a + b)*g^2 + (a - b)*x^4)*Sqrt[1 - x^4/g^2]), x], x, Sq
rt[g*Cos[e + f*x]]/Sqrt[1 + Sin[e + f*x]]], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(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)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1213

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rubi steps

\begin{align*} \int \frac{\sqrt{e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx &=\left (\sqrt{e \cot (c+d x)} \sqrt{e \tan (c+d x)}\right ) \int \frac{1}{(a+b \cos (c+d x)) \sqrt{e \cot (c+d x)}} \, dx\\ &=\frac{\left (\sqrt{-\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \int \frac{\sqrt{\sin (c+d x)}}{\sqrt{-\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{\sqrt{\sin (c+d x)}}\\ &=\frac{\left (\sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \int \frac{\sqrt{\sin (c+d x)}}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{\sqrt{\sin (c+d x)}}\\ &=\frac{\left (4 \sqrt{2} \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^4} \left (a+b+(a-b) x^4\right )} \, dx,x,\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )}{d \sqrt{\sin (c+d x)}}\\ &=\frac{\left (2 \sqrt{2} \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{a+b}-\sqrt{-a+b} x^2\right ) \sqrt{1-x^4}} \, dx,x,\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )}{\sqrt{-a+b} d \sqrt{\sin (c+d x)}}-\frac{\left (2 \sqrt{2} \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{a+b}+\sqrt{-a+b} x^2\right ) \sqrt{1-x^4}} \, dx,x,\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )}{\sqrt{-a+b} d \sqrt{\sin (c+d x)}}\\ &=\frac{\left (2 \sqrt{2} \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+x^2} \left (\sqrt{a+b}-\sqrt{-a+b} x^2\right )} \, dx,x,\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )}{\sqrt{-a+b} d \sqrt{\sin (c+d x)}}-\frac{\left (2 \sqrt{2} \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+x^2} \left (\sqrt{a+b}+\sqrt{-a+b} x^2\right )} \, dx,x,\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )}{\sqrt{-a+b} d \sqrt{\sin (c+d x)}}\\ &=-\frac{2 \sqrt{2} \sqrt{\cos (c+d x)} \Pi \left (-\frac{\sqrt{-a+b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )\right |-1\right ) \sqrt{e \tan (c+d x)}}{\sqrt{-a+b} \sqrt{a+b} d \sqrt{\sin (c+d x)}}+\frac{2 \sqrt{2} \sqrt{\cos (c+d x)} \Pi \left (\frac{\sqrt{-a+b}}{\sqrt{a+b}};\left .\sin ^{-1}\left (\frac{\sqrt{\sin (c+d x)}}{\sqrt{1+\cos (c+d x)}}\right )\right |-1\right ) \sqrt{e \tan (c+d x)}}{\sqrt{-a+b} \sqrt{a+b} d \sqrt{\sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 3.02794, size = 363, normalized size = 1.78 \[ \frac{2 \sqrt{e \tan (c+d x)} \left (a \sqrt{\sec ^2(c+d x)}+b\right ) \left (\frac{b \tan ^{\frac{3}{2}}(c+d x) F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};-\tan ^2(c+d x),-\frac{a^2 \tan ^2(c+d x)}{a^2-b^2}\right )}{3 \left (b^2-a^2\right )}+\frac{-2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )+\log \left (-\sqrt{2} \sqrt{a} \sqrt [4]{a^2-b^2} \sqrt{\tan (c+d x)}+\sqrt{a^2-b^2}+a \tan (c+d x)\right )-\log \left (\sqrt{2} \sqrt{a} \sqrt [4]{a^2-b^2} \sqrt{\tan (c+d x)}+\sqrt{a^2-b^2}+a \tan (c+d x)\right )}{4 \sqrt{2} \sqrt{a} \sqrt [4]{a^2-b^2}}\right )}{d \sqrt{\tan (c+d x)} \sqrt{\sec ^2(c+d x)} (a+b \cos (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(b + a*Sqrt[Sec[c + d*x]^2])*Sqrt[e*Tan[c + d*x]]*((-2*ArcTan[1 - (Sqrt[2]*Sqrt[a]*Sqrt[Tan[c + d*x]])/(a^2
- b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[a]*Sqrt[Tan[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[Tan[c + d*x]] + a*Tan[c + d*x]] - Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[a]
*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + a*Tan[c + d*x]])/(4*Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)) + (b*AppellF1[3
/4, 1/2, 1, 7/4, -Tan[c + d*x]^2, -((a^2*Tan[c + d*x]^2)/(a^2 - b^2))]*Tan[c + d*x]^(3/2))/(3*(-a^2 + b^2))))/
(d*(a + b*Cos[c + d*x])*Sqrt[Sec[c + d*x]^2]*Sqrt[Tan[c + d*x]])

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Maple [B]  time = 0.585, size = 546, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(e*tan(d*x + c))/(b*cos(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*tan(d*x+c))^(1/2)/(a+b*cos(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 \tan{\left (c + d x \right )}}}{a + b \cos{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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