3.602 \(\int \frac{(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=218 \[ \frac{10 a \left (3 a^2+2 b^2\right ) \tan (e+f x)}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^2(e+f x) \left (2 b \left (7 a^2+2 b^2\right )-a \left (9 a^2-b^2\right ) \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac{10 a \left (3 a^2+2 b^2\right ) \sec ^2(e+f x)^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}} \]

[Out]

(10*a*(3*a^2 + 2*b^2)*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(3/4))/(77*d^4*f*(d*Sec[e + f*x])^
(3/2)) + (10*a*(3*a^2 + 2*b^2)*Tan[e + f*x])/(77*d^4*f*(d*Sec[e + f*x])^(3/2)) - (2*Cos[e + f*x]^4*(b - a*Tan[
e + f*x])*(a + b*Tan[e + f*x])^2)/(11*d^4*f*(d*Sec[e + f*x])^(3/2)) - (2*Cos[e + f*x]^2*(2*b*(7*a^2 + 2*b^2) -
a*(9*a^2 - b^2)*Tan[e + f*x]))/(77*d^4*f*(d*Sec[e + f*x])^(3/2))

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Rubi [A]  time = 0.166802, antiderivative size = 218, 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, 199, 231} \[ \frac{10 a \left (3 a^2+2 b^2\right ) \tan (e+f x)}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^2(e+f x) \left (2 b \left (7 a^2+2 b^2\right )-a \left (9 a^2-b^2\right ) \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac{10 a \left (3 a^2+2 b^2\right ) \sec ^2(e+f x)^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(11/2),x]

[Out]

(10*a*(3*a^2 + 2*b^2)*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(3/4))/(77*d^4*f*(d*Sec[e + f*x])^
(3/2)) + (10*a*(3*a^2 + 2*b^2)*Tan[e + f*x])/(77*d^4*f*(d*Sec[e + f*x])^(3/2)) - (2*Cos[e + f*x]^4*(b - a*Tan[
e + f*x])*(a + b*Tan[e + f*x])^2)/(11*d^4*f*(d*Sec[e + f*x])^(3/2)) - (2*Cos[e + f*x]^2*(2*b*(7*a^2 + 2*b^2) -
a*(9*a^2 - b^2)*Tan[e + f*x]))/(77*d^4*f*(d*Sec[e + f*x])^(3/2))

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*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))^{11/2}} \, dx &=\frac{\sec ^2(e+f x)^{3/4} \operatorname{Subst}\left (\int \frac{(a+x)^3}{\left (1+\frac{x^2}{b^2}\right )^{15/4}} \, dx,x,b \tan (e+f x)\right )}{b d^4 f (d \sec (e+f x))^{3/2}}\\ &=-\frac{2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}}+\frac{\left (2 b \sec ^2(e+f x)^{3/4}\right ) \operatorname{Subst}\left (\int \frac{(a+x) \left (\frac{1}{2} \left (4+\frac{9 a^2}{b^2}\right )+\frac{5 a x}{2 b^2}\right )}{\left (1+\frac{x^2}{b^2}\right )^{11/4}} \, dx,x,b \tan (e+f x)\right )}{11 d^4 f (d \sec (e+f x))^{3/2}}\\ &=-\frac{2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^2(e+f x) \left (2 b \left (7 a^2+2 b^2\right )-a \left (9 a^2-b^2\right ) \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac{\left (15 a \left (2+\frac{3 a^2}{b^2}\right ) b \sec ^2(e+f x)^{3/4}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{x^2}{b^2}\right )^{7/4}} \, dx,x,b \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}\\ &=\frac{10 a \left (3 a^2+2 b^2\right ) \tan (e+f x)}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^2(e+f x) \left (2 b \left (7 a^2+2 b^2\right )-a \left (9 a^2-b^2\right ) \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac{\left (5 a \left (2+\frac{3 a^2}{b^2}\right ) b \sec ^2(e+f x)^{3/4}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}\\ &=\frac{10 a \left (3 a^2+2 b^2\right ) F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sec ^2(e+f x)^{3/4}}{77 d^4 f (d \sec (e+f x))^{3/2}}+\frac{10 a \left (3 a^2+2 b^2\right ) \tan (e+f x)}{77 d^4 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{11 d^4 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^2(e+f x) \left (2 b \left (7 a^2+2 b^2\right )-a \left (9 a^2-b^2\right ) \tan (e+f x)\right )}{77 d^4 f (d \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 6.43815, size = 296, normalized size = 1.36 \[ \frac{\sec ^3(e+f x) (a+b \tan (e+f x))^3 \left (\frac{a \left (347 a^2+103 b^2\right ) \sin (2 (e+f x))}{1232}+\frac{1}{308} a \left (16 a^2-15 b^2\right ) \sin (4 (e+f x))+\frac{1}{176} a \left (a^2-3 b^2\right ) \sin (6 (e+f x))-\frac{b \left (315 a^2+71 b^2\right ) \cos (2 (e+f x))}{1232}-\frac{1}{616} b \left (63 a^2+b^2\right ) \cos (4 (e+f x))-\frac{1}{176} b \left (3 a^2-b^2\right ) \cos (6 (e+f x))-\frac{1}{616} b \left (105 a^2+31 b^2\right )\right )}{f (d \sec (e+f x))^{11/2} (a \cos (e+f x)+b \sin (e+f x))^3}+\frac{10 a \left (3 a^2+2 b^2\right ) F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) (a+b \tan (e+f x))^3}{77 f \cos ^{\frac{5}{2}}(e+f x) (d \sec (e+f x))^{11/2} (a \cos (e+f x)+b \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(11/2),x]

