3.4.23 \(\int \frac {\sec ^4(x)}{(a+b \sin ^2(x))^2} \, dx\) [323]

Optimal. Leaf size=96 \[ \frac {b^2 (6 a+b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^{7/2}}+\frac {(a+3 b) \tan (x)}{(a+b)^3}+\frac {\tan ^3(x)}{3 (a+b)^2}+\frac {b^3 \tan (x)}{2 a (a+b)^3 \left (a+(a+b) \tan ^2(x)\right )} \]

[Out]

1/2*b^2*(6*a+b)*arctan((a+b)^(1/2)*tan(x)/a^(1/2))/a^(3/2)/(a+b)^(7/2)+(a+3*b)*tan(x)/(a+b)^3+1/3*tan(x)^3/(a+
b)^2+1/2*b^3*tan(x)/a/(a+b)^3/(a+(a+b)*tan(x)^2)

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Rubi [A]
time = 0.09, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3270, 398, 393, 211} \begin {gather*} \frac {b^2 (6 a+b) \text {ArcTan}\left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^{7/2}}+\frac {b^3 \tan (x)}{2 a (a+b)^3 \left ((a+b) \tan ^2(x)+a\right )}+\frac {\tan ^3(x)}{3 (a+b)^2}+\frac {(a+3 b) \tan (x)}{(a+b)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[x]^4/(a + b*Sin[x]^2)^2,x]

[Out]

(b^2*(6*a + b)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^(7/2)) + ((a + 3*b)*Tan[x])/(a + b)^3
+ Tan[x]^3/(3*(a + b)^2) + (b^3*Tan[x])/(2*a*(a + b)^3*(a + (a + b)*Tan[x]^2))

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 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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

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Mathematica [A]
time = 0.70, size = 97, normalized size = 1.01 \begin {gather*} \frac {1}{6} \left (\frac {3 b^2 (6 a+b) \tan ^{-1}\left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{a^{3/2} (a+b)^{7/2}}+\frac {\frac {3 b^3 \sin (2 x)}{a (2 a+b-b \cos (2 x))}+4 a \tan (x)+16 b \tan (x)+2 (a+b) \sec ^2(x) \tan (x)}{(a+b)^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b^2*(6*a + b)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(a^(3/2)*(a + b)^(7/2)) + ((3*b^3*Sin[2*x])/(a*(2*a +
b - b*Cos[2*x])) + 4*a*Tan[x] + 16*b*Tan[x] + 2*(a + b)*Sec[x]^2*Tan[x])/(a + b)^3)/6

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Maple [A]
time = 0.37, size = 110, normalized size = 1.15

method	result	size
default	\(\frac {\frac {a \left (\tan ^{3}\left (x \right )\right )}{3}+\frac {b \left (\tan ^{3}\left (x \right )\right )}{3}+\tan \left (x \right ) a +3 \tan \left (x \right ) b}{\left (a^{2}+2 a b +b^{2}\right ) \left (a +b \right )}+\frac {b^{2} \left (\frac {b \tan \left (x \right )}{2 a \left (a \left (\tan ^{2}\left (x \right )\right )+b \left (\tan ^{2}\left (x \right )\right )+a \right )}+\frac {\left (6 a +b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (x \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 a \sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{3}}\)	\(110\)
risch	\(\frac {i \left (-18 a \,b^{2} {\mathrm e}^{8 i x}-3 b^{3} {\mathrm e}^{8 i x}+36 a^{2} b \,{\mathrm e}^{6 i x}-30 a \,b^{2} {\mathrm e}^{6 i x}-6 b^{3} {\mathrm e}^{6 i x}+48 a^{3} {\mathrm e}^{4 i x}+164 a^{2} b \,{\mathrm e}^{4 i x}+26 a \,b^{2} {\mathrm e}^{4 i x}+16 a^{3} {\mathrm e}^{2 i x}+60 a^{2} b \,{\mathrm e}^{2 i x}-10 a \,b^{2} {\mathrm e}^{2 i x}+6 b^{3} {\mathrm e}^{2 i x}-4 a^{2} b -16 a \,b^{2}+3 b^{3}\right )}{3 \left ({\mathrm e}^{2 i x}+1\right )^{3} \left (a +b \right )^{3} a \left (-b \,{\mathrm e}^{4 i x}+4 a \,{\mathrm e}^{2 i x}+2 b \,{\mathrm e}^{2 i x}-b \right )}-\frac {3 b^{2} \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{3}}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{3} a}+\frac {3 b^{2} \ln \left ({\mathrm e}^{2 i x}-\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{3}}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i x}-\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{3} a}\)	\(551\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a^2+2*a*b+b^2)/(a+b)*(1/3*a*tan(x)^3+1/3*b*tan(x)^3+tan(x)*a+3*tan(x)*b)+b^2/(a+b)^3*(1/2/a*b*tan(x)/(a*tan
(x)^2+b*tan(x)^2+a)+1/2*(6*a+b)/a/(a*(a+b))^(1/2)*arctan((a+b)*tan(x)/(a*(a+b))^(1/2)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (82) = 164\).
time = 0.49, size = 170, normalized size = 1.77 \begin {gather*} \frac {b^{3} \tan \left (x\right )}{2 \, {\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3} + {\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \tan \left (x\right )^{2}\right )}} + \frac {{\left (6 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{2 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {{\left (a + b\right )} \tan \left (x\right )^{3} + 3 \, {\left (a + 3 \, b\right )} \tan \left (x\right )}{3 \, {\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)^4/(a+b*sin(x)^2)^2,x, algorithm="maxima")

