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

Optimal. Leaf size=176 \[ \frac {2 b (d \sec (e+f x))^{3/2} \left (10 \left (9 a^2-2 b^2\right )+33 a b \tan (e+f x)\right )}{105 f}+\frac {2 a \left (5 a^2-6 b^2\right ) \sin (e+f x) \cos (e+f x) (d \sec (e+f x))^{3/2}}{5 f}-\frac {2 a \left (5 a^2-6 b^2\right ) (d \sec (e+f x))^{3/2} E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{5 f \sec ^2(e+f x)^{3/4}}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f} \]

[Out]

-2/5*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)))*EllipticE(sin(1/2*arcta
n(tan(f*x+e))),2^(1/2))*(d*sec(f*x+e))^(3/2)/f/(sec(f*x+e)^2)^(3/4)+2/5*a*(5*a^2-6*b^2)*cos(f*x+e)*(d*sec(f*x+
e))^(3/2)*sin(f*x+e)/f+2/7*b*(d*sec(f*x+e))^(3/2)*(a+b*tan(f*x+e))^2/f+2/105*b*(d*sec(f*x+e))^(3/2)*(90*a^2-20
*b^2+33*a*b*tan(f*x+e))/f

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Rubi [A]  time = 0.15, antiderivative size = 176, 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, 227, 196} \[ \frac {2 b (d \sec (e+f x))^{3/2} \left (10 \left (9 a^2-2 b^2\right )+33 a b \tan (e+f x)\right )}{105 f}+\frac {2 a \left (5 a^2-6 b^2\right ) \sin (e+f x) \cos (e+f x) (d \sec (e+f x))^{3/2}}{5 f}-\frac {2 a \left (5 a^2-6 b^2\right ) (d \sec (e+f x))^{3/2} E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{5 f \sec ^2(e+f x)^{3/4}}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*(5*a^2 - 6*b^2)*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(d*Sec[e + f*x])^(3/2))/(5*f*(Sec[e + f*x]^2)^(3/4)
) + (2*a*(5*a^2 - 6*b^2)*Cos[e + f*x]*(d*Sec[e + f*x])^(3/2)*Sin[e + f*x])/(5*f) + (2*b*(d*Sec[e + f*x])^(3/2)
*(a + b*Tan[e + f*x])^2)/(7*f) + (2*b*(d*Sec[e + f*x])^(3/2)*(10*(9*a^2 - 2*b^2) + 33*a*b*Tan[e + f*x]))/(105*
f)

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]

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 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))^{3/2} (a+b \tan (e+f x))^3 \, dx &=\frac {(d \sec (e+f x))^{3/2} \operatorname {Subst}\left (\int \frac {(a+x)^3}{\sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{3/4}}\\ &=\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f}+\frac {\left (2 b (d \sec (e+f x))^{3/2}\right ) \operatorname {Subst}\left (\int \frac {(a+x) \left (\frac {1}{2} \left (-4+\frac {7 a^2}{b^2}\right )+\frac {11 a x}{2 b^2}\right )}{\sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{7 f \sec ^2(e+f x)^{3/4}}\\ &=\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f}+\frac {2 b (d \sec (e+f x))^{3/2} \left (10 \left (9 a^2-2 b^2\right )+33 a b \tan (e+f x)\right )}{105 f}-\frac {\left (a \left (6-\frac {5 a^2}{b^2}\right ) b (d \sec (e+f x))^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{5 f \sec ^2(e+f x)^{3/4}}\\ &=\frac {2 a \left (5 a^2-6 b^2\right ) \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{5 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f}+\frac {2 b (d \sec (e+f x))^{3/2} \left (10 \left (9 a^2-2 b^2\right )+33 a b \tan (e+f x)\right )}{105 f}+\frac {\left (a \left (6-\frac {5 a^2}{b^2}\right ) b (d \sec (e+f x))^{3/2}\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 )}{5 f \sec ^2(e+f x)^{3/4}}\\ &=-\frac {2 a \left (5 a^2-6 b^2\right ) E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{5 f \sec ^2(e+f x)^{3/4}}+\frac {2 a \left (5 a^2-6 b^2\right ) \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{5 f}+\frac {2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2}{7 f}+\frac {2 b (d \sec (e+f x))^{3/2} \left (10 \left (9 a^2-2 b^2\right )+33 a b \tan (e+f x)\right )}{105 f}\\ \end {align*}

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Mathematica [A]  time = 1.79, size = 155, normalized size = 0.88 \[ -\frac {d \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \left (70 b \left (b^2-3 a^2\right ) \cos ^2(e+f x)+42 a \left (5 a^2-6 b^2\right ) \cos ^{\frac {7}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )-42 a \left (5 a^2-6 b^2\right ) \sin (e+f x) \cos ^3(e+f x)-3 b^2 (21 a \sin (2 (e+f x))+10 b)\right )}{105 f (a \cos (e+f x)+b \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/105*(d*Sqrt[d*Sec[e + f*x]]*(70*b*(-3*a^2 + b^2)*Cos[e + f*x]^2 + 42*a*(5*a^2 - 6*b^2)*Cos[e + f*x]^(7/2)*E
llipticE[(e + f*x)/2, 2] - 42*a*(5*a^2 - 6*b^2)*Cos[e + f*x]^3*Sin[e + f*x] - 3*b^2*(10*b + 21*a*Sin[2*(e + f*
x)]))*(a + b*Tan[e + f*x])^3)/(f*(a*Cos[e + f*x] + b*Sin[e + f*x])^3)

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fricas [F]  time = 1.28, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{3} d \sec \left (f x + e\right ) \tan \left (f x + e\right )^{3} + 3 \, a b^{2} d \sec \left (f x + e\right ) \tan \left (f x + e\right )^{2} + 3 \, a^{2} b d \sec \left (f x + e\right ) \tan \left (f x + e\right ) + a^{3} d \sec \left (f x + e\right )\right )} \sqrt {d \sec \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.01, size = 759, normalized size = 4.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105/f*(1+cos(f*x+e))^2*(-1+cos(f*x+e))^2*(-126*I*cos(f*x+e)^3*sin(f*x+e)*(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*b^2+126*I*cos(f*x+e)^4*sin(f*x+e)*(1/(1+c
os(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a*b^2-126*I*cos(
f*x+e)^4*sin(f*x+e)*(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*b^2-105*I*cos(f*x+e)^4*sin(f*x+e)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*Elli
pticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a^3+126*I*cos(f*x+e)^3*sin(f*x+e)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(
1+cos(f*x+e)))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a*b^2-105*I*cos(f*x+e)^3*sin(f*x+e)*(1/(1+cos(f
*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a^3+105*I*cos(f*x+e)
^3*sin(f*x+e)*(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+105*I*cos(f*x+e)^4*sin(f*x+e)*(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+105*cos(f*x+e)^4*a^3-126*cos(f*x+e)^4*a*b^2-105*a^3*cos(f*x+e)^3+189*a*b^2*c
os(f*x+e)^3-105*a^2*cos(f*x+e)^2*b*sin(f*x+e)+35*cos(f*x+e)^2*sin(f*x+e)*b^3-63*a*cos(f*x+e)*b^2-15*sin(f*x+e)
*b^3)*(d/cos(f*x+e))^(3/2)/cos(f*x+e)^2/sin(f*x+e)^5

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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))^(3/2)*(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 \[ \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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