3.600 \(\int \frac{(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=170 \[ -\frac{2 \left (2 b \left (3 a^2+2 b^2\right )-a \left (5 a^2+3 b^2\right ) \tan (e+f x)\right )}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac{2 a \left (5 a^2+6 b^2\right ) \sec ^2(e+f x)^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{21 d^2 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{7 d^2 f (d \sec (e+f x))^{3/2}} \]

[Out]

(2*a*(5*a^2 + 6*b^2)*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(3/4))/(21*d^2*f*(d*Sec[e + f*x])^(
3/2)) - (2*Cos[e + f*x]^2*(b - a*Tan[e + f*x])*(a + b*Tan[e + f*x])^2)/(7*d^2*f*(d*Sec[e + f*x])^(3/2)) - (2*(
2*b*(3*a^2 + 2*b^2) - a*(5*a^2 + 3*b^2)*Tan[e + f*x]))/(21*d^2*f*(d*Sec[e + f*x])^(3/2))

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Rubi [A]  time = 0.142954, antiderivative size = 170, 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, 739, 778, 231} \[ -\frac{2 \left (2 b \left (3 a^2+2 b^2\right )-a \left (5 a^2+3 b^2\right ) \tan (e+f x)\right )}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac{2 a \left (5 a^2+6 b^2\right ) \sec ^2(e+f x)^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{21 d^2 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{7 d^2 f (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(7/2),x]

[Out]

(2*a*(5*a^2 + 6*b^2)*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(3/4))/(21*d^2*f*(d*Sec[e + f*x])^(
3/2)) - (2*Cos[e + f*x]^2*(b - a*Tan[e + f*x])*(a + b*Tan[e + f*x])^2)/(7*d^2*f*(d*Sec[e + f*x])^(3/2)) - (2*(
2*b*(3*a^2 + 2*b^2) - a*(5*a^2 + 3*b^2)*Tan[e + f*x]))/(21*d^2*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 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))^{7/2}} \, 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 )^{11/4}} \, dx,x,b \tan (e+f x)\right )}{b d^2 f (d \sec (e+f x))^{3/2}}\\ &=-\frac{2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{7 d^2 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{5 a^2}{b^2}\right )+\frac{a x}{2 b^2}\right )}{\left (1+\frac{x^2}{b^2}\right )^{7/4}} \, dx,x,b \tan (e+f x)\right )}{7 d^2 f (d \sec (e+f x))^{3/2}}\\ &=-\frac{2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{7 d^2 f (d \sec (e+f x))^{3/2}}-\frac{2 \left (2 b \left (3 a^2+2 b^2\right )-a \left (5 a^2+3 b^2\right ) \tan (e+f x)\right )}{21 d^2 f (d \sec (e+f x))^{3/2}}+\frac{\left (a \left (6+\frac{5 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 )}{21 d^2 f (d \sec (e+f x))^{3/2}}\\ &=\frac{2 a \left (5 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}}{21 d^2 f (d \sec (e+f x))^{3/2}}-\frac{2 \cos ^2(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{7 d^2 f (d \sec (e+f x))^{3/2}}-\frac{2 \left (2 b \left (3 a^2+2 b^2\right )-a \left (5 a^2+3 b^2\right ) \tan (e+f x)\right )}{21 d^2 f (d \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 2.44114, size = 150, normalized size = 0.88 \[ \frac{\sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)} \left (4 \left (5 a^3+6 a b^2\right ) F\left (\left .\frac{1}{2} (e+f x)\right |2\right )+\sqrt{\cos (e+f x)} \left (-b \left (27 a^2+19 b^2\right ) \cos (e+f x)+\left (3 b^3-9 a^2 b\right ) \cos (3 (e+f x))+2 a \sin (e+f x) \left (3 \left (a^2-3 b^2\right ) \cos (2 (e+f x))+13 a^2+3 b^2\right )\right )\right )}{42 d^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(7/2),x]

[Out]

(Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]*(4*(5*a^3 + 6*a*b^2)*EllipticF[(e + f*x)/2, 2] + Sqrt[Cos[e + f*x]]*(
-(b*(27*a^2 + 19*b^2)*Cos[e + f*x]) + (-9*a^2*b + 3*b^3)*Cos[3*(e + f*x)] + 2*a*(13*a^2 + 3*b^2 + 3*(a^2 - 3*b
^2)*Cos[2*(e + f*x)])*Sin[e + f*x])))/(42*d^4*f)

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Maple [C]  time = 0.304, size = 391, normalized size = 2.3 \begin{align*} -{\frac{2}{21\,f \left ( \cos \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 ) \cos \left ( fx+e \right ){a}^{3}-6\,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{b}^{2}+9\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}b-3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{3}-3\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}{a}^{3}+9\,\sin \left ( fx+e \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 ){a}^{3}-6\,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{b}^{2}+7\,{b}^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{2}-5\,\cos \left ( fx+e \right ){a}^{3}\sin \left ( fx+e \right ) -6\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) a{b}^{2} \right ) \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{7}{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))^(7/2),x)

[Out]

-2/21/f*(-5*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)*a^3-6*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*EllipticF(I*(cos(
f*x+e)-1)/sin(f*x+e),I)*a*b^2+9*cos(f*x+e)^4*a^2*b-3*cos(f*x+e)^4*b^3-3*sin(f*x+e)*cos(f*x+e)^3*a^3+9*sin(f*x+
e)*cos(f*x+e)^3*a*b^2-5*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^3-6*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*b^2+7*b^3*cos(f*x+e)^2-5*cos(f*x+e)*a^3*sin(f*x+e)-6*sin(f*x+e)*cos(f*x+e)*a*b^2)/(d/cos(f*x+e
))^(7/2)/cos(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((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(7/2),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^{4} \sec \left (f x + e\right )^{4}}, 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))^(7/2),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^4*
sec(f*x + e)^4), 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))**(7/2),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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(7/2), x)