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

Optimal. Leaf size=178 \[ -\frac{2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt{d \sec (e+f x)}}-\frac{2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt{d \sec (e+f x)}}+\frac{2 a \left (a^2-6 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{f \sqrt{d \sec (e+f x)}}-\frac{2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt{d \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.140579, antiderivative size = 178, 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, 780, 227, 196} \[ -\frac{2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt{d \sec (e+f x)}}-\frac{2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt{d \sec (e+f x)}}+\frac{2 a \left (a^2-6 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{f \sqrt{d \sec (e+f x)}}-\frac{2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 1.89124, size = 130, normalized size = 0.73 \[ \frac{d (a+b \tan (e+f x))^3 \left (6 a \left (a^2-6 b^2\right ) \cos ^{\frac{3}{2}}(e+f x) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )+b \left (\left (3 b^2-9 a^2\right ) \cos (2 (e+f x))-9 a^2+9 a b \sin (2 (e+f x))+5 b^2\right )\right )}{3 f (d \sec (e+f x))^{3/2} (a \cos (e+f x)+b \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.346, size = 5006, normalized size = 28.1 \begin{align*} \text{output 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))^(1/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\sqrt{d \sec \left (f x + e\right )}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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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 \sec \left (f x + e\right )}, 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))^(1/2),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*se
c(f*x + e)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{3}}{\sqrt{d \sec{\left (e + f x \right )}}}\, dx \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))**(1/2),x)

[Out]

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

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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}}{\sqrt{d \sec \left (f x + e\right )}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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