3.584 \(\int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=94 \[ \frac {6 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}} \]

[Out]

-2/5*b/f/(d*sec(f*x+e))^(5/2)+2/5*a*sin(f*x+e)/d/f/(d*sec(f*x+e))^(3/2)+6/5*a*(cos(1/2*e+1/2*f*x)^2)^(1/2)/cos
(1/2*e+1/2*f*x)*EllipticE(sin(1/2*e+1/2*f*x),2^(1/2))/d^2/f/cos(f*x+e)^(1/2)/(d*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.07, 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 = {3486, 3769, 3771, 2639} \[ \frac {6 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(5/2),x]

[Out]

(-2*b)/(5*f*(d*Sec[e + f*x])^(5/2)) + (6*a*EllipticE[(e + f*x)/2, 2])/(5*d^2*f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e
+ f*x]]) + (2*a*Sin[e + f*x])/(5*d*f*(d*Sec[e + f*x])^(3/2))

Rule 2639

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

Rule 3486

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 3769

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 3771

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))^{5/2}} \, dx &=-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}+a \int \frac {1}{(d \sec (e+f x))^{5/2}} \, dx\\ &=-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}+\frac {2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}+\frac {(3 a) \int \frac {1}{\sqrt {d \sec (e+f x)}} \, dx}{5 d^2}\\ &=-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}+\frac {2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}+\frac {(3 a) \int \sqrt {\cos (e+f x)} \, dx}{5 d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}\\ &=-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}+\frac {6 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 74, normalized size = 0.79 \[ \frac {2 \sqrt {d \sec (e+f x)} \left (\cos ^2(e+f x) (a \sin (e+f x)-b \cos (e+f x))+3 a \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )\right )}{5 d^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(5/2),x]

[Out]

(2*Sqrt[d*Sec[e + f*x]]*(3*a*Sqrt[Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2] + Cos[e + f*x]^2*(-(b*Cos[e + f*x])
+ a*Sin[e + f*x])))/(5*d^3*f)

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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \sec \left (f x + e\right )} {\left (b \tan \left (f x + e\right ) + a\right )}}{d^{3} \sec \left (f x + e\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(5/2), x)

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maple [C]  time = 0.81, size = 345, normalized size = 3.67 \[ -\frac {2 \left (-3 i \cos \left (f x +e \right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a \sin \left (f x +e \right )+3 i \cos \left (f x +e \right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a \sin \left (f x +e \right )-3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a +3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a +a \left (\cos ^{4}\left (f x +e \right )\right )+\left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) b +2 \left (\cos ^{2}\left (f x +e \right )\right ) a -3 a \cos \left (f x +e \right )\right )}{5 f \cos \left (f x +e \right )^{3} \sin \left (f x +e \right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x)

[Out]

-2/5/f*(-3*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)*sin(f*x+e)*EllipticF(I*(-1+
cos(f*x+e))/sin(f*x+e),I)*a+3*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*cos(f*x+e)*sin(f*x+
e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a-3*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*
sin(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a+3*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e))
)^(1/2)*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a+a*cos(f*x+e)^4+cos(f*x+e)^3*sin(f*x+e)*b+2*cos(
f*x+e)^2*a-3*a*cos(f*x+e))/cos(f*x+e)^3/sin(f*x+e)/(d/cos(f*x+e))^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {tan}\left (e+f\,x\right )}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(5/2),x)

[Out]

int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \tan {\left (e + f x \right )}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))**(5/2),x)

[Out]

Integral((a + b*tan(e + f*x))/(d*sec(e + f*x))**(5/2), x)

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