3.601 \(\int \frac{(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=176 \[ -\frac{2 \cos ^2(e+f x) \left (2 b \left (5 a^2+2 b^2\right )-a \left (7 a^2+b^2\right ) \tan (e+f x)\right )}{45 d^4 f \sqrt{d \sec (e+f x)}}+\frac{2 a \left (7 a^2+6 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{15 d^4 f \sqrt{d \sec (e+f x)}}-\frac{2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{9 d^4 f \sqrt{d \sec (e+f x)}} \]

[Out]

(2*a*(7*a^2 + 6*b^2)*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(1/4))/(15*d^4*f*Sqrt[d*Sec[e + f*x
]]) - (2*Cos[e + f*x]^4*(b - a*Tan[e + f*x])*(a + b*Tan[e + f*x])^2)/(9*d^4*f*Sqrt[d*Sec[e + f*x]]) - (2*Cos[e
+ f*x]^2*(2*b*(5*a^2 + 2*b^2) - a*(7*a^2 + b^2)*Tan[e + f*x]))/(45*d^4*f*Sqrt[d*Sec[e + f*x]])

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Rubi [A]  time = 0.139476, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3512, 739, 778, 196} \[ -\frac{2 \cos ^2(e+f x) \left (2 b \left (5 a^2+2 b^2\right )-a \left (7 a^2+b^2\right ) \tan (e+f x)\right )}{45 d^4 f \sqrt{d \sec (e+f x)}}+\frac{2 a \left (7 a^2+6 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{15 d^4 f \sqrt{d \sec (e+f x)}}-\frac{2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{9 d^4 f \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(9/2),x]

[Out]

(2*a*(7*a^2 + 6*b^2)*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(1/4))/(15*d^4*f*Sqrt[d*Sec[e + f*x
]]) - (2*Cos[e + f*x]^4*(b - a*Tan[e + f*x])*(a + b*Tan[e + f*x])^2)/(9*d^4*f*Sqrt[d*Sec[e + f*x]]) - (2*Cos[e
+ f*x]^2*(2*b*(5*a^2 + 2*b^2) - a*(7*a^2 + b^2)*Tan[e + f*x]))/(45*d^4*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 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))^{9/2}} \, 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 )^{13/4}} \, dx,x,b \tan (e+f x)\right )}{b d^4 f \sqrt{d \sec (e+f x)}}\\ &=-\frac{2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{9 d^4 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{7 a^2}{b^2}\right )+\frac{3 a x}{2 b^2}\right )}{\left (1+\frac{x^2}{b^2}\right )^{9/4}} \, dx,x,b \tan (e+f x)\right )}{9 d^4 f \sqrt{d \sec (e+f x)}}\\ &=-\frac{2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{9 d^4 f \sqrt{d \sec (e+f x)}}-\frac{2 \cos ^2(e+f x) \left (2 b \left (5 a^2+2 b^2\right )-a \left (7 a^2+b^2\right ) \tan (e+f x)\right )}{45 d^4 f \sqrt{d \sec (e+f x)}}+\frac{\left (a \left (6+\frac{7 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 )}{15 d^4 f \sqrt{d \sec (e+f x)}}\\ &=\frac{2 a \left (7 a^2+6 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)}}{15 d^4 f \sqrt{d \sec (e+f x)}}-\frac{2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{9 d^4 f \sqrt{d \sec (e+f x)}}-\frac{2 \cos ^2(e+f x) \left (2 b \left (5 a^2+2 b^2\right )-a \left (7 a^2+b^2\right ) \tan (e+f x)\right )}{45 d^4 f \sqrt{d \sec (e+f x)}}\\ \end{align*}