[Out]

1/2*b^3*tan(x)/(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3 + (a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*tan(x)^2
) + 1/2*(6*a*b^2 + b^3)*arctan((a + b)*tan(x)/sqrt((a + b)*a))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*sqrt((a +
b)*a)) + 1/3*((a + b)*tan(x)^3 + 3*(a + 3*b)*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. 284 vs. \(2 (82) = 164\).
time = 0.44, size = 653, normalized size = 6.80 \begin {gather*} \left [-\frac {3 \, {\left ({\left (6 \, a b^{3} + b^{4}\right )} \cos \left (x\right )^{5} - {\left (6 \, a^{2} b^{2} + 7 \, a b^{3} + b^{4}\right )} \cos \left (x\right )^{3}\right )} \sqrt {-a^{2} - a b} \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 (2 \, a^{5} + 6 \, a^{4} b + 6 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - {\left (4 \, a^{4} b + 20 \, a^{3} b^{2} + 13 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (x\right )^{4} + 2 \, {\left (2 \, a^{5} + 11 \, a^{4} b + 16 \, a^{3} b^{2} + 7 \, a^{2} b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{24 \, {\left ({\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} \cos \left (x\right )^{5} - {\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} \cos \left (x\right )^{3}\right )}}, -\frac {3 \, {\left ({\left (6 \, a b^{3} + b^{4}\right )} \cos \left (x\right )^{5} - {\left (6 \, a^{2} b^{2} + 7 \, a b^{3} + b^{4}\right )} \cos \left (x\right )^{3}\right )} \sqrt {a^{2} + a b} \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 ) + 2 \, {\left (2 \, a^{5} + 6 \, a^{4} b + 6 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - {\left (4 \, a^{4} b + 20 \, a^{3} b^{2} + 13 \, a^{2} b^{3} - 3 \, a b^{4}\right )} \cos \left (x\right )^{4} + 2 \, {\left (2 \, a^{5} + 11 \, a^{4} b + 16 \, a^{3} b^{2} + 7 \, a^{2} b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{12 \, {\left ({\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} \cos \left (x\right )^{5} - {\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} \cos \left (x\right )^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(3*((6*a*b^3 + b^4)*cos(x)^5 - (6*a^2*b^2 + 7*a*b^3 + b^4)*cos(x)^3)*sqrt(-a^2 - a*b)*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*(2*a^5 + 6*a^4
*b + 6*a^3*b^2 + 2*a^2*b^3 - (4*a^4*b + 20*a^3*b^2 + 13*a^2*b^3 - 3*a*b^4)*cos(x)^4 + 2*(2*a^5 + 11*a^4*b + 16
*a^3*b^2 + 7*a^2*b^3)*cos(x)^2)*sin(x))/((a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*cos(x)^5 - (a^7
+ 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*cos(x)^3), -1/12*(3*((6*a*b^3 + b^4)*cos(x)^5 - (6
*a^2*b^2 + 7*a*b^3 + b^4)*cos(x)^3)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(x)^2 - a - b)/(sqrt(a^2 + a*b)*c
os(x)*sin(x))) + 2*(2*a^5 + 6*a^4*b + 6*a^3*b^2 + 2*a^2*b^3 - (4*a^4*b + 20*a^3*b^2 + 13*a^2*b^3 - 3*a*b^4)*co
s(x)^4 + 2*(2*a^5 + 11*a^4*b + 16*a^3*b^2 + 7*a^2*b^3)*cos(x)^2)*sin(x))/((a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a
^3*b^4 + a^2*b^5)*cos(x)^5 - (a^7 + 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*cos(x)^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{4}{\left (x \right )}}{\left (a + b \sin ^{2}{\left (x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (82) = 164\).
time = 0.48, size = 270, normalized size = 2.81 \begin {gather*} \frac {b^{3} \tan \left (x\right )}{2 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} {\left (a \tan \left (x\right )^{2} + b \tan \left (x\right )^{2} + a\right )}} + \frac {{\left (6 \, a b^{2} + b^{3}\right )} {\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 )}}{2 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \sqrt {a^{2} + a b}} + \frac {a^{4} \tan \left (x\right )^{3} + 4 \, a^{3} b \tan \left (x\right )^{3} + 6 \, a^{2} b^{2} \tan \left (x\right )^{3} + 4 \, a b^{3} \tan \left (x\right )^{3} + b^{4} \tan \left (x\right )^{3} + 3 \, a^{4} \tan \left (x\right ) + 18 \, a^{3} b \tan \left (x\right ) + 36 \, a^{2} b^{2} \tan \left (x\right ) + 30 \, a b^{3} \tan \left (x\right ) + 9 \, b^{4} \tan \left (x\right )}{3 \, {\left (a^{6} + 6 \, a^{5} b + 15 \, a^{4} b^{2} + 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} + 6 \, a b^{5} + b^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*tan(x)/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*(a*tan(x)^2 + b*tan(x)^2 + a)) + 1/2*(6*a*b^2 + b^3)*(pi*f
loor(x/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(x) + b*tan(x))/sqrt(a^2 + a*b)))/((a^4 + 3*a^3*b + 3*a^2*b^2 +
a*b^3)*sqrt(a^2 + a*b)) + 1/3*(a^4*tan(x)^3 + 4*a^3*b*tan(x)^3 + 6*a^2*b^2*tan(x)^3 + 4*a*b^3*tan(x)^3 + b^4*
tan(x)^3 + 3*a^4*tan(x) + 18*a^3*b*tan(x) + 36*a^2*b^2*tan(x) + 30*a*b^3*tan(x) + 9*b^4*tan(x))/(a^6 + 6*a^5*b
+ 15*a^4*b^2 + 20*a^3*b^3 + 15*a^2*b^4 + 6*a*b^5 + b^6)

