3.3.32 \(\int \sec (e+f x) \sqrt {d \tan (e+f x)} \, dx\) [232]

Optimal. Leaf size=75 \[ -\frac {2 \cos (e+f x) E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {d \tan (e+f x)}}{f \sqrt {\sin (2 e+2 f x)}}+\frac {2 \cos (e+f x) (d \tan (e+f x))^{3/2}}{d f} \]

[Out]

2*cos(f*x+e)*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*EllipticE(cos(e+1/4*Pi+f*x),2^(1/2))*(d*tan(f*x+e))
^(1/2)/f/sin(2*f*x+2*e)^(1/2)+2*cos(f*x+e)*(d*tan(f*x+e))^(3/2)/d/f

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Rubi [A]
time = 0.06, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2693, 2695, 2652, 2719} \begin {gather*} \frac {2 \cos (e+f x) (d \tan (e+f x))^{3/2}}{d f}-\frac {2 \cos (e+f x) E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (e+f x)}}{f \sqrt {\sin (2 e+2 f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*Sqrt[d*Tan[e + f*x]],x]

[Out]

(-2*Cos[e + f*x]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Tan[e + f*x]])/(f*Sqrt[Sin[2*e + 2*f*x]]) + (2*Cos[e + f*
x]*(d*Tan[e + f*x])^(3/2))/(d*f)

Rule 2652

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a*Sin[e +
f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]), Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2693

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^2*(a*Sec[e
+ f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Dist[a^2*((m - 2)/(m + n - 1)), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2695

Int[Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]]/sec[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[Sqrt[Cos[e + f*x]]*(Sqrt[b*
Tan[e + f*x]]/Sqrt[Sin[e + f*x]]), Int[Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]], x], x] /; FreeQ[{b, e, f}, x]

Rule 2719

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

Rubi steps

\begin {align*} \int \sec (e+f x) \sqrt {d \tan (e+f x)} \, dx &=\frac {2 \cos (e+f x) (d \tan (e+f x))^{3/2}}{d f}-2 \int \cos (e+f x) \sqrt {d \tan (e+f x)} \, dx\\ &=\frac {2 \cos (e+f x) (d \tan (e+f x))^{3/2}}{d f}-\frac {\left (2 \sqrt {\cos (e+f x)} \sqrt {d \tan (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \sqrt {\sin (e+f x)} \, dx}{\sqrt {\sin (e+f x)}}\\ &=\frac {2 \cos (e+f x) (d \tan (e+f x))^{3/2}}{d f}-\frac {\left (2 \cos (e+f x) \sqrt {d \tan (e+f x)}\right ) \int \sqrt {\sin (2 e+2 f x)} \, dx}{\sqrt {\sin (2 e+2 f x)}}\\ &=-\frac {2 \cos (e+f x) E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {d \tan (e+f x)}}{f \sqrt {\sin (2 e+2 f x)}}+\frac {2 \cos (e+f x) (d \tan (e+f x))^{3/2}}{d f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.30, size = 61, normalized size = 0.81 \begin {gather*} -\frac {2 \left (-3+2 \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(e+f x)\right ) \sqrt {\sec ^2(e+f x)}\right ) \sin (e+f x) \sqrt {d \tan (e+f x)}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]*Sqrt[d*Tan[e + f*x]],x]

[Out]

(-2*(-3 + 2*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[e + f*x]^2]*Sqrt[Sec[e + f*x]^2])*Sin[e + f*x]*Sqrt[d*Tan[e
+ f*x]])/(3*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(512\) vs. \(2(94)=188\).
time = 0.29, size = 513, normalized size = 6.84

method	result	size
default	\(\frac {\sqrt {\frac {d \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )+1\right )^{2} \left (\cos \left (f x +e \right )-1\right )^{2} \left (2 \sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticE \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )-\sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+2 \sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticE \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )-\sqrt {\frac {\cos \left (f x +e \right )-1}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )-\cos \left (f x +e \right ) \sqrt {2}+\sqrt {2}\right ) \sqrt {2}}{f \sin \left (f x +e \right )^{5}}\)	\(513\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/f*(d*sin(f*x+e)/cos(f*x+e))^(1/2)*(cos(f*x+e)+1)^2*(cos(f*x+e)-1)^2*(2*((cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((-
1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticE((-(-1+cos(
f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*cos(f*x+e)-((cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)
+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f
*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*cos(f*x+e)+2*((cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e
))/sin(f*x+e))^(1/2)*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticE((-(-1+cos(f*x+e)-sin(f*x+e))/sin
(f*x+e))^(1/2),1/2*2^(1/2))-((cos(f*x+e)-1)/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-
(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1
/2))-cos(f*x+e)*2^(1/2)+2^(1/2))/sin(f*x+e)^5*2^(1/2)

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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(sec(f*x+e)*(d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*tan(f*x + e))*sec(f*x + e), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {d \tan {\left (e + f x \right )}} \sec {\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(d*tan(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(d*tan(e + f*x))*sec(e + f*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(sec(f*x+e)*(d*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*tan(f*x + e))*sec(f*x + e), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\cos \left (e+f\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^(1/2)/cos(e + f*x),x)

[Out]

int((d*tan(e + f*x))^(1/2)/cos(e + f*x), x)

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