3.6.96 \(\int \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx\) [596]

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

[Out]

2*a*(a^2-2*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*arctan(tan
(f*x+e))),2^(1/2))*(d*sec(f*x+e))^(1/2)/f/(sec(f*x+e)^2)^(1/4)+2/5*b*(d*sec(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2/f
+2/5*b*(d*sec(f*x+e))^(1/2)*(14*a^2-4*b^2+3*a*b*tan(f*x+e))/f

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Rubi [A]
time = 0.08, antiderivative size = 129, 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 = {3593, 757, 794, 237} \begin {gather*} \frac {2 a \left (a^2-2 b^2\right ) \sqrt {d \sec (e+f x)} F\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right )}{f \sqrt [4]{\sec ^2(e+f x)}}+\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 b \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2}{5 f} \end {gather*}

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 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 757

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 794

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 3593

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 \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^3 \, dx &=\frac {\sqrt {d \sec (e+f x)} \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 2.21, size = 132, normalized size = 1.02 \begin {gather*} -\frac {2 \sqrt {d \sec (e+f x)} \left (5 b \left (-3 a^2+b^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) (2 b+5 a \sin (2 (e+f x)))\right ) (a+b \tan (e+f x))^3}{5 f (a \cos (e+f x)+b \sin (e+f x))^3} \end {gather*}

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] Result contains complex when optimal does not.
time = 1.12, size = 373, normalized size = 2.89

method	result	size
default	\(\frac {2 \left (\cos \left (f x +e \right )+1\right )^{2} \left (\cos \left (f x +e \right )-1\right )^{2} \left (5 i \left (\cos ^{3}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a^{3}-10 i \left (\cos ^{3}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a \,b^{2}+5 i \left (\cos ^{2}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a^{3}-10 i \left (\cos ^{2}\left (f x +e \right )\right ) \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, a \,b^{2}+15 a^{2} \left (\cos ^{2}\left (f x +e \right )\right ) b -5 b^{3} \left (\cos ^{2}\left (f x +e \right )\right )+5 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}+b^{3}\right ) \sqrt {\frac {d}{\cos \left (f x +e \right )}}}{5 f \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4}}\)	\(373\)

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,method=_RETURNVERBOSE)

[Out]

2/5/f*(cos(f*x+e)+1)^2*(cos(f*x+e)-1)^2*(5*I*cos(f*x+e)^3*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(1/(cos(f*x
+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a^3-10*I*cos(f*x+e)^3*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)
*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2+5*I*cos(f*x+e)^2*EllipticF(I*(cos(f*x+e)-1)/
sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a^3-10*I*cos(f*x+e)^2*EllipticF(I*(co
s(f*x+e)-1)/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2+15*a^2*cos(f*x+e)^2
*b-5*b^3*cos(f*x+e)^2+5*cos(f*x+e)*sin(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 175, normalized size = 1.36 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (i \, a^{3} - 2 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 \, \sqrt {2} {\left (-i \, a^{3} + 2 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (5 \, a b^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b^{3} + 5 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, f \cos \left (f x + e\right )^{2}} \end {gather*}

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]

-1/5*(5*sqrt(2)*(I*a^3 - 2*I*a*b^2)*sqrt(d)*cos(f*x + e)^2*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x
+ e)) + 5*sqrt(2)*(-I*a^3 + 2*I*a*b^2)*sqrt(d)*cos(f*x + e)^2*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin
(f*x + e)) - 2*(5*a*b^2*cos(f*x + e)*sin(f*x + e) + b^3 + 5*(3*a^2*b - b^3)*cos(f*x + e)^2)*sqrt(d/cos(f*x + e
)))/(f*cos(f*x + e)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {d \sec {\left (e + f x \right )}} \left (a + b \tan {\left (e + f x \right )}\right )^{3}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {\frac {d}{\cos \left (e+f\,x\right )}}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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