3.4.13 \(\int \frac {\sec ^6(x)}{a+b \sin ^2(x)} \, dx\) [313]

Optimal. Leaf size=87 \[ \frac {b^3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{7/2}}+\frac {\left (a^2+3 a b+3 b^2\right ) \tan (x)}{(a+b)^3}+\frac {(2 a+3 b) \tan ^3(x)}{3 (a+b)^2}+\frac {\tan ^5(x)}{5 (a+b)} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 87, 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 = {3270, 398, 211} \begin {gather*} \frac {\left (a^2+3 a b+3 b^2\right ) \tan (x)}{(a+b)^3}+\frac {b^3 \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{7/2}}+\frac {\tan ^5(x)}{5 (a+b)}+\frac {(2 a+3 b) \tan ^3(x)}{3 (a+b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[x]^6/(a + b*Sin[x]^2),x]

[Out]

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

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 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 3270

Int[cos[(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/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[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 \sin ^2(x)} \, dx &=\text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{a+(a+b) x^2} \, dx,x,\tan (x)\right )\\ &=\text {Subst}\left (\int \left (\frac {a^2+3 a b+3 b^2}{(a+b)^3}+\frac {(2 a+3 b) x^2}{(a+b)^2}+\frac {x^4}{a+b}+\frac {b^3}{(a+b)^3 \left (a+(a+b) x^2\right )}\right ) \, dx,x,\tan (x)\right )\\ &=\frac {\left (a^2+3 a b+3 b^2\right ) \tan (x)}{(a+b)^3}+\frac {(2 a+3 b) \tan ^3(x)}{3 (a+b)^2}+\frac {\tan ^5(x)}{5 (a+b)}+\frac {b^3 \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (x)\right )}{(a+b)^3}\\ &=\frac {b^3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{7/2}}+\frac {\left (a^2+3 a b+3 b^2\right ) \tan (x)}{(a+b)^3}+\frac {(2 a+3 b) \tan ^3(x)}{3 (a+b)^2}+\frac {\tan ^5(x)}{5 (a+b)}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 90, normalized size = 1.03 \begin {gather*} \frac {b^3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{7/2}}+\frac {\left (8 a^2+26 a b+33 b^2+\left (4 a^2+13 a b+9 b^2\right ) \sec ^2(x)+3 (a+b)^2 \sec ^4(x)\right ) \tan (x)}{15 (a+b)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^3*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^(7/2)) + ((8*a^2 + 26*a*b + 33*b^2 + (4*a^2 + 13*a
*b + 9*b^2)*Sec[x]^2 + 3*(a + b)^2*Sec[x]^4)*Tan[x])/(15*(a + b)^3)

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Maple [A]
time = 0.35, size = 109, normalized size = 1.25

