3.7.1 \(\int \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}} \, dx\) [601]

Optimal. Leaf size=176 \[ \frac {2 a \left (7 a^2+6 b^2\right ) E\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{15 d^4 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{9 d^4 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^2(e+f x) \left (2 b \left (5 a^2+2 b^2\right )-a \left (7 a^2+b^2\right ) \tan (e+f x)\right )}{45 d^4 f \sqrt {d \sec (e+f x)}} \]

[Out]

2/15*a*(7*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))*(sec(f*x+e)^2)^(1/4)/d^4/f/(d*sec(f*x+e))^(1/2)-2/9*cos(f*x+e)^4*(b-a*tan(f*x+e))*(a+b
*tan(f*x+e))^2/d^4/f/(d*sec(f*x+e))^(1/2)-2/45*cos(f*x+e)^2*(2*b*(5*a^2+2*b^2)-a*(7*a^2+b^2)*tan(f*x+e))/d^4/f
/(d*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 176, 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, 753, 792, 202} \begin {gather*} \frac {2 a \left (7 a^2+6 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right )}{15 d^4 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^2(e+f x) \left (2 b \left (5 a^2+2 b^2\right )-a \left (7 a^2+b^2\right ) \tan (e+f x)\right )}{45 d^4 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{9 d^4 f \sqrt {d \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(9/2),x]

[Out]

(2*a*(7*a^2 + 6*b^2)*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(1/4))/(15*d^4*f*Sqrt[d*Sec[e + f*x
]]) - (2*Cos[e + f*x]^4*(b - a*Tan[e + f*x])*(a + b*Tan[e + f*x])^2)/(9*d^4*f*Sqrt[d*Sec[e + f*x]]) - (2*Cos[e
+ f*x]^2*(2*b*(5*a^2 + 2*b^2) - a*(7*a^2 + b^2)*Tan[e + f*x]))/(45*d^4*f*Sqrt[d*Sec[e + f*x]])

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 753

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 792

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 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 \frac {(a+b \tan (e+f x))^3}{(d \sec (e+f x))^{9/2}} \, dx &=\frac {\sqrt [4]{\sec ^2(e+f x)} \text {Subst}\left (\int \frac {(a+x)^3}{\left (1+\frac {x^2}{b^2}\right )^{13/4}} \, dx,x,b \tan (e+f x)\right )}{b d^4 f \sqrt {d \sec (e+f x)}}\\ &=-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{9 d^4 f \sqrt {d \sec (e+f x)}}+\frac {\left (2 b \sqrt [4]{\sec ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {(a+x) \left (\frac {1}{2} \left (4+\frac {7 a^2}{b^2}\right )+\frac {3 a x}{2 b^2}\right )}{\left (1+\frac {x^2}{b^2}\right )^{9/4}} \, dx,x,b \tan (e+f x)\right )}{9 d^4 f \sqrt {d \sec (e+f x)}}\\ &=-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{9 d^4 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^2(e+f x) \left (2 b \left (5 a^2+2 b^2\right )-a \left (7 a^2+b^2\right ) \tan (e+f x)\right )}{45 d^4 f \sqrt {d \sec (e+f x)}}+\frac {\left (a \left (6+\frac {7 a^2}{b^2}\right ) b \sqrt [4]{\sec ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{15 d^4 f \sqrt {d \sec (e+f x)}}\\ &=\frac {2 a \left (7 a^2+6 b^2\right ) E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{15 d^4 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^4(e+f x) (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{9 d^4 f \sqrt {d \sec (e+f x)}}-\frac {2 \cos ^2(e+f x) \left (2 b \left (5 a^2+2 b^2\right )-a \left (7 a^2+b^2\right ) \tan (e+f x)\right )}{45 d^4 f \sqrt {d \sec (e+f x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(372\) vs. \(2(176)=352\).
time = 6.43, size = 372, normalized size = 2.11 \begin {gather*} \frac {\sec ^{\frac {3}{2}}(e+f x) \left (\frac {2 \left (56 a^3+48 a b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\sqrt {\cos (e+f x)} \sqrt {\sec (e+f x)}}-\frac {2 \left (15 a^2 b+7 b^3\right ) \sin ^2(e+f x)}{\sqrt {1-\cos ^2(e+f x)} \sqrt {\sec (e+f x)} \sqrt {\cos ^2(e+f x) \left (-1+\sec ^2(e+f x)\right )}}\right ) (a+b \tan (e+f x))^3}{120 f (d \sec (e+f x))^{9/2} (a \cos (e+f x)+b \sin (e+f x))^3}+\frac {\sec ^2(e+f x) \left (-\frac {1}{90} b \left (15 a^2+4 b^2\right ) \cos (e+f x)-\frac {1}{360} b \left (75 a^2+11 b^2\right ) \cos (3 (e+f x))-\frac {1}{72} b \left (3 a^2-b^2\right ) \cos (5 (e+f x))+\frac {1}{180} a \left (19 a^2-3 b^2\right ) \sin (e+f x)+\frac {1}{360} a \left (43 a^2-21 b^2\right ) \sin (3 (e+f x))+\frac {1}{72} a \left (a^2-3 b^2\right ) \sin (5 (e+f x))\right ) (a+b \tan (e+f x))^3}{f (d \sec (e+f x))^{9/2} (a \cos (e+f x)+b \sin (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^3/(d*Sec[e + f*x])^(9/2),x]