[Out]

(10*a*(3*a^2 + 2*b^2)*EllipticF[(e + f*x)/2, 2]*(a + b*Tan[e + f*x])^3)/(77*f*Cos[e + f*x]^(5/2)*(d*Sec[e + f*
x])^(11/2)*(a*Cos[e + f*x] + b*Sin[e + f*x])^3) + (Sec[e + f*x]^3*(-(b*(105*a^2 + 31*b^2))/616 - (b*(315*a^2 +
71*b^2)*Cos[2*(e + f*x)])/1232 - (b*(63*a^2 + b^2)*Cos[4*(e + f*x)])/616 - (b*(3*a^2 - b^2)*Cos[6*(e + f*x)])
/176 + (a*(347*a^2 + 103*b^2)*Sin[2*(e + f*x)])/1232 + (a*(16*a^2 - 15*b^2)*Sin[4*(e + f*x)])/308 + (a*(a^2 -
3*b^2)*Sin[6*(e + f*x)])/176)*(a + b*Tan[e + f*x])^3)/(f*(d*Sec[e + f*x])^(11/2)*(a*Cos[e + f*x] + b*Sin[e + f
*x])^3)

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Maple [C]  time = 0.387, size = 430, normalized size = 2. \begin{align*}{\frac{2}{77\,f \left ( \cos \left ( fx+e \right ) \right ) ^{6}} \left ( -21\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{2}b+7\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{b}^{3}+7\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}\sin \left ( fx+e \right ){a}^{3}-21\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}\sin \left ( fx+e \right ) a{b}^{2}+15\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \cos \left ( fx+e \right ){a}^{3}+10\,i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \cos \left ( fx+e \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}a{b}^{2}+15\,i\sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ){a}^{3}+10\,i{\it EllipticF} \left ({\frac{i \left ( \cos \left ( fx+e \right ) -1 \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}a{b}^{2}-11\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{3}+9\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}{a}^{3}+6\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}a{b}^{2}+15\,\cos \left ( fx+e \right ){a}^{3}\sin \left ( fx+e \right ) +10\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) a{b}^{2} \right ) \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{11}{2}}}} \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))^(11/2),x)

[Out]

2/77/f*(-21*cos(f*x+e)^6*a^2*b+7*cos(f*x+e)^6*b^3+7*cos(f*x+e)^5*sin(f*x+e)*a^3-21*cos(f*x+e)^5*sin(f*x+e)*a*b
^2+15*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)*co
s(f*x+e)*a^3+10*I*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(co
s(f*x+e)+1))^(1/2)*a*b^2+15*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^3+10*I*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(co
s(f*x+e)+1))^(1/2)*a*b^2-11*cos(f*x+e)^4*b^3+9*sin(f*x+e)*cos(f*x+e)^3*a^3+6*sin(f*x+e)*cos(f*x+e)^3*a*b^2+15*
cos(f*x+e)*a^3*sin(f*x+e)+10*sin(f*x+e)*cos(f*x+e)*a*b^2)/cos(f*x+e)^6/(d/cos(f*x+e))^(11/2)

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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))^(11/2),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^{6} \sec \left (f x + e\right )^{6}}, 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))^(11/2),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^6*
sec(f*x + e)^6), 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))**(11/2),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{11}{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))^(11/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(11/2), x)