3.599 \(\int \frac{(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=204 \[ -\frac{6 a \left (a^2+2 b^2\right ) \tan (e+f x)}{5 d^2 f \sqrt{d \sec (e+f x)}}-\frac{2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \tan (e+f x)\right )}{5 d^2 f \sqrt{d \sec (e+f x)}}+\frac{6 a \left (a^2+2 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{5 d^2 f \sqrt{d \sec (e+f x)}}-\frac{2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt{d \sec (e+f x)}} \]

[Out]

(6*a*(a^2 + 2*b^2)*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(1/4))/(5*d^2*f*Sqrt[d*Sec[e + f*x]])
- (6*a*(a^2 + 2*b^2)*Tan[e + f*x])/(5*d^2*f*Sqrt[d*Sec[e + f*x]]) - (2*Cos[e + f*x]^2*(b - a*Tan[e + f*x])*(a
+ b*Tan[e + f*x])^2)/(5*d^2*f*Sqrt[d*Sec[e + f*x]]) - (2*(2*b*(a^2 + 2*b^2) - a*(3*a^2 + 5*b^2)*Tan[e + f*x])
)/(5*d^2*f*Sqrt[d*Sec[e + f*x]])

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Rubi [A]  time = 0.157326, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3512, 739, 778, 227, 196} \[ -\frac{6 a \left (a^2+2 b^2\right ) \tan (e+f x)}{5 d^2 f \sqrt{d \sec (e+f x)}}-\frac{2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \tan (e+f x)\right )}{5 d^2 f \sqrt{d \sec (e+f x)}}+\frac{6 a \left (a^2+2 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{5 d^2 f \sqrt{d \sec (e+f x)}}-\frac{2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(5/2),x]

[Out]

(6*a*(a^2 + 2*b^2)*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(1/4))/(5*d^2*f*Sqrt[d*Sec[e + f*x]])
- (6*a*(a^2 + 2*b^2)*Tan[e + f*x])/(5*d^2*f*Sqrt[d*Sec[e + f*x]]) - (2*Cos[e + f*x]^2*(b - a*Tan[e + f*x])*(a
+ b*Tan[e + f*x])^2)/(5*d^2*f*Sqrt[d*Sec[e + f*x]]) - (2*(2*b*(a^2 + 2*b^2) - a*(3*a^2 + 5*b^2)*Tan[e + f*x])
)/(5*d^2*f*Sqrt[d*Sec[e + f*x]])

Rule 3512

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

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rubi steps

\begin{align*} \int \frac{(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{5/2}} \, dx &=\frac{\sqrt [4]{\sec ^2(e+f x)} \operatorname{Subst}\left (\int \frac{(a+x)^3}{\left (1+\frac{x^2}{b^2}\right )^{9/4}} \, dx,x,b \tan (e+f x)\right )}{b d^2 f \sqrt{d \sec (e+f x)}}\\ &=-\frac{2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt{d \sec (e+f x)}}+\frac{\left (2 b \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a+x) \left (\frac{1}{2} \left (4+\frac{3 a^2}{b^2}\right )-\frac{a x}{2 b^2}\right )}{\left (1+\frac{x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{5 d^2 f \sqrt{d \sec (e+f x)}}\\ &=-\frac{2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt{d \sec (e+f x)}}-\frac{2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \tan (e+f x)\right )}{5 d^2 f \sqrt{d \sec (e+f x)}}-\frac{\left (3 a \left (2+\frac{a^2}{b^2}\right ) b \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{5 d^2 f \sqrt{d \sec (e+f x)}}\\ &=-\frac{6 a \left (a^2+2 b^2\right ) \tan (e+f x)}{5 d^2 f \sqrt{d \sec (e+f x)}}-\frac{2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt{d \sec (e+f x)}}-\frac{2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \tan (e+f x)\right )}{5 d^2 f \sqrt{d \sec (e+f x)}}+\frac{\left (3 a \left (2+\frac{a^2}{b^2}\right ) b \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{5 d^2 f \sqrt{d \sec (e+f x)}}\\ &=\frac{6 a \left (a^2+2 b^2\right ) E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{5 d^2 f \sqrt{d \sec (e+f x)}}-\frac{6 a \left (a^2+2 b^2\right ) \tan (e+f x)}{5 d^2 f \sqrt{d \sec (e+f x)}}-\frac{2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{5 d^2 f \sqrt{d \sec (e+f x)}}-\frac{2 \left (2 b \left (a^2+2 b^2\right )-a \left (3 a^2+5 b^2\right ) \tan (e+f x)\right )}{5 d^2 f \sqrt{d \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.4614, size = 150, normalized size = 0.74 \[ \frac{\sqrt{d \sec (e+f x)} \left (-b \left (9 a^2+17 b^2\right ) \cos (e+f x)+12 a \left (a^2+2 b^2\right ) \sqrt{\cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )-3 a^2 b \cos (3 (e+f x))+a^3 \sin (e+f x)+a^3 \sin (3 (e+f x))-3 a b^2 \sin (e+f x)-3 a b^2 \sin (3 (e+f x))+b^3 \cos (3 (e+f x))\right )}{10 d^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(5/2),x]

