3.1.41 \(\int \frac {\sec ^6(x)}{a+b \cos ^2(x)} \, dx\) [41]

Optimal. Leaf size=79 \[ \frac {b^3 \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{a^{7/2} \sqrt {a+b}}+\frac {\left (a^2-a b+b^2\right ) \tan (x)}{a^3}+\frac {(2 a-b) \tan ^3(x)}{3 a^2}+\frac {\tan ^5(x)}{5 a} \]

[Out]

b^3*arctan(cot(x)*(a+b)^(1/2)/a^(1/2))/a^(7/2)/(a+b)^(1/2)+(a^2-a*b+b^2)*tan(x)/a^3+1/3*(2*a-b)*tan(x)^3/a^2+1
/5*tan(x)^5/a

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Rubi [A]
time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3266, 472, 211} \begin {gather*} \frac {b^3 \text {ArcTan}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{a^{7/2} \sqrt {a+b}}+\frac {(2 a-b) \tan ^3(x)}{3 a^2}+\frac {\left (a^2-a b+b^2\right ) \tan (x)}{a^3}+\frac {\tan ^5(x)}{5 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[x]^6/(a + b*Cos[x]^2),x]

[Out]

(b^3*ArcTan[(Sqrt[a + b]*Cot[x])/Sqrt[a]])/(a^(7/2)*Sqrt[a + b]) + ((a^2 - a*b + b^2)*Tan[x])/a^3 + ((2*a - b)
*Tan[x]^3)/(3*a^2) + Tan[x]^5/(5*a)

Rule 211

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

Rule 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 3266

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p +
1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\sec ^6(x)}{a+b \cos ^2(x)} \, dx &=-\text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^6 \left (a+(a+b) x^2\right )} \, dx,x,\cot (x)\right )\\ &=-\text {Subst}\left (\int \left (\frac {1}{a x^6}+\frac {2 a-b}{a^2 x^4}+\frac {a^2-a b+b^2}{a^3 x^2}+\frac {b^3}{a^3 \left (-a-(a+b) x^2\right )}\right ) \, dx,x,\cot (x)\right )\\ &=\frac {\left (a^2-a b+b^2\right ) \tan (x)}{a^3}+\frac {(2 a-b) \tan ^3(x)}{3 a^2}+\frac {\tan ^5(x)}{5 a}-\frac {b^3 \text {Subst}\left (\int \frac {1}{-a-(a+b) x^2} \, dx,x,\cot (x)\right )}{a^3}\\ &=\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {a+b} \cot (x)}{\sqrt {a}}\right )}{a^{7/2} \sqrt {a+b}}+\frac {\left (a^2-a b+b^2\right ) \tan (x)}{a^3}+\frac {(2 a-b) \tan ^3(x)}{3 a^2}+\frac {\tan ^5(x)}{5 a}\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 80, normalized size = 1.01 \begin {gather*} -\frac {b^3 \text {ArcTan}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a+b}}\right )}{a^{7/2} \sqrt {a+b}}+\frac {\left (8 a^2-10 a b+15 b^2+a (4 a-5 b) \sec ^2(x)+3 a^2 \sec ^4(x)\right ) \tan (x)}{15 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^6/(a + b*Cos[x]^2),x]

[Out]

-((b^3*ArcTan[(Sqrt[a]*Tan[x])/Sqrt[a + b]])/(a^(7/2)*Sqrt[a + b])) + ((8*a^2 - 10*a*b + 15*b^2 + a*(4*a - 5*b
)*Sec[x]^2 + 3*a^2*Sec[x]^4)*Tan[x])/(15*a^3)

