3.6.93 \(\int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{9/2}} \, dx\) [593]

Optimal. Leaf size=184 \[ -\frac {10 a b}{63 f (d \sec (e+f x))^{9/2}}+\frac {2 \left (7 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 d^4 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}+\frac {2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{45 d^3 f (d \sec (e+f x))^{3/2}}-\frac {2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}} \]

[Out]

-10/63*a*b/f/(d*sec(f*x+e))^(9/2)+2/63*(7*a^2+2*b^2)*sin(f*x+e)/d/f/(d*sec(f*x+e))^(7/2)+2/45*(7*a^2+2*b^2)*si
n(f*x+e)/d^3/f/(d*sec(f*x+e))^(3/2)+2/15*(7*a^2+2*b^2)*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Ellipti
cE(sin(1/2*f*x+1/2*e),2^(1/2))/d^4/f/cos(f*x+e)^(1/2)/(d*sec(f*x+e))^(1/2)-2/7*b*(a+b*tan(f*x+e))/f/(d*sec(f*x
+e))^(9/2)

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Rubi [A]
time = 0.14, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3589, 3567, 3854, 3856, 2719} \begin {gather*} \frac {2 \left (7 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 d^4 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{45 d^3 f (d \sec (e+f x))^{3/2}}+\frac {2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}-\frac {10 a b}{63 f (d \sec (e+f x))^{9/2}}-\frac {2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*a*b)/(63*f*(d*Sec[e + f*x])^(9/2)) + (2*(7*a^2 + 2*b^2)*EllipticE[(e + f*x)/2, 2])/(15*d^4*f*Sqrt[Cos[e +
f*x]]*Sqrt[d*Sec[e + f*x]]) + (2*(7*a^2 + 2*b^2)*Sin[e + f*x])/(63*d*f*(d*Sec[e + f*x])^(7/2)) + (2*(7*a^2 +
2*b^2)*Sin[e + f*x])/(45*d^3*f*(d*Sec[e + f*x])^(3/2)) - (2*b*(a + b*Tan[e + f*x]))/(7*f*(d*Sec[e + f*x])^(9/2
))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 3567

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3589

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[b*(d*Sec
[e + f*x])^m*((a + b*Tan[e + f*x])/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]

Rule 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \frac {(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{9/2}} \, dx &=-\frac {2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}-\frac {2}{7} \int \frac {-\frac {7 a^2}{2}-b^2-\frac {5}{2} a b \tan (e+f x)}{(d \sec (e+f x))^{9/2}} \, dx\\ &=-\frac {10 a b}{63 f (d \sec (e+f x))^{9/2}}-\frac {2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}-\frac {1}{7} \left (-7 a^2-2 b^2\right ) \int \frac {1}{(d \sec (e+f x))^{9/2}} \, dx\\ &=-\frac {10 a b}{63 f (d \sec (e+f x))^{9/2}}+\frac {2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}-\frac {2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}+\frac {\left (7 a^2+2 b^2\right ) \int \frac {1}{(d \sec (e+f x))^{5/2}} \, dx}{9 d^2}\\ &=-\frac {10 a b}{63 f (d \sec (e+f x))^{9/2}}+\frac {2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}+\frac {2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{45 d^3 f (d \sec (e+f x))^{3/2}}-\frac {2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}+\frac {\left (7 a^2+2 b^2\right ) \int \frac {1}{\sqrt {d \sec (e+f x)}} \, dx}{15 d^4}\\ &=-\frac {10 a b}{63 f (d \sec (e+f x))^{9/2}}+\frac {2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}+\frac {2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{45 d^3 f (d \sec (e+f x))^{3/2}}-\frac {2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}+\frac {\left (7 a^2+2 b^2\right ) \int \sqrt {\cos (e+f x)} \, dx}{15 d^4 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}\\ &=-\frac {10 a b}{63 f (d \sec (e+f x))^{9/2}}+\frac {2 \left (7 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{15 d^4 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{63 d f (d \sec (e+f x))^{7/2}}+\frac {2 \left (7 a^2+2 b^2\right ) \sin (e+f x)}{45 d^3 f (d \sec (e+f x))^{3/2}}-\frac {2 b (a+b \tan (e+f x))}{7 f (d \sec (e+f x))^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 3.08, size = 126, normalized size = 0.68 \begin {gather*} \frac {\frac {48 \left (7 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\sqrt {\cos (e+f x)}}+4 \cos (e+f x) \left (-30 a b \cos (e+f x)-10 a b \cos (3 (e+f x))+2 \left (19 a^2-b^2+5 \left (a^2-b^2\right ) \cos (2 (e+f x))\right ) \sin (e+f x)\right )}{360 d^4 f \sqrt {d \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((48*(7*a^2 + 2*b^2)*EllipticE[(e + f*x)/2, 2])/Sqrt[Cos[e + f*x]] + 4*Cos[e + f*x]*(-30*a*b*Cos[e + f*x] - 10
*a*b*Cos[3*(e + f*x)] + 2*(19*a^2 - b^2 + 5*(a^2 - b^2)*Cos[2*(e + f*x)])*Sin[e + f*x]))/(360*d^4*f*Sqrt[d*Sec
[e + f*x]])