method	result	size
default	\(\frac {\frac {a^{2} \left (\tan ^{5}\left (x \right )\right )}{5}+\frac {2 a b \left (\tan ^{5}\left (x \right )\right )}{5}+\frac {b^{2} \left (\tan ^{5}\left (x \right )\right )}{5}+\frac {2 a^{2} \left (\tan ^{3}\left (x \right )\right )}{3}+\frac {5 a b \left (\tan ^{3}\left (x \right )\right )}{3}+b^{2} \left (\tan ^{3}\left (x \right )\right )+a^{2} \tan \left (x \right )+3 a b \tan \left (x \right )+3 b^{2} \tan \left (x \right )}{\left (a +b \right )^{3}}+\frac {b^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (x \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{3} \sqrt {a \left (a +b \right )}}\)	\(109\)
risch	\(\frac {2 i \left (15 b^{2} {\mathrm e}^{8 i x}+30 a b \,{\mathrm e}^{6 i x}+90 b^{2} {\mathrm e}^{6 i x}+80 a^{2} {\mathrm e}^{4 i x}+230 a b \,{\mathrm e}^{4 i x}+240 b^{2} {\mathrm e}^{4 i x}+40 \,{\mathrm e}^{2 i x} a^{2}+130 b \,{\mathrm e}^{2 i x} a +150 b^{2} {\mathrm e}^{2 i x}+8 a^{2}+26 a b +33 b^{2}\right )}{15 \left (a +b \right )^{3} \left ({\mathrm e}^{2 i x}+1\right )^{5}}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{3}}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i x}-\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{3}}\)	\(295\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 126, normalized size = 1.45 \begin {gather*} \frac {b^{3} \arctan \left (\frac {{\left (a + b\right )} \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (x\right )^{5} + 5 \, {\left (2 \, a^{2} + 5 \, a b + 3 \, b^{2}\right )} \tan \left (x\right )^{3} + 15 \, {\left (a^{2} + 3 \, a b + 3 \, b^{2}\right )} \tan \left (x\right )}{15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (75) = 150\).
time = 0.47, size = 459, normalized size = 5.28 \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} + 5 \, a b + b^{2}\right )} \cos \left (x\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{3} - {\left (a + b\right )} \cos \left (x\right )\right )} \sqrt {-a^{2} - a b} \sin \left (x\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (x\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 4 \, {\left ({\left (8 \, a^{4} + 34 \, a^{3} b + 59 \, a^{2} b^{2} + 33 \, a b^{3}\right )} \cos \left (x\right )^{4} + 3 \, a^{4} + 9 \, a^{3} b + 9 \, a^{2} b^{2} + 3 \, a b^{3} + {\left (4 \, a^{4} + 17 \, a^{3} b + 22 \, a^{2} b^{2} + 9 \, a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{60 \, {\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\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 - b}{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} + 34 \, a^{3} b + 59 \, a^{2} b^{2} + 33 \, a b^{3}\right )} \cos \left (x\right )^{4} + 3 \, a^{4} + 9 \, a^{3} b + 9 \, a^{2} b^{2} + 3 \, a b^{3} + {\left (4 \, a^{4} + 17 \, a^{3} b + 22 \, a^{2} b^{2} + 9 \, a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{30 \, {\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\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*sin(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 + 5*a*b + b^2)*cos(x)^
2 + 4*((2*a + b)*cos(x)^3 - (a + b)*cos(x))*sqrt(-a^2 - a*b)*sin(x) + a^2 + 2*a*b + b^2)/(b^2*cos(x)^4 - 2*(a*
b + b^2)*cos(x)^2 + a^2 + 2*a*b + b^2)) - 4*((8*a^4 + 34*a^3*b + 59*a^2*b^2 + 33*a*b^3)*cos(x)^4 + 3*a^4 + 9*a
^3*b + 9*a^2*b^2 + 3*a*b^3 + (4*a^4 + 17*a^3*b + 22*a^2*b^2 + 9*a*b^3)*cos(x)^2)*sin(x))/((a^5 + 4*a^4*b + 6*a
^3*b^2 + 4*a^2*b^3 + a*b^4)*cos(x)^5), -1/30*(15*sqrt(a^2 + a*b)*b^3*arctan(1/2*((2*a + b)*cos(x)^2 - a - b)/(
sqrt(a^2 + a*b)*cos(x)*sin(x)))*cos(x)^5 - 2*((8*a^4 + 34*a^3*b + 59*a^2*b^2 + 33*a*b^3)*cos(x)^4 + 3*a^4 + 9*
a^3*b + 9*a^2*b^2 + 3*a*b^3 + (4*a^4 + 17*a^3*b + 22*a^2*b^2 + 9*a*b^3)*cos(x)^2)*sin(x))/((a^5 + 4*a^4*b + 6*
a^3*b^2 + 4*a^2*b^3 + a*b^4)*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 \sin ^{2}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (75) = 150\).
time = 0.44, size = 254, normalized size = 2.92 \begin {gather*} \frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (x\right ) + b \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )} b^{3}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a^{2} + a b}} + \frac {3 \, a^{4} \tan \left (x\right )^{5} + 12 \, a^{3} b \tan \left (x\right )^{5} + 18 \, a^{2} b^{2} \tan \left (x\right )^{5} + 12 \, a b^{3} \tan \left (x\right )^{5} + 3 \, b^{4} \tan \left (x\right )^{5} + 10 \, a^{4} \tan \left (x\right )^{3} + 45 \, a^{3} b \tan \left (x\right )^{3} + 75 \, a^{2} b^{2} \tan \left (x\right )^{3} + 55 \, a b^{3} \tan \left (x\right )^{3} + 15 \, b^{4} \tan \left (x\right )^{3} + 15 \, a^{4} \tan \left (x\right ) + 75 \, a^{3} b \tan \left (x\right ) + 150 \, a^{2} b^{2} \tan \left (x\right ) + 135 \, a b^{3} \tan \left (x\right ) + 45 \, b^{4} \tan \left (x\right )}{15 \, {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(pi*floor(x/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(x) + b*tan(x))/sqrt(a^2 + a*b)))*b^3/((a^3 + 3*a^2*b + 3*
a*b^2 + b^3)*sqrt(a^2 + a*b)) + 1/15*(3*a^4*tan(x)^5 + 12*a^3*b*tan(x)^5 + 18*a^2*b^2*tan(x)^5 + 12*a*b^3*tan(
x)^5 + 3*b^4*tan(x)^5 + 10*a^4*tan(x)^3 + 45*a^3*b*tan(x)^3 + 75*a^2*b^2*tan(x)^3 + 55*a*b^3*tan(x)^3 + 15*b^4
*tan(x)^3 + 15*a^4*tan(x) + 75*a^3*b*tan(x) + 150*a^2*b^2*tan(x) + 135*a*b^3*tan(x) + 45*b^4*tan(x))/(a^5 + 5*
a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)

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Mupad [B]
time = 13.99, size = 121, normalized size = 1.39 \begin {gather*} \frac {{\mathrm {tan}\left (x\right )}^5}{5\,\left (a+b\right )}-{\mathrm {tan}\left (x\right )}^3\,\left (\frac {a}{3\,{\left (a+b\right )}^2}-\frac {1}{a+b}\right )+\mathrm {tan}\left (x\right )\,\left (\frac {3}{a+b}+\frac {a\,\left (\frac {a}{{\left (a+b\right )}^2}-\frac {3}{a+b}\right )}{a+b}\right )+\frac {b^3\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (x\right )\,\left (2\,a+2\,b\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{7/2}}\right )}{\sqrt {a}\,{\left (a+b\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^6*(a + b*sin(x)^2)),x)

[Out]

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

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