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]

2/21*a*(5*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*arcta
n(tan(f*x+e))),2^(1/2))*(sec(f*x+e)^2)^(3/4)/d^2/f/(d*sec(f*x+e))^(3/2)-2/7*cos(f*x+e)^2*(b-a*tan(f*x+e))*(a+b
*tan(f*x+e))^2/d^2/f/(d*sec(f*x+e))^(3/2)-2/21*(2*b*(3*a^2+2*b^2)-a*(5*a^2+3*b^2)*tan(f*x+e))/d^2/f/(d*sec(f*x
+e))^(3/2)

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Rubi [A]  time = 0.14, antiderivative size = 170, 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, 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 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 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 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))^{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*}

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Mathematica [A]  time = 2.60, size = 150, normalized size = 0.88 \[ \frac {\sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \left (4 \left (5 a^3+6 a b^2\right ) F\left (\left .\frac {1}{2} (e+f x)\right |2\right )+\sqrt {\cos (e+f x)} \left (\left (3 b^3-9 a^2 b\right ) \cos (3 (e+f x))-b \left (27 a^2+19 b^2\right ) \cos (e+f x)+2 a \sin (e+f x) \left (3 \left (a^2-3 b^2\right ) \cos (2 (e+f x))+13 a^2+3 b^2\right )\right )\right )}{42 d^4 f} \]

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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fricas [F]  time = 2.38, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{3} \tan \left (f x + e\right )^{3} + 3 \, a b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} b \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \sec \left (f x + e\right )}}{d^{4} \sec \left (f x + e\right )^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.98, size = 391, normalized size = 2.30 \[ -\frac {2 \left (-5 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) a^{3}-6 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) a \,b^{2}+9 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2} b -3 \left (\cos ^{4}\left (f x +e \right )\right ) b^{3}-3 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a^{3}+9 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a \,b^{2}-5 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a^{3}-6 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) a \,b^{2}+7 b^{3} \left (\cos ^{2}\left (f x +e \right )\right )-5 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{3}-6 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}\right )}{21 f \cos \left (f x +e \right )^{4} \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21/f*(-5*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*cos(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x
+e)/(1+cos(f*x+e)))^(1/2)*a*b^2+9*cos(f*x+e)^4*a^2*b-3*cos(f*x+e)^4*b^3-3*cos(f*x+e)^3*sin(f*x+e)*a^3+9*cos(f*
x+e)^3*sin(f*x+e)*a*b^2-5*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*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+c
os(f*x+e)))^(1/2)*a*b^2+7*b^3*cos(f*x+e)^2-5*cos(f*x+e)*sin(f*x+e)*a^3-6*cos(f*x+e)*sin(f*x+e)*a*b^2)/cos(f*x+
e)^4/(d/cos(f*x+e))^(7/2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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