3.313 \(\int \frac{\sec ^6(x)}{a+b \sin ^2(x)} \, dx\)

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

[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))

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Rubi [A]  time = 0.113406, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3191, 390, 205} \[ \frac{\left (a^2+3 a b+3 b^2\right ) \tan (x)}{(a+b)^3}+\frac{b^3 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^{7/2}}+\frac{\tan ^5(x)}{5 (a+b)}+\frac{(2 a+3 b) \tan ^3(x)}{3 (a+b)^2} \]

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 3191

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]

Rule 390

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 205

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

Rubi steps

\begin{align*} \int \frac{\sec ^6(x)}{a+b \sin ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{a+(a+b) x^2} \, dx,x,\tan (x)\right )\\ &=\operatorname{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 \operatorname{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*}

Mathematica [A]  time = 0.370925, size = 90, normalized size = 1.03 \[ \frac{\tan (x) \left (\left (4 a^2+13 a b+9 b^2\right ) \sec ^2(x)+8 a^2+3 (a+b)^2 \sec ^4(x)+26 a b+33 b^2\right )}{15 (a+b)^3}+\frac{b^3 \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^{7/2}} \]

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 = 0.089, size = 147, normalized size = 1.7 \begin{align*}{\frac{ \left ( \tan \left ( x \right ) \right ) ^{5}{a}^{2}}{5\, \left ( a+b \right ) ^{3}}}+{\frac{2\, \left ( \tan \left ( x \right ) \right ) ^{5}ab}{5\, \left ( a+b \right ) ^{3}}}+{\frac{ \left ( \tan \left ( x \right ) \right ) ^{5}{b}^{2}}{5\, \left ( a+b \right ) ^{3}}}+{\frac{2\, \left ( \tan \left ( x \right ) \right ) ^{3}{a}^{2}}{3\, \left ( a+b \right ) ^{3}}}+{\frac{5\, \left ( \tan \left ( x \right ) \right ) ^{3}ab}{3\, \left ( a+b \right ) ^{3}}}+{\frac{ \left ( \tan \left ( x \right ) \right ) ^{3}{b}^{2}}{ \left ( a+b \right ) ^{3}}}+{\frac{{a}^{2}\tan \left ( x \right ) }{ \left ( a+b \right ) ^{3}}}+3\,{\frac{ab\tan \left ( x \right ) }{ \left ( a+b \right ) ^{3}}}+3\,{\frac{{b}^{2}\tan \left ( x \right ) }{ \left ( a+b \right ) ^{3}}}+{\frac{{b}^{3}}{ \left ( a+b \right ) ^{3}}\arctan \left ({ \left ( a+b \right ) \tan \left ( x \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \right ){\frac{1}{\sqrt{a \left ( a+b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/(a+b)^3*tan(x)^5*a^2+2/5/(a+b)^3*tan(x)^5*a*b+1/5/(a+b)^3*tan(x)^5*b^2+2/3/(a+b)^3*tan(x)^3*a^2+5/3/(a+b)^
3*tan(x)^3*a*b+1/(a+b)^3*tan(x)^3*b^2+1/(a+b)^3*a^2*tan(x)+3/(a+b)^3*a*b*tan(x)+3/(a+b)^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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.32889, size = 1098, normalized size = 12.62 \begin{align*} \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{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.11407, size = 343, normalized size = 3.94 \begin{align*} \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{align*}

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)