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Maple [C] Result contains complex when optimal does not.
time = 0.58, size = 697, normalized size = 3.79

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^{2}}{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 ) b^{2}}{15}+\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^{2}}{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 ) b^{2}}{15}-\frac {2 \left (\cos ^{6}\left (f x +e \right )\right ) a^{2}}{9}+\frac {2 \left (\cos ^{6}\left (f x +e \right )\right ) b^{2}}{9}-\frac {4 \left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a b}{9}-\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^{2}}{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 ) b^{2}}{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 ) \cos \left (f x +e \right ) \sin \left (f x +e \right ) b^{2}}{15}-\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^{2}}{15}-\frac {4 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2}}{45}-\frac {14 \left (\cos ^{4}\left (f x +e \right )\right ) b^{2}}{45}-\frac {28 a^{2} \left (\cos ^{2}\left (f x +e \right )\right )}{45}-\frac {8 b^{2} \left (\cos ^{2}\left (f x +e \right )\right )}{45}+\frac {14 \cos \left (f x +e \right ) a^{2}}{15}+\frac {4 \cos \left (f x +e \right ) b^{2}}{15}}{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}}}\)	\(697\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45/f*(6*I*sin(f*x+e)*(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)/s
in(f*x+e),I)*b^2+21*I*sin(f*x+e)*(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)*a^2-21*I*cos(f*x+e)*sin(f*x+e)*(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)*a^2+21*I*cos(f*x+e)*sin(f*x+e)*(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)*a^2-5*cos(f*x+e)^6*a^2+5*cos(f*x+e)^6*b^2-10*
cos(f*x+e)^5*sin(f*x+e)*a*b-6*I*sin(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Elliptic
E(I*(cos(f*x+e)-1)/sin(f*x+e),I)*b^2-21*I*sin(f*x+e)*(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)*a^2+6*I*cos(f*x+e)*sin(f*x+e)*(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)*b^2-6*I*cos(f*x+e)*sin(f*x+e)*(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)*b^2-2*cos(f*x+e)^4*a^2-7*co
s(f*x+e)^4*b^2-14*a^2*cos(f*x+e)^2-4*b^2*cos(f*x+e)^2+21*cos(f*x+e)*a^2+6*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))^2/(d*sec(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^2/(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 = 0.15, size = 173, normalized size = 0.94 \begin {gather*} -\frac {3 \, \sqrt {2} {\left (-7 i \, a^{2} - 2 i \, 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^{2} + 2 i \, 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 (10 \, a b \cos \left (f x + e\right )^{5} - {\left (5 \, {\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{4} + {\left (7 \, a^{2} + 2 \, 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))^2/(d*sec(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

-1/45*(3*sqrt(2)*(-7*I*a^2 - 2*I*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^2 + 2*I*b^2)*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co
s(f*x + e) - I*sin(f*x + e))) + 2*(10*a*b*cos(f*x + e)^5 - (5*(a^2 - b^2)*cos(f*x + e)^4 + (7*a^2 + 2*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))**2/(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))^2/(d*sec(f*x+e))^(9/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^2/(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 )}^2}{{\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))^2/(d/cos(e + f*x))^(9/2),x)

[Out]

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

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