Mathematica [B]  time = 6.37808, size = 372, normalized size = 2.11 \[ \frac{\sec ^2(e+f x) (a+b \tan (e+f x))^3 \left (\frac{1}{180} a \left (19 a^2-3 b^2\right ) \sin (e+f x)+\frac{1}{360} a \left (43 a^2-21 b^2\right ) \sin (3 (e+f x))+\frac{1}{72} a \left (a^2-3 b^2\right ) \sin (5 (e+f x))-\frac{1}{90} b \left (15 a^2+4 b^2\right ) \cos (e+f x)-\frac{1}{360} b \left (75 a^2+11 b^2\right ) \cos (3 (e+f x))-\frac{1}{72} b \left (3 a^2-b^2\right ) \cos (5 (e+f x))\right )}{f (d \sec (e+f x))^{9/2} (a \cos (e+f x)+b \sin (e+f x))^3}+\frac{\sec ^{\frac{3}{2}}(e+f x) (a+b \tan (e+f x))^3 \left (\frac{2 \left (56 a^3+48 a b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{\sqrt{\cos (e+f x)} \sqrt{\sec (e+f x)}}-\frac{2 \left (15 a^2 b+7 b^3\right ) \sin ^2(e+f x)}{\sqrt{1-\cos ^2(e+f x)} \sqrt{\sec (e+f x)} \sqrt{\cos ^2(e+f x) \left (\sec ^2(e+f x)-1\right )}}\right )}{120 f (d \sec (e+f x))^{9/2} (a \cos (e+f x)+b \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(9/2),x]

[Out]

(Sec[e + f*x]^(3/2)*((2*(56*a^3 + 48*a*b^2)*EllipticE[(e + f*x)/2, 2])/(Sqrt[Cos[e + f*x]]*Sqrt[Sec[e + f*x]])
- (2*(15*a^2*b + 7*b^3)*Sin[e + f*x]^2)/(Sqrt[1 - Cos[e + f*x]^2]*Sqrt[Sec[e + f*x]]*Sqrt[Cos[e + f*x]^2*(-1
+ Sec[e + f*x]^2)]))*(a + b*Tan[e + f*x])^3)/(120*f*(d*Sec[e + f*x])^(9/2)*(a*Cos[e + f*x] + b*Sin[e + f*x])^3
) + (Sec[e + f*x]^2*(-(b*(15*a^2 + 4*b^2)*Cos[e + f*x])/90 - (b*(75*a^2 + 11*b^2)*Cos[3*(e + f*x)])/360 - (b*(
3*a^2 - b^2)*Cos[5*(e + f*x)])/72 + (a*(19*a^2 - 3*b^2)*Sin[e + f*x])/180 + (a*(43*a^2 - 21*b^2)*Sin[3*(e + f*
x)])/360 + (a*(a^2 - 3*b^2)*Sin[5*(e + f*x)])/72)*(a + b*Tan[e + f*x])^3)/(f*(d*Sec[e + f*x])^(9/2)*(a*Cos[e +
f*x] + b*Sin[e + f*x])^3)

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Maple [C]  time = 0.384, size = 745, normalized size = 4.2 \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))^(9/2),x)

[Out]

-2/45/f*(-21*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*sin(f*x+e)*EllipticF(I*(c
os(f*x+e)-1)/sin(f*x+e),I)*a^3+18*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*sin(
f*x+e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a*b^2+18*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1)
)^(1/2)*sin(f*x+e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a*b^2+21*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(c
os(f*x+e)+1))^(1/2)*sin(f*x+e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a^3+5*cos(f*x+e)^6*a^3-15*cos(f*x+e)^6
*a*b^2+15*cos(f*x+e)^5*sin(f*x+e)*a^2*b-5*cos(f*x+e)^5*sin(f*x+e)*b^3-18*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e
)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a*b^2+21*I*(1/(cos(f*x+e)+1))^(1/2
)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*sin(f*x+e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a^3-18*I*(1
/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*sin(f*x+e)*EllipticF(I*(cos(f*x+e)-1)/sin(
f*x+e),I)*a*b^2-21*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)*EllipticF(I*(cos(f*
x+e)-1)/sin(f*x+e),I)*a^3+2*cos(f*x+e)^4*a^3+21*cos(f*x+e)^4*a*b^2+9*cos(f*x+e)^3*sin(f*x+e)*b^3+14*cos(f*x+e)
^2*a^3+12*cos(f*x+e)^2*a*b^2-21*a^3*cos(f*x+e)-18*a*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))^3/(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^{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^{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))^3/(d*sec(f*x+e))^(9/2),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^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))**3/(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 )}^{3}}{\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))^3/(d*sec(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(9/2), x)