3.594 \(\int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx\)

Optimal. Leaf size=198 \[ \frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}+\frac {2 a d^2 \left (7 a^2-6 b^2\right ) \tan (e+f x) \sqrt {d \sec (e+f x)}}{21 f}+\frac {2 a d^2 \left (7 a^2-6 b^2\right ) \sqrt {d \sec (e+f x)} F\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f} \]

[Out]

2/21*a*(7*a^2-6*b^2)*d^2*(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(1/2*arctan(tan(f*x+e)))*EllipticF(sin(1/2*a
rctan(tan(f*x+e))),2^(1/2))*(d*sec(f*x+e))^(1/2)/f/(sec(f*x+e)^2)^(1/4)+2/21*a*(7*a^2-6*b^2)*d^2*(d*sec(f*x+e)
)^(1/2)*tan(f*x+e)/f+2/9*b*d^2*sec(f*x+e)^2*(d*sec(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2/f+2/315*b*d^2*sec(f*x+e)^2
*(d*sec(f*x+e))^(1/2)*(154*a^2-28*b^2+65*a*b*tan(f*x+e))/f

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Rubi [A]  time = 0.15, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3512, 743, 780, 195, 231} \[ \frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}+\frac {2 a d^2 \left (7 a^2-6 b^2\right ) \tan (e+f x) \sqrt {d \sec (e+f x)}}{21 f}+\frac {2 a d^2 \left (7 a^2-6 b^2\right ) \sqrt {d \sec (e+f x)} F\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/2)*(a + b*Tan[e + f*x])^3,x]

[Out]

(2*a*(7*a^2 - 6*b^2)*d^2*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*Sqrt[d*Sec[e + f*x]])/(21*f*(Sec[e + f*x]^2)^(1/
4)) + (2*a*(7*a^2 - 6*b^2)*d^2*Sqrt[d*Sec[e + f*x]]*Tan[e + f*x])/(21*f) + (2*b*d^2*Sec[e + f*x]^2*Sqrt[d*Sec[
e + f*x]]*(a + b*Tan[e + f*x])^2)/(9*f) + (2*b*d^2*Sec[e + f*x]^2*Sqrt[d*Sec[e + f*x]]*(14*(11*a^2 - 2*b^2) +
65*a*b*Tan[e + f*x]))/(315*f)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 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 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 (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))^3 \, dx &=\frac {\left (d^2 \sqrt {d \sec (e+f x)}\right ) \operatorname {Subst}\left (\int (a+x)^3 \sqrt [4]{1+\frac {x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {\left (2 b d^2 \sqrt {d \sec (e+f x)}\right ) \operatorname {Subst}\left (\int (a+x) \left (\frac {1}{2} \left (-4+\frac {9 a^2}{b^2}\right )+\frac {13 a x}{2 b^2}\right ) \sqrt [4]{1+\frac {x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{9 f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}-\frac {\left (a \left (6-\frac {7 a^2}{b^2}\right ) b d^2 \sqrt {d \sec (e+f x)}\right ) \operatorname {Subst}\left (\int \sqrt [4]{1+\frac {x^2}{b^2}} \, dx,x,b \tan (e+f x)\right )}{7 f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt {d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}-\frac {\left (a \left (6-\frac {7 a^2}{b^2}\right ) b d^2 \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 )}{21 f \sqrt [4]{\sec ^2(e+f x)}}\\ &=\frac {2 a \left (7 a^2-6 b^2\right ) d^2 F\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt {d \sec (e+f x)}}{21 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 a \left (7 a^2-6 b^2\right ) d^2 \sqrt {d \sec (e+f x)} \tan (e+f x)}{21 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{9 f}+\frac {2 b d^2 \sec ^2(e+f x) \sqrt {d \sec (e+f x)} \left (14 \left (11 a^2-2 b^2\right )+65 a b \tan (e+f x)\right )}{315 f}\\ \end {align*}

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Mathematica [A]  time = 1.70, size = 157, normalized size = 0.79 \[ -\frac {2 d (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^3 \left (63 b \left (b^2-3 a^2\right ) \cos ^2(e+f x)-15 a \left (7 a^2-6 b^2\right ) \cos ^{\frac {9}{2}}(e+f x) F\left (\left .\frac {1}{2} (e+f x)\right |2\right )-15 a \left (7 a^2-6 b^2\right ) \sin (e+f x) \cos ^3(e+f x)-\frac {5}{2} b^2 (27 a \sin (2 (e+f x))+14 b)\right )}{315 f (a \cos (e+f x)+b \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(5/2)*(a + b*Tan[e + f*x])^3,x]

[Out]

(-2*d*(d*Sec[e + f*x])^(3/2)*(63*b*(-3*a^2 + b^2)*Cos[e + f*x]^2 - 15*a*(7*a^2 - 6*b^2)*Cos[e + f*x]^(9/2)*Ell
ipticF[(e + f*x)/2, 2] - 15*a*(7*a^2 - 6*b^2)*Cos[e + f*x]^3*Sin[e + f*x] - (5*b^2*(14*b + 27*a*Sin[2*(e + f*x
)]))/2)*(a + b*Tan[e + f*x])^3)/(315*f*(a*Cos[e + f*x] + b*Sin[e + f*x])^3)

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{3} d^{2} \sec \left (f x + e\right )^{2} \tan \left (f x + e\right )^{3} + 3 \, a b^{2} d^{2} \sec \left (f x + e\right )^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} b d^{2} \sec \left (f x + e\right )^{2} \tan \left (f x + e\right ) + a^{3} d^{2} \sec \left (f x + e\right )^{2}\right )} \sqrt {d \sec \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))^3,x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/2)*(b*tan(f*x + e) + a)^3, x)

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maple [C]  time = 1.05, size = 414, normalized size = 2.09 \[ \frac {2 \left (1+\cos \left (f x +e \right )\right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2} \left (105 i \left (\cos ^{5}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a^{3}-90 i \left (\cos ^{5}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a \,b^{2}+105 i \left (\cos ^{4}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a^{3}-90 i \left (\cos ^{4}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, a \,b^{2}+105 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a^{3}-90 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a \,b^{2}+189 a^{2} \left (\cos ^{2}\left (f x +e \right )\right ) b -63 b^{3} \left (\cos ^{2}\left (f x +e \right )\right )+135 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}+35 b^{3}\right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}{315 f \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))^3,x)

[Out]

2/315/f*(1+cos(f*x+e))^2*(-1+cos(f*x+e))^2*(105*I*cos(f*x+e)^5*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^3-90*I*cos(f*x+e)^5*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+105*I*cos(f*x+e)^4*EllipticF(I*(-1+co
s(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^3-90*I*cos(f*x+e)^4*Ellip
ticF(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+105*cos(
f*x+e)^3*sin(f*x+e)*a^3-90*cos(f*x+e)^3*sin(f*x+e)*a*b^2+189*a^2*cos(f*x+e)^2*b-63*b^3*cos(f*x+e)^2+135*cos(f*
x+e)*sin(f*x+e)*a*b^2+35*b^3)*(d/cos(f*x+e))^(5/2)/cos(f*x+e)^2/sin(f*x+e)^4

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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((d*sec(f*x+e))^(5/2)*(a+b*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(5/2)*(a + b*tan(e + f*x))^3,x)

[Out]

int((d/cos(e + f*x))^(5/2)*(a + b*tan(e + f*x))^3, 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((d*sec(f*x+e))**(5/2)*(a+b*tan(f*x+e))**3,x)

[Out]

Timed out

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