3.593 \(\int \frac{(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=184 \[ \frac{2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{45 d^3 f (d \sec (e+f x))^{3/2}}+\frac{2 \left (7 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{15 d^4 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}-\frac{10 a b}{63 f (d \sec (e+f x))^{9/2}}-\frac{2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}} \]

[Out]

(-10*a*b)/(63*f*(d*Sec[e + f*x])^(9/2)) + (2*(7*a^2 + 2*b^2)*EllipticE[(e + f*x)/2, 2])/(15*d^4*f*Sqrt[Cos[e +
f*x]]*Sqrt[d*Sec[e + f*x]]) + (2*(7*a^2 + 2*b^2)*Sin[e + f*x])/(63*d*f*(d*Sec[e + f*x])^(7/2)) + (2*(7*a^2 +
2*b^2)*Sin[e + f*x])/(45*d^3*f*(d*Sec[e + f*x])^(3/2)) - (2*b*(a + b*Tan[e + f*x]))/(7*f*(d*Sec[e + f*x])^(9/2
))

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Rubi [A]  time = 0.191764, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3508, 3486, 3769, 3771, 2639} \[ \frac{2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{45 d^3 f (d \sec (e+f x))^{3/2}}+\frac{2 \left (7 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{15 d^4 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}-\frac{10 a b}{63 f (d \sec (e+f x))^{9/2}}-\frac{2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(9/2),x]

[Out]

(-10*a*b)/(63*f*(d*Sec[e + f*x])^(9/2)) + (2*(7*a^2 + 2*b^2)*EllipticE[(e + f*x)/2, 2])/(15*d^4*f*Sqrt[Cos[e +
f*x]]*Sqrt[d*Sec[e + f*x]]) + (2*(7*a^2 + 2*b^2)*Sin[e + f*x])/(63*d*f*(d*Sec[e + f*x])^(7/2)) + (2*(7*a^2 +
2*b^2)*Sin[e + f*x])/(45*d^3*f*(d*Sec[e + f*x])^(3/2)) - (2*b*(a + b*Tan[e + f*x]))/(7*f*(d*Sec[e + f*x])^(9/2
))

Rule 3508

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

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]

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]

Rubi steps

\begin{align*} \int \frac{(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{9/2}} \, dx &=-\frac{2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}-\frac{2}{7} \int \frac{-\frac{7 a^2}{2}-b^2-\frac{5}{2} a b \tan (e+f x)}{(d \sec (e+f x))^{9/2}} \, dx\\ &=-\frac{10 a b}{63 f (d \sec (e+f x))^{9/2}}-\frac{2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}-\frac{1}{7} \left (-7 a^2-2 b^2\right ) \int \frac{1}{(d \sec (e+f x))^{9/2}} \, dx\\ &=-\frac{10 a b}{63 f (d \sec (e+f x))^{9/2}}+\frac{2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}-\frac{2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}+\frac{\left (7 a^2+2 b^2\right ) \int \frac{1}{(d \sec (e+f x))^{5/2}} \, dx}{9 d^2}\\ &=-\frac{10 a b}{63 f (d \sec (e+f x))^{9/2}}+\frac{2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}+\frac{2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{45 d^3 f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}+\frac{\left (7 a^2+2 b^2\right ) \int \frac{1}{\sqrt{d \sec (e+f x)}} \, dx}{15 d^4}\\ &=-\frac{10 a b}{63 f (d \sec (e+f x))^{9/2}}+\frac{2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}+\frac{2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{45 d^3 f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}+\frac{\left (7 a^2+2 b^2\right ) \int \sqrt{\cos (e+f x)} \, dx}{15 d^4 \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}\\ &=-\frac{10 a b}{63 f (d \sec (e+f x))^{9/2}}+\frac{2 \left (7 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{15 d^4 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}+\frac{2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{45 d^3 f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}\\ \end{align*}

Mathematica [A]  time = 2.84275, size = 126, normalized size = 0.68 \[ \frac{4 \cos (e+f x) \left (2 \sin (e+f x) \left (5 \left (a^2-b^2\right ) \cos (2 (e+f x))+19 a^2-b^2\right )-30 a b \cos (e+f x)-10 a b \cos (3 (e+f x))\right )+\frac{48 \left (7 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{\sqrt{\cos (e+f x)}}}{360 d^4 f \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^2/(d*Sec[e + f*x])^(9/2),x]

[Out]

((48*(7*a^2 + 2*b^2)*EllipticE[(e + f*x)/2, 2])/Sqrt[Cos[e + f*x]] + 4*Cos[e + f*x]*(-30*a*b*Cos[e + f*x] - 10
*a*b*Cos[3*(e + f*x)] + 2*(19*a^2 - b^2 + 5*(a^2 - b^2)*Cos[2*(e + f*x)])*Sin[e + f*x]))/(360*d^4*f*Sqrt[d*Sec
[e + f*x]])

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Maple [C]  time = 0.352, size = 697, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(9/2),x)

[Out]

-2/45/f*(5*cos(f*x+e)^6*a^2-5*cos(f*x+e)^6*b^2+10*cos(f*x+e)^5*sin(f*x+e)*a*b-21*I*(cos(f*x+e)/(cos(f*x+e)+1))
^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)*a^2-6*I*(cos(f*x+e)/(cos(f
*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)*b^2+21*I*(cos(f*x
+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*sin(f*x
+e)*a^2+6*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I
)*sin(f*x+e)*b^2-6*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin
(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*b^2+6*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*EllipticE(
I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*b^2+21*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)
+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)*a^2-21*I*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(
cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*a^2+2*cos(f*x+e)^4*a^2+7*c
os(f*x+e)^4*b^2+14*a^2*cos(f*x+e)^2+4*b^2*cos(f*x+e)^2-21*cos(f*x+e)*a^2-6*cos(f*x+e)*b^2)/cos(f*x+e)^5/sin(f*
x+e)/(d/cos(f*x+e))^(9/2)

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Maxima [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((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}\right )} \sqrt{d \sec \left (f x + e\right )}}{d^{5} \sec \left (f x + e\right )^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

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

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Sympy [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((a+b*tan(f*x+e))**2/(d*sec(f*x+e))**(9/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^2/(d*sec(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(9/2), x)