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Maple [A]
time = 0.24, size = 78, normalized size = 0.99

method	result	size
default	\(\frac {\frac {\left (\tan ^{5}\left (x \right )\right ) a^{2}}{5}+\frac {2 a^{2} \left (\tan ^{3}\left (x \right )\right )}{3}-\frac {a b \left (\tan ^{3}\left (x \right )\right )}{3}+a^{2} \tan \left (x \right )-a b \tan \left (x \right )+b^{2} \tan \left (x \right )}{a^{3}}-\frac {b^{3} \arctan \left (\frac {a \tan \left (x \right )}{\sqrt {\left (a +b \right ) a}}\right )}{a^{3} \sqrt {\left (a +b \right ) a}}\)	\(78\)
risch	\(\frac {2 i \left (15 b^{2} {\mathrm e}^{8 i x}-30 a b \,{\mathrm e}^{6 i x}+60 b^{2} {\mathrm e}^{6 i x}+80 a^{2} {\mathrm e}^{4 i x}-70 a b \,{\mathrm e}^{4 i x}+90 b^{2} {\mathrm e}^{4 i x}+40 a^{2} {\mathrm e}^{2 i x}-50 b \,{\mathrm e}^{2 i x} a +60 b^{2} {\mathrm e}^{2 i x}+8 a^{2}-10 a b +15 b^{2}\right )}{15 a^{3} \left ({\mathrm e}^{2 i x}+1\right )^{5}}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i x}+\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, a^{3}}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, a^{3}}\)	\(287\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^6/(a+b*cos(x)^2),x,method=_RETURNVERBOSE)

[Out]

1/a^3*(1/5*tan(x)^5*a^2+2/3*a^2*tan(x)^3-1/3*a*b*tan(x)^3+a^2*tan(x)-a*b*tan(x)+b^2*tan(x))-b^3/a^3/((a+b)*a)^
(1/2)*arctan(a*tan(x)/((a+b)*a)^(1/2))

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Maxima [A]
time = 0.47, size = 74, normalized size = 0.94 \begin {gather*} -\frac {b^{3} \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{3}} + \frac {3 \, a^{2} \tan \left (x\right )^{5} + 5 \, {\left (2 \, a^{2} - a b\right )} \tan \left (x\right )^{3} + 15 \, {\left (a^{2} - a b + b^{2}\right )} \tan \left (x\right )}{15 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^6/(a+b*cos(x)^2),x, algorithm="maxima")

[Out]

-b^3*arctan(a*tan(x)/sqrt((a + b)*a))/(sqrt((a + b)*a)*a^3) + 1/15*(3*a^2*tan(x)^5 + 5*(2*a^2 - a*b)*tan(x)^3
+ 15*(a^2 - a*b + b^2)*tan(x))/a^3

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (67) = 134\).
time = 0.48, size = 348, normalized size = 4.41 \begin {gather*} \left [-\frac {15 \, \sqrt {-a^{2} - a b} b^{3} \cos \left (x\right )^{5} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \, {\left (4 \, a^{2} + 3 \, a b\right )} \cos \left (x\right )^{2} - 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{3} - a \cos \left (x\right )\right )} \sqrt {-a^{2} - a b} \sin \left (x\right ) + a^{2}}{b^{2} \cos \left (x\right )^{4} + 2 \, a b \cos \left (x\right )^{2} + a^{2}}\right ) - 4 \, {\left ({\left (8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3}\right )} \cos \left (x\right )^{4} + 3 \, a^{4} + 3 \, a^{3} b + {\left (4 \, a^{4} - a^{3} b - 5 \, a^{2} b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{60 \, {\left (a^{5} + a^{4} b\right )} \cos \left (x\right )^{5}}, \frac {15 \, \sqrt {a^{2} + a b} b^{3} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a}{2 \, \sqrt {a^{2} + a b} \cos \left (x\right ) \sin \left (x\right )}\right ) \cos \left (x\right )^{5} + 2 \, {\left ({\left (8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3}\right )} \cos \left (x\right )^{4} + 3 \, a^{4} + 3 \, a^{3} b + {\left (4 \, a^{4} - a^{3} b - 5 \, a^{2} b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{30 \, {\left (a^{5} + a^{4} b\right )} \cos \left (x\right )^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^6/(a+b*cos(x)^2),x, algorithm="fricas")

