3.1.21 \(\int \frac {\sqrt {e \tan (c+d x)}}{a+b \cos (c+d x)} \, dx\) [21]

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

[Out]

-2*EllipticPi(sin(d*x+c)^(1/2)/(1+cos(d*x+c))^(1/2),-(-a+b)^(1/2)/(a+b)^(1/2),I)*2^(1/2)*cos(d*x+c)^(1/2)*(e*t
an(d*x+c))^(1/2)/d/(-a+b)^(1/2)/(a+b)^(1/2)/sin(d*x+c)^(1/2)+2*EllipticPi(sin(d*x+c)^(1/2)/(1+cos(d*x+c))^(1/2
),(-a+b)^(1/2)/(a+b)^(1/2),I)*2^(1/2)*cos(d*x+c)^(1/2)*(e*tan(d*x+c))^(1/2)/d/(-a+b)^(1/2)/(a+b)^(1/2)/sin(d*x
+c)^(1/2)

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Rubi [A]
time = 0.39, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2812, 2809, 2985, 2984, 504, 1227, 551} \begin {gather*} \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 .\text {ArcSin}\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 .\text {ArcSin}\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)}} \end {gather*}

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 504

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), Int[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 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], 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])

Rule 1227

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 2809

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 2812

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 2984

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 2985

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]

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 12.89, size = 363, normalized size = 1.78 \begin {gather*} \frac {2 \left (b+a \sqrt {\sec ^2(c+d x)}\right ) \sqrt {e \tan (c+d x)} \left (\frac {-2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+a \tan (c+d x)\right )-\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+a \tan (c+d x)\right )}{4 \sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2}}+\frac {b 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 ) \tan ^{\frac {3}{2}}(c+d x)}{3 \left (-a^2+b^2\right )}\right )}{d (a+b \cos (c+d x)) \sqrt {\sec ^2(c+d x)} \sqrt {\tan (c+d x)}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(545\) vs. \(2(164)=328\).
time = 0.96, size = 546, normalized size = 2.68

method	result	size
default	\(\frac {\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {e \sin \left (d x +c \right )}{\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (\sqrt {-a^{2}+b^{2}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )-\sqrt {-a^{2}+b^{2}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )-a \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {a -b}{a -b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b -a \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, -\frac {a -b}{-a +b +\sqrt {-\left (a -b \right ) \left (a +b \right )}}, \frac {\sqrt {2}}{2}\right ) b \right ) \sqrt {2}}{d \sin \left (d x +c \right )^{3} \left (\sqrt {-a^{2}+b^{2}}-a +b \right ) \left (\sqrt {-a^{2}+b^{2}}+a -b \right )}\)	\(546\)

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,method=_RETURNVERBOSE)

[Out]

1/d*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(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+cos(d*x+c))^2*(-1+cos(d*x+c))*((-a^2+b^2)^(1/2)*Ellipti
cPi(((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*El
lipticPi(((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*2^(1/2)/((
-a^2+b^2)^(1/2)-a+b)/((-a^2+b^2)^(1/2)+a-b)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {e \tan {\left (c + d x \right )}}}{a + b \cos {\left (c + d x \right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}}{a+b\,\cos \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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