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

Optimal. Leaf size=129 \[ \frac{2 b \sqrt{d \sec (e+f x)} \left (2 \left (7 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{5 f}+\frac{2 a \left (a^2-2 b^2\right ) \sqrt{d \sec (e+f x)} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{f \sqrt [4]{\sec ^2(e+f x)}}+\frac{2 b \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f} \]

[Out]

(2*a*(a^2 - 2*b^2)*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*Sqrt[d*Sec[e + f*x]])/(f*(Sec[e + f*x]^2)^(1/4)) + (2*
b*Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])^2)/(5*f) + (2*b*Sqrt[d*Sec[e + f*x]]*(2*(7*a^2 - 2*b^2) + 3*a*b*Ta
n[e + f*x]))/(5*f)

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Rubi [A]  time = 0.112089, antiderivative size = 129, 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, 743, 780, 231} \[ \frac{2 b \sqrt{d \sec (e+f x)} \left (2 \left (7 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{5 f}+\frac{2 a \left (a^2-2 b^2\right ) \sqrt{d \sec (e+f x)} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{f \sqrt [4]{\sec ^2(e+f x)}}+\frac{2 b \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(a^2 - 2*b^2)*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*Sqrt[d*Sec[e + f*x]])/(f*(Sec[e + f*x]^2)^(1/4)) + (2*
b*Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])^2)/(5*f) + (2*b*Sqrt[d*Sec[e + f*x]]*(2*(7*a^2 - 2*b^2) + 3*a*b*Ta
n[e + f*x]))/(5*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 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 \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx &=\frac{\sqrt{d \sec (e+f x)} \operatorname{Subst}\left (\int \frac{(a+x)^3}{\left (1+\frac{x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac{2 b \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f}+\frac{\left (2 b \sqrt{d \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a+x) \left (\frac{1}{2} \left (-4+\frac{5 a^2}{b^2}\right )+\frac{9 a x}{2 b^2}\right )}{\left (1+\frac{x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{5 f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac{2 b \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f}+\frac{2 b \sqrt{d \sec (e+f x)} \left (2 \left (7 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{5 f}-\frac{\left (a \left (2-\frac{a^2}{b^2}\right ) b \sqrt{d \sec (e+f x)}\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 )}{f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac{2 a \left (a^2-2 b^2\right ) F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt{d \sec (e+f x)}}{f \sqrt [4]{\sec ^2(e+f x)}}+\frac{2 b \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f}+\frac{2 b \sqrt{d \sec (e+f x)} \left (2 \left (7 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{5 f}\\ \end{align*}

Mathematica [A]  time = 2.03126, size = 132, normalized size = 1.02 \[ -\frac{2 \sqrt{d \sec (e+f x)} (a+b \tan (e+f x))^3 \left (5 b \left (b^2-3 a^2\right ) \cos ^3(e+f x)-5 a \left (a^2-2 b^2\right ) \cos ^{\frac{7}{2}}(e+f x) F\left (\left .\frac{1}{2} (e+f x)\right |2\right )-\frac{1}{2} b^2 \cos (e+f x) (5 a \sin (2 (e+f x))+2 b)\right )}{5 f (a \cos (e+f x)+b \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[d*Sec[e + f*x]]*(5*b*(-3*a^2 + b^2)*Cos[e + f*x]^3 - 5*a*(a^2 - 2*b^2)*Cos[e + f*x]^(7/2)*EllipticF[(
e + f*x)/2, 2] - (b^2*Cos[e + f*x]*(2*b + 5*a*Sin[2*(e + f*x)]))/2)*(a + b*Tan[e + f*x])^3)/(5*f*(a*Cos[e + f*
x] + b*Sin[e + f*x])^3)

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Maple [C]  time = 0.327, size = 373, normalized size = 2.9 \begin{align*}{\frac{2\, \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2} \left ( \cos \left ( fx+e \right ) -1 \right ) ^{2}}{5\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{4}} \left ( 5\,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 ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}{a}^{3}-10\,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 ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}a{b}^{2}+5\,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 ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{3}-10\,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 ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}a{b}^{2}+15\,{a}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{2}b-5\,{b}^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{2}+5\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) a{b}^{2}+{b}^{3} \right ) \sqrt{{\frac{d}{\cos \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(1/2)*(a+b*tan(f*x+e))^3,x)

[Out]

2/5/f*(cos(f*x+e)+1)^2*(cos(f*x+e)-1)^2*(5*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Ellipt
icF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^3*a^3-10*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)*cos(f*x+e)^3*a*b^2+5*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(co
s(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^2*a^3-10*I*(1/(cos(f*x+e)+1))^(1/2)*(co
s(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)^2*a*b^2+15*a^2*cos(f*x+e)^2
*b-5*b^3*cos(f*x+e)^2+5*sin(f*x+e)*cos(f*x+e)*a*b^2+b^3)*(d/cos(f*x+e))^(1/2)/cos(f*x+e)^2/sin(f*x+e)^4

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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))^(1/2)*(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} \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 )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(1/2)*(a+b*tan(f*x+e))^3,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)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(1/2)*(a+b*tan(f*x+e))**3,x)

[Out]

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

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

[Out]

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