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

Optimal. Leaf size=146 \[ -\frac {2 b \sec ^2(e+f x) \left (2 \left (a^2-2 b^2\right )+a b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}}+\frac {2 a \left (a^2+6 b^2\right ) \sec ^2(e+f x)^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{3 f (d \sec (e+f x))^{3/2}} \]

[Out]

2/3*a*(a^2+6*b^2)*(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(1/2*arctan(tan(f*x+e)))*EllipticF(sin(1/2*arctan(t
an(f*x+e))),2^(1/2))*(sec(f*x+e)^2)^(3/4)/f/(d*sec(f*x+e))^(3/2)-2/3*(b-a*tan(f*x+e))*(a+b*tan(f*x+e))^2/f/(d*
sec(f*x+e))^(3/2)-2/3*b*sec(f*x+e)^2*(2*a^2-4*b^2+a*b*tan(f*x+e))/f/(d*sec(f*x+e))^(3/2)

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Rubi [A]  time = 0.12, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3512, 739, 780, 231} \[ -\frac {2 b \sec ^2(e+f x) \left (2 \left (a^2-2 b^2\right )+a b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}}+\frac {2 a \left (a^2+6 b^2\right ) \sec ^2(e+f x)^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{3 f (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(a^2 + 6*b^2)*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(3/4))/(3*f*(d*Sec[e + f*x])^(3/2)) -
(2*(b - a*Tan[e + f*x])*(a + b*Tan[e + f*x])^2)/(3*f*(d*Sec[e + f*x])^(3/2)) - (2*b*Sec[e + f*x]^2*(2*(a^2 -
2*b^2) + a*b*Tan[e + f*x]))/(3*f*(d*Sec[e + f*x])^(3/2))

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]

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 780

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

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]

Rubi steps

\begin {align*} \int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{3/2}} \, 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 )^{7/4}} \, dx,x,b \tan (e+f x)\right )}{b f (d \sec (e+f x))^{3/2}}\\ &=-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{3 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 {a^2}{b^2}\right )-\frac {3 a x}{2 b^2}\right )}{\left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}}\\ &=-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (a^2-2 b^2\right )+a b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}}+\frac {\left (a \left (6+\frac {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 )}{3 f (d \sec (e+f x))^{3/2}}\\ &=\frac {2 a \left (a^2+6 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}}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{3 f (d \sec (e+f x))^{3/2}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (a^2-2 b^2\right )+a b \tan (e+f x)\right )}{3 f (d \sec (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 1.32, size = 117, normalized size = 0.80 \[ \frac {\sec ^2(e+f x) \left (a^3 \sin (2 (e+f x))+\left (b^3-3 a^2 b\right ) \cos (2 (e+f x))+2 a \left (a^2+6 b^2\right ) \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )-3 a^2 b-3 a b^2 \sin (2 (e+f x))+7 b^3\right )}{3 f (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^2*(-3*a^2*b + 7*b^3 + (-3*a^2*b + b^3)*Cos[2*(e + f*x)] + 2*a*(a^2 + 6*b^2)*Sqrt[Cos[e + f*x]]*E
llipticF[(e + f*x)/2, 2] + a^3*Sin[2*(e + f*x)] - 3*a*b^2*Sin[2*(e + f*x)]))/(3*f*(d*Sec[e + f*x])^(3/2))

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{3} \tan \left (f x + e\right )^{3} + 3 \, a b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} b \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \sec \left (f x + e\right )}}{d^{2} \sec \left (f x + e\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.00, size = 342, normalized size = 2.34 \[ \frac {\frac {2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) a^{3}}{3}+4 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) a \,b^{2}+\frac {2 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a^{3}}{3}+4 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a \,b^{2}-2 a^{2} \left (\cos ^{2}\left (f x +e \right )\right ) b +\frac {2 b^{3} \left (\cos ^{2}\left (f x +e \right )\right )}{3}+\frac {2 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{3}}{3}-2 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}+2 b^{3}}{f \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \cos \left (f x +e \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(3/2),x)

[Out]

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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