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Mupad [B]
time = 14.30, size = 176, normalized size = 1.83 \begin {gather*} \frac {{\mathrm {tan}\left (x\right )}^3}{3\,{\left (a+b\right )}^2}-\mathrm {tan}\left (x\right )\,\left (\frac {2\,a}{{\left (a+b\right )}^3}-\frac {3}{{\left (a+b\right )}^2}\right )+\frac {b^3\,\mathrm {tan}\left (x\right )}{2\,a\,\left ({\mathrm {tan}\left (x\right )}^2\,\left (a^4+4\,a^3\,b+6\,a^2\,b^2+4\,a\,b^3+b^4\right )+a\,b^3+3\,a^3\,b+a^4+3\,a^2\,b^2\right )}+\frac {b^2\,\mathrm {atan}\left (\frac {b^2\,\mathrm {tan}\left (x\right )\,\left (6\,a+b\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}\,\left (b^3+6\,a\,b^2\right )}\right )\,\left (6\,a+b\right )}{2\,a^{3/2}\,{\left (a+b\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^4*(a + b*sin(x)^2)^2),x)

[Out]

tan(x)^3/(3*(a + b)^2) - tan(x)*((2*a)/(a + b)^3 - 3/(a + b)^2) + (b^3*tan(x))/(2*a*(tan(x)^2*(4*a*b^3 + 4*a^3
*b + a^4 + b^4 + 6*a^2*b^2) + a*b^3 + 3*a^3*b + a^4 + 3*a^2*b^2)) + (b^2*atan((b^2*tan(x)*(6*a + b)*(2*a + 2*b
)*(3*a*b^2 + 3*a^2*b + a^3 + b^3))/(2*a^(1/2)*(a + b)^(7/2)*(6*a*b^2 + b^3)))*(6*a + b))/(2*a^(3/2)*(a + b)^(7
/2))

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