\(\int \frac {\sec ^4(x)}{a+b \sin ^2(x)} \, dx\) [311]

Optimal result

Integrand size = 15, antiderivative size = 59 \[ \int \frac {\sec ^4(x)}{a+b \sin ^2(x)} \, dx=\frac {b^2 \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{5/2}}+\frac {(a+2 b) \tan (x)}{(a+b)^2}+\frac {\tan ^3(x)}{3 (a+b)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3270, 398, 211} \[ \int \frac {\sec ^4(x)}{a+b \sin ^2(x)} \, dx=\frac {b^2 \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{5/2}}+\frac {\tan ^3(x)}{3 (a+b)}+\frac {(a+2 b) \tan (x)}{(a+b)^2} \]

[In]

Int[Sec[x]^4/(a + b*Sin[x]^2),x]

[Out]

(b^2*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^(5/2)) + ((a + 2*b)*Tan[x])/(a + b)^2 + Tan[x]^3/(
3*(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*} \text {integral}& = \text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{a+(a+b) x^2} \, dx,x,\tan (x)\right ) \\ & = \text {Subst}\left (\int \left (\frac {a+2 b}{(a+b)^2}+\frac {x^2}{a+b}+\frac {b^2}{(a+b)^2 \left (a+(a+b) x^2\right )}\right ) \, dx,x,\tan (x)\right ) \\ & = \frac {(a+2 b) \tan (x)}{(a+b)^2}+\frac {\tan ^3(x)}{3 (a+b)}+\frac {b^2 \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (x)\right )}{(a+b)^2} \\ & = \frac {b^2 \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{5/2}}+\frac {(a+2 b) \tan (x)}{(a+b)^2}+\frac {\tan ^3(x)}{3 (a+b)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^4(x)}{a+b \sin ^2(x)} \, dx=\frac {b^2 \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^{5/2}}+\frac {\left (2 a+5 b+(a+b) \sec ^2(x)\right ) \tan (x)}{3 (a+b)^2} \]

[In]

Integrate[Sec[x]^4/(a + b*Sin[x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {\frac {a \left (\tan ^{3}\left (x \right )\right )}{3}+\frac {b \left (\tan ^{3}\left (x \right )\right )}{3}+\tan \left (x \right ) a +2 \tan \left (x \right ) b}{\left (a +b \right )^{2}}+\frac {b^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (x \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{2} \sqrt {a \left (a +b \right )}}\)	\(62\)
risch	\(\frac {2 i \left (3 b \,{\mathrm e}^{4 i x}+6 a \,{\mathrm e}^{2 i x}+12 b \,{\mathrm e}^{2 i x}+2 a +5 b \right )}{3 \left ({\mathrm e}^{2 i x}+1\right )^{3} \left (a +b \right )^{2}}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2}}+\frac {b^{2} \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2}}\)	\(224\)

[In]

int(sec(x)^4/(a+b*sin(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (49) = 98\).

Time = 0.32 (sec) , antiderivative size = 343, normalized size of antiderivative = 5.81 \[ \int \frac {\sec ^4(x)}{a+b \sin ^2(x)} \, dx=\left [-\frac {3 \, \sqrt {-a^{2} - a b} b^{2} \cos \left (x\right )^{3} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (x\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{3} - {\left (a + b\right )} \cos \left (x\right )\right )} \sqrt {-a^{2} - a b} \sin \left (x\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (x\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 4 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2} + {\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{12 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (x\right )^{3}}, -\frac {3 \, \sqrt {a^{2} + a b} b^{2} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (x\right ) \sin \left (x\right )}\right ) \cos \left (x\right )^{3} - 2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2} + {\left (2 \, a^{3} + 7 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (x\right )^{3}}\right ] \]

[In]

integrate(sec(x)^4/(a+b*sin(x)^2),x, algorithm="fricas")

[Out]

[-1/12*(3*sqrt(-a^2 - a*b)*b^2*cos(x)^3*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*(a^3 + 2*a^2*b + a*b^2 + (2*a^3 + 7*a^2*b + 5*a*b^2)*cos(x)^2)*sin(
x))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(x)^3), -1/6*(3*sqrt(a^2 + a*b)*b^2*arctan(1/2*((2*a + b)*cos(x)^2
- a - b)/(sqrt(a^2 + a*b)*cos(x)*sin(x)))*cos(x)^3 - 2*(a^3 + 2*a^2*b + a*b^2 + (2*a^3 + 7*a^2*b + 5*a*b^2)*c
os(x)^2)*sin(x))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*cos(x)^3)]

Sympy [F]

\[ \int \frac {\sec ^4(x)}{a+b \sin ^2(x)} \, dx=\int \frac {\sec ^{4}{\left (x \right )}}{a + b \sin ^{2}{\left (x \right )}}\, dx \]

[In]

integrate(sec(x)**4/(a+b*sin(x)**2),x)

[Out]

Integral(sec(x)**4/(a + b*sin(x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.22 \[ \int \frac {\sec ^4(x)}{a+b \sin ^2(x)} \, dx=\frac {b^{2} \arctan \left (\frac {{\left (a + b\right )} \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {{\left (a + b\right )} \tan \left (x\right )^{3} + 3 \, {\left (a + 2 \, b\right )} \tan \left (x\right )}{3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} \]

[In]

integrate(sec(x)^4/(a+b*sin(x)^2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (49) = 98\).

Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.27 \[ \int \frac {\sec ^4(x)}{a+b \sin ^2(x)} \, dx=\frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (x\right ) + b \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )} b^{2}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a^{2} + a b}} + \frac {a^{2} \tan \left (x\right )^{3} + 2 \, a b \tan \left (x\right )^{3} + b^{2} \tan \left (x\right )^{3} + 3 \, a^{2} \tan \left (x\right ) + 9 \, a b \tan \left (x\right ) + 6 \, b^{2} \tan \left (x\right )}{3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} \]

[In]

integrate(sec(x)^4/(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^2/((a^2 + 2*a*b + b^2)
*sqrt(a^2 + a*b)) + 1/3*(a^2*tan(x)^3 + 2*a*b*tan(x)^3 + b^2*tan(x)^3 + 3*a^2*tan(x) + 9*a*b*tan(x) + 6*b^2*ta
n(x))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3)

Mupad [B] (verification not implemented)

Time = 14.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.31 \[ \int \frac {\sec ^4(x)}{a+b \sin ^2(x)} \, dx=\frac {{\mathrm {tan}\left (x\right )}^3}{3\,\left (a+b\right )}-\mathrm {tan}\left (x\right )\,\left (\frac {a}{{\left (a+b\right )}^2}-\frac {2}{a+b}\right )+\frac {b^2\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (x\right )\,\left (2\,a+2\,b\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{5/2}}\right )}{\sqrt {a}\,{\left (a+b\right )}^{5/2}} \]

[In]

int(1/(cos(x)^4*(a + b*sin(x)^2)),x)

[Out]

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