[Out]

(Sqrt[d*Sec[e + f*x]]*(-(b*(9*a^2 + 17*b^2)*Cos[e + f*x]) - 3*a^2*b*Cos[3*(e + f*x)] + b^3*Cos[3*(e + f*x)] +
12*a*(a^2 + 2*b^2)*Sqrt[Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2] + a^3*Sin[e + f*x] - 3*a*b^2*Sin[e + f*x] + a^
3*Sin[3*(e + f*x)] - 3*a*b^2*Sin[3*(e + f*x)]))/(10*d^3*f)

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Maple [C]  time = 0.314, size = 1920, normalized size = 9.4 \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))^3/(d*sec(f*x+e))^(5/2),x)

[Out]

1/10/f*(cos(f*x+e)+1)*(cos(f*x+e)-1)^2*(12*cos(f*x+e)^5*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a*b^2+4*cos(f*x+e
)^4*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*b^3+12*cos(f*x+e)^4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a
*b^2+4*cos(f*x+e)^3*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*b^3-36*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+
e)+1)^2)^(1/2)*a*b^2-20*cos(f*x+e)^2*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*b^3-12*cos(f*x+e)^2*(-cos
(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a*b^2-20*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*b^3+24*cos
(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a*b^2+5*cos(f*x+e)*ln(-2*(2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*
cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)*b^3*sin(f*x+e)-
5*cos(f*x+e)*ln(-(2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(cos(f*x+e)+
1)^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)*b^3*sin(f*x+e)-12*I*cos(f*x+e)^2*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)
+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),
I)*a^3+24*I*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(c
os(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a^3-24*I*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(cos(
f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f
*x+e),I)*a^3+24*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(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^2*sin(f*x+e)-24*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(1/(co
s(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a*b^2*sin(f*x+e)
+12*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+
e)^2*sin(f*x+e)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a^3+48*I*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+
e)+1)^2)^(1/2)*(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^2-48*I*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e
)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a*b^2+24*I*cos(f*x+e)^2*sin(f*x+e)*(-cos(f*x+
e)/(cos(f*x+e)+1)^2)^(1/2)*(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^2-24*I*cos(f*x+e)^2*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1
/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a*b^2+12*cos(f*x+e)*(-cos(f*x+e
)/(cos(f*x+e)+1)^2)^(1/2)*a^3-4*cos(f*x+e)^5*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a^3-4*cos(f*x+e)^4*(-cos(f*x
+e)/(cos(f*x+e)+1)^2)^(1/2)*a^3-8*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a^3+4*cos(f*x+e)^2*(-cos(f
*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a^3-12*cos(f*x+e)^4*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a^2*b-12*cos
(f*x+e)^3*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*a^2*b+12*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(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^3*sin(f*x+e)
-12*I*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Elliptic
E(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a^3*sin(f*x+e))/cos(f*x+e)^3/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/sin(f*x+e)^
5/(d/cos(f*x+e))^(5/2)

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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))^3/(d*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{3} \tan \left (f x + e\right )^{3} + 3 \, a b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} b \tan \left (f x + e\right ) + a^{3}\right )} \sqrt{d \sec \left (f x + e\right )}}{d^{3} \sec \left (f x + e\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral((b^3*tan(f*x + e)^3 + 3*a*b^2*tan(f*x + e)^2 + 3*a^2*b*tan(f*x + e) + a^3)*sqrt(d*sec(f*x + e))/(d^3*
sec(f*x + e)^3), 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))**3/(d*sec(f*x+e))**(5/2),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 )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(5/2), x)