[Out]

[-1/60*(15*sqrt(-a^2 - a*b)*b^3*cos(x)^5*log(((8*a^2 + 8*a*b + b^2)*cos(x)^4 - 2*(4*a^2 + 3*a*b)*cos(x)^2 - 4*
((2*a + b)*cos(x)^3 - a*cos(x))*sqrt(-a^2 - a*b)*sin(x) + a^2)/(b^2*cos(x)^4 + 2*a*b*cos(x)^2 + a^2)) - 4*((8*
a^4 - 2*a^3*b + 5*a^2*b^2 + 15*a*b^3)*cos(x)^4 + 3*a^4 + 3*a^3*b + (4*a^4 - a^3*b - 5*a^2*b^2)*cos(x)^2)*sin(x
))/((a^5 + a^4*b)*cos(x)^5), 1/30*(15*sqrt(a^2 + a*b)*b^3*arctan(1/2*((2*a + b)*cos(x)^2 - a)/(sqrt(a^2 + a*b)
*cos(x)*sin(x)))*cos(x)^5 + 2*((8*a^4 - 2*a^3*b + 5*a^2*b^2 + 15*a*b^3)*cos(x)^4 + 3*a^4 + 3*a^3*b + (4*a^4 -
a^3*b - 5*a^2*b^2)*cos(x)^2)*sin(x))/((a^5 + a^4*b)*cos(x)^5)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{6}{\left (x \right )}}{a + b \cos ^{2}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**6/(a+b*cos(x)**2),x)

[Out]

Integral(sec(x)**6/(a + b*cos(x)**2), x)

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Giac [A]
time = 0.41, size = 104, normalized size = 1.32 \begin {gather*} -\frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )} b^{3}}{\sqrt {a^{2} + a b} a^{3}} + \frac {3 \, a^{4} \tan \left (x\right )^{5} + 10 \, a^{4} \tan \left (x\right )^{3} - 5 \, a^{3} b \tan \left (x\right )^{3} + 15 \, a^{4} \tan \left (x\right ) - 15 \, a^{3} b \tan \left (x\right ) + 15 \, a^{2} b^{2} \tan \left (x\right )}{15 \, a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^6/(a+b*cos(x)^2),x, algorithm="giac")

[Out]

-(pi*floor(x/pi + 1/2)*sgn(a) + arctan(a*tan(x)/sqrt(a^2 + a*b)))*b^3/(sqrt(a^2 + a*b)*a^3) + 1/15*(3*a^4*tan(
x)^5 + 10*a^4*tan(x)^3 - 5*a^3*b*tan(x)^3 + 15*a^4*tan(x) - 15*a^3*b*tan(x) + 15*a^2*b^2*tan(x))/a^5

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Mupad [B]
time = 2.30, size = 84, normalized size = 1.06 \begin {gather*} \frac {{\mathrm {tan}\left (x\right )}^5}{5\,a}-{\mathrm {tan}\left (x\right )}^3\,\left (\frac {a+b}{3\,a^2}-\frac {1}{a}\right )+\mathrm {tan}\left (x\right )\,\left (\frac {3}{a}+\frac {\left (a+b\right )\,\left (\frac {a+b}{a^2}-\frac {3}{a}\right )}{a}\right )-\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a}\,\mathrm {tan}\left (x\right )}{\sqrt {a+b}}\right )}{a^{7/2}\,\sqrt {a+b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^6*(a + b*cos(x)^2)),x)

[Out]

tan(x)^5/(5*a) - tan(x)^3*((a + b)/(3*a^2) - 1/a) + tan(x)*(3/a + ((a + b)*((a + b)/a^2 - 3/a))/a) - (b^3*atan
((a^(1/2)*tan(x))/(a + b)^(1/2)))/(a^(7/2)*(a + b)^(1/2))

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