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

Optimal. Leaf size=176 \[ \frac{2 b (d \sec (e+f x))^{3/2} \left (10 \left (9 a^2-2 b^2\right )+33 a b \tan (e+f x)\right )}{105 f}+\frac{2 a \left (5 a^2-6 b^2\right ) \sin (e+f x) \cos (e+f x) (d \sec (e+f x))^{3/2}}{5 f}-\frac{2 a \left (5 a^2-6 b^2\right ) (d \sec (e+f x))^{3/2} E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{5 f \sec ^2(e+f x)^{3/4}}+\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f} \]

[Out]

(-2*a*(5*a^2 - 6*b^2)*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(d*Sec[e + f*x])^(3/2))/(5*f*(Sec[e + f*x]^2)^(3/4)
) + (2*a*(5*a^2 - 6*b^2)*Cos[e + f*x]*(d*Sec[e + f*x])^(3/2)*Sin[e + f*x])/(5*f) + (2*b*(d*Sec[e + f*x])^(3/2)
*(a + b*Tan[e + f*x])^2)/(7*f) + (2*b*(d*Sec[e + f*x])^(3/2)*(10*(9*a^2 - 2*b^2) + 33*a*b*Tan[e + f*x]))/(105*
f)

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Rubi [A]  time = 0.14558, antiderivative size = 176, 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, 743, 780, 227, 196} \[ \frac{2 b (d \sec (e+f x))^{3/2} \left (10 \left (9 a^2-2 b^2\right )+33 a b \tan (e+f x)\right )}{105 f}+\frac{2 a \left (5 a^2-6 b^2\right ) \sin (e+f x) \cos (e+f x) (d \sec (e+f x))^{3/2}}{5 f}-\frac{2 a \left (5 a^2-6 b^2\right ) (d \sec (e+f x))^{3/2} E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{5 f \sec ^2(e+f x)^{3/4}}+\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(5*a^2 - 6*b^2)*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(d*Sec[e + f*x])^(3/2))/(5*f*(Sec[e + f*x]^2)^(3/4)
) + (2*a*(5*a^2 - 6*b^2)*Cos[e + f*x]*(d*Sec[e + f*x])^(3/2)*Sin[e + f*x])/(5*f) + (2*b*(d*Sec[e + f*x])^(3/2)
*(a + b*Tan[e + f*x])^2)/(7*f) + (2*b*(d*Sec[e + f*x])^(3/2)*(10*(9*a^2 - 2*b^2) + 33*a*b*Tan[e + f*x]))/(105*
f)

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 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && 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 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 (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \, dx &=\frac{(d \sec (e+f x))^{3/2} \operatorname{Subst}\left (\int \frac{(a+x)^3}{\sqrt [4]{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{3/4}}\\ &=\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f}+\frac{\left (2 b (d \sec (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{(a+x) \left (\frac{1}{2} \left (-4+\frac{7 a^2}{b^2}\right )+\frac{11 a x}{2 b^2}\right )}{\sqrt [4]{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{7 f \sec ^2(e+f x)^{3/4}}\\ &=\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f}+\frac{2 b (d \sec (e+f x))^{3/2} \left (10 \left (9 a^2-2 b^2\right )+33 a b \tan (e+f x)\right )}{105 f}-\frac{\left (a \left (6-\frac{5 a^2}{b^2}\right ) b (d \sec (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{5 f \sec ^2(e+f x)^{3/4}}\\ &=\frac{2 a \left (5 a^2-6 b^2\right ) \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{5 f}+\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f}+\frac{2 b (d \sec (e+f x))^{3/2} \left (10 \left (9 a^2-2 b^2\right )+33 a b \tan (e+f x)\right )}{105 f}+\frac{\left (a \left (6-\frac{5 a^2}{b^2}\right ) b (d \sec (e+f x))^{3/2}\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 f \sec ^2(e+f x)^{3/4}}\\ &=-\frac{2 a \left (5 a^2-6 b^2\right ) E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{5 f \sec ^2(e+f x)^{3/4}}+\frac{2 a \left (5 a^2-6 b^2\right ) \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{5 f}+\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f}+\frac{2 b (d \sec (e+f x))^{3/2} \left (10 \left (9 a^2-2 b^2\right )+33 a b \tan (e+f x)\right )}{105 f}\\ \end{align*}

Mathematica [A]  time = 1.78392, size = 155, normalized size = 0.88 \[ -\frac{d \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^3 \left (70 b \left (b^2-3 a^2\right ) \cos ^2(e+f x)-42 a \left (5 a^2-6 b^2\right ) \sin (e+f x) \cos ^3(e+f x)+42 a \left (5 a^2-6 b^2\right ) \cos ^{\frac{7}{2}}(e+f x) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )-3 b^2 (21 a \sin (2 (e+f x))+10 b)\right )}{105 f (a \cos (e+f x)+b \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*Sqrt[d*Sec[e + f*x]]*(70*b*(-3*a^2 + b^2)*Cos[e + f*x]^2 + 42*a*(5*a^2 - 6*b^2)*Cos[e + f*x]^(7/2)*Ellipti
cE[(e + f*x)/2, 2] - 42*a*(5*a^2 - 6*b^2)*Cos[e + f*x]^3*Sin[e + f*x] - 3*b^2*(10*b + 21*a*Sin[2*(e + f*x)]))*
(a + b*Tan[e + f*x])^3)/(105*f*(a*Cos[e + f*x] + b*Sin[e + f*x])^3)

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Maple [C]  time = 0.34, size = 759, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105/f*(cos(f*x+e)+1)^2*(cos(f*x+e)-1)^2*(-105*I*cos(f*x+e)^4*sin(f*x+e)*(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^3-105*I*cos(f*x+e)^3*sin(f*x+e)*(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+126*I*cos(f*x+e)^4
*sin(f*x+e)*(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^2-126*I*cos(f*x+e)^3*sin(f*x+e)*(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^2-126*I*cos(f*x+e)^4*sin(f*x+e)*(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+105*I*cos(f*x+e)^4*sin(f*x+e)*(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+126*I*cos(f*x+e)^3*sin(f*x
+e)*(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+
105*I*cos(f*x+e)^3*sin(f*x+e)*(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^3-105*cos(f*x+e)^4*a^3+126*cos(f*x+e)^4*a*b^2+105*a^3*cos(f*x+e)^3-189*a*b^2*cos(f*x+e)^
3+105*a^2*cos(f*x+e)^2*b*sin(f*x+e)-35*cos(f*x+e)^2*sin(f*x+e)*b^3+63*a*cos(f*x+e)*b^2+15*sin(f*x+e)*b^3)*(d/c
os(f*x+e))^(3/2)/cos(f*x+e)^2/sin(f*x+e)^5

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^3*d*sec(f*x + e)*tan(f*x + e)^3 + 3*a*b^2*d*sec(f*x + e)*tan(f*x + e)^2 + 3*a^2*b*d*sec(f*x + e)*t
an(f*x + e) + a^3*d*sec(f*x + e))*sqrt(d*sec(f*x + e)), 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((d*sec(f*x+e))**(3/2)*(a+b*tan(f*x+e))**3,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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