[Out]

(Sec[e + f*x]^(3/2)*((2*(56*a^3 + 48*a*b^2)*EllipticE[(e + f*x)/2, 2])/(Sqrt[Cos[e + f*x]]*Sqrt[Sec[e + f*x]])
- (2*(15*a^2*b + 7*b^3)*Sin[e + f*x]^2)/(Sqrt[1 - Cos[e + f*x]^2]*Sqrt[Sec[e + f*x]]*Sqrt[Cos[e + f*x]^2*(-1
+ Sec[e + f*x]^2)]))*(a + b*Tan[e + f*x])^3)/(120*f*(d*Sec[e + f*x])^(9/2)*(a*Cos[e + f*x] + b*Sin[e + f*x])^3
) + (Sec[e + f*x]^2*(-1/90*(b*(15*a^2 + 4*b^2)*Cos[e + f*x]) - (b*(75*a^2 + 11*b^2)*Cos[3*(e + f*x)])/360 - (b
*(3*a^2 - b^2)*Cos[5*(e + f*x)])/72 + (a*(19*a^2 - 3*b^2)*Sin[e + f*x])/180 + (a*(43*a^2 - 21*b^2)*Sin[3*(e +
f*x)])/360 + (a*(a^2 - 3*b^2)*Sin[5*(e + f*x)])/72)*(a + b*Tan[e + f*x])^3)/(f*(d*Sec[e + f*x])^(9/2)*(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 = 0.68, size = 745, normalized size = 4.23

method	result	size
default	\(\frac {\frac {14 i \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}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{3}}{15}-\frac {4 i \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}}\, \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a \,b^{2}}{5}-\frac {14 i \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}}\, \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a^{3}}{15}-\frac {4 i \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}}\, \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}}{5}-\frac {2 \left (\cos ^{6}\left (f x +e \right )\right ) a^{3}}{9}+\frac {2 \left (\cos ^{6}\left (f x +e \right )\right ) a \,b^{2}}{3}-\frac {2 \left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a^{2} b}{3}+\frac {2 \left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right ) b^{3}}{9}+\frac {4 i \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}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}}{5}-\frac {14 i \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}}\, \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{3}}{15}+\frac {4 i \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}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a \,b^{2}}{5}+\frac {14 i \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}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a^{3}}{15}-\frac {4 \left (\cos ^{4}\left (f x +e \right )\right ) a^{3}}{45}-\frac {14 \left (\cos ^{4}\left (f x +e \right )\right ) a \,b^{2}}{15}-\frac {2 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) b^{3}}{5}-\frac {28 \left (\cos ^{2}\left (f x +e \right )\right ) a^{3}}{45}-\frac {8 \left (\cos ^{2}\left (f x +e \right )\right ) a \,b^{2}}{15}+\frac {14 a^{3} \cos \left (f x +e \right )}{15}+\frac {4 a \cos \left (f x +e \right ) b^{2}}{5}}{f \cos \left (f x +e \right )^{5} \sin \left (f x +e \right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {9}{2}}}\)	\(745\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/45/f*(21*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)*sin(f*x+e)*a^3-21*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/s
in(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*a^3+18*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Ellipti
cF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*a*b^2-21*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f
*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)*a^3-5*cos(f*x+e)^6*a^3+15*cos(f*x+e)^6*a*b
^2-15*cos(f*x+e)^5*sin(f*x+e)*a^2*b+5*cos(f*x+e)^5*sin(f*x+e)*b^3-18*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(c
os(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*sin(f*x+e)*a*b^2-18*I*(1/(cos(f*x+e)+1))^(1/2)*(c
os(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*a*b^2+21*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)*sin
(f*x+e)*a^3+18*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)*sin(f*x+e)*a*b^2-2*cos(f*x+e)^4*a^3-21*cos(f*x+e)^4*a*b^2-9*cos(f*x+e)^3*sin(f*x+e)*b^3-14*cos(f*x+e)^2
*a^3-12*cos(f*x+e)^2*a*b^2+21*a^3*cos(f*x+e)+18*a*cos(f*x+e)*b^2)/cos(f*x+e)^5/sin(f*x+e)/(d/cos(f*x+e))^(9/2)

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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((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(9/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.16, size = 201, normalized size = 1.14 \begin {gather*} -\frac {3 \, \sqrt {2} {\left (-7 i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {2} {\left (7 i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (9 \, b^{3} \cos \left (f x + e\right )^{3} + 5 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{5} - {\left (5 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + {\left (7 \, a^{3} + 6 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{45 \, d^{5} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

-1/45*(3*sqrt(2)*(-7*I*a^3 - 6*I*a*b^2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e)
+ I*sin(f*x + e))) + 3*sqrt(2)*(7*I*a^3 + 6*I*a*b^2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0
, cos(f*x + e) - I*sin(f*x + e))) + 2*(9*b^3*cos(f*x + e)^3 + 5*(3*a^2*b - b^3)*cos(f*x + e)^5 - (5*(a^3 - 3*a
*b^2)*cos(f*x + e)^4 + (7*a^3 + 6*a*b^2)*cos(f*x + e)^2)*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^5*f)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))**3/(d*sec(f*x+e))**(9/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3879 deep

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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((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^3/(d*sec(f*x + e))^(9/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(9/2),x)

[Out]

int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(9/2), x)

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