3.6.83 \(\int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx\) [583]

Optimal. Leaf size=94 \[ -\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}+\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f}+\frac {2 a \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}} \]

[Out]

-2/3*b/f/(d*sec(f*x+e))^(3/2)+2/3*a*sin(f*x+e)/d/f/(d*sec(f*x+e))^(1/2)+2/3*a*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos
(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(d*sec(f*x+e))^(1/2)/d^2/f

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Rubi [A]
time = 0.06, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3567, 3854, 3856, 2720} \begin {gather*} \frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f}+\frac {2 a \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(3/2),x]

[Out]

(-2*b)/(3*f*(d*Sec[e + f*x])^(3/2)) + (2*a*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/
(3*d^2*f) + (2*a*Sin[e + f*x])/(3*d*f*Sqrt[d*Sec[e + f*x]])

Rule 2720

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

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx &=-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}+a \int \frac {1}{(d \sec (e+f x))^{3/2}} \, dx\\ &=-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}+\frac {2 a \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}+\frac {a \int \sqrt {d \sec (e+f x)} \, dx}{3 d^2}\\ &=-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}+\frac {2 a \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}+\frac {\left (a \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{3 d^2}\\ &=-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}+\frac {2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f}+\frac {2 a \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 69, normalized size = 0.73 \begin {gather*} -\frac {\sqrt {d \sec (e+f x)} \left (b+b \cos (2 (e+f x))-2 a \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )-a \sin (2 (e+f x))\right )}{3 d^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(3/2),x]

[Out]

-1/3*(Sqrt[d*Sec[e + f*x]]*(b + b*Cos[2*(e + f*x)] - 2*a*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2] - a*Sin[
2*(e + f*x)]))/(d^2*f)

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Maple [C] Result contains complex when optimal does not.
time = 0.90, size = 172, normalized size = 1.83

method	result	size
default	\(\frac {\frac {2 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) a}{3}+\frac {2 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) a}{3}-\frac {2 b \left (\cos ^{2}\left (f x +e \right )\right )}{3}+\frac {2 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a}{3}}{f \cos \left (f x +e \right )^{2} \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}\)	\(172\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/3/f*(I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*c
os(f*x+e)*a+I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e)
,I)*a-b*cos(f*x+e)^2+cos(f*x+e)*sin(f*x+e)*a)/cos(f*x+e)^2/(d/cos(f*x+e))^(3/2)

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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((a+b*tan(f*x+e))/(d*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 111, normalized size = 1.18 \begin {gather*} \frac {-i \, \sqrt {2} a \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + i \, \sqrt {2} a \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (b \cos \left (f x + e\right )^{2} - a \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, d^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/3*(-I*sqrt(2)*a*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) + I*sqrt(2)*a*sqrt(d)*weie
rstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*(b*cos(f*x + e)^2 - a*cos(f*x + e)*sin(f*x + e))*sqr
t(d/cos(f*x + e)))/(d^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \tan {\left (e + f x \right )}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))**(3/2),x)

[Out]

Integral((a + b*tan(e + f*x))/(d*sec(e + f*x))**(3/2), 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((a+b*tan(f*x+e))/(d*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {tan}\left (e+f\,x\right )}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(3/2),x)

[Out]

int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(3/2), x)

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