3.312 \(\int \frac {\sec ^5(x)}{a+b \sin ^2(x)} \, dx\)

Optimal. Leaf size=93 \[ \frac {\left (3 a^2+10 a b+15 b^2\right ) \tanh ^{-1}(\sin (x))}{8 (a+b)^3}+\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^3}+\frac {\tan (x) \sec ^3(x)}{4 (a+b)}+\frac {(3 a+7 b) \tan (x) \sec (x)}{8 (a+b)^2} \]

[Out]

1/8*(3*a^2+10*a*b+15*b^2)*arctanh(sin(x))/(a+b)^3+b^(5/2)*arctan(sin(x)*b^(1/2)/a^(1/2))/(a+b)^3/a^(1/2)+1/8*(
3*a+7*b)*sec(x)*tan(x)/(a+b)^2+1/4*sec(x)^3*tan(x)/(a+b)

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Rubi [A]  time = 0.14, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3190, 414, 527, 522, 206, 205} \[ \frac {\left (3 a^2+10 a b+15 b^2\right ) \tanh ^{-1}(\sin (x))}{8 (a+b)^3}+\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^3}+\frac {\tan (x) \sec ^3(x)}{4 (a+b)}+\frac {(3 a+7 b) \tan (x) \sec (x)}{8 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^5/(a + b*Sin[x]^2),x]

[Out]

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

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]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 3190

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

Rubi steps

\begin {align*} \int \frac {\sec ^5(x)}{a+b \sin ^2(x)} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\sin (x)\right )\\ &=\frac {\sec ^3(x) \tan (x)}{4 (a+b)}+\frac {\operatorname {Subst}\left (\int \frac {3 a+4 b+3 b x^2}{\left (1-x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\sin (x)\right )}{4 (a+b)}\\ &=\frac {(3 a+7 b) \sec (x) \tan (x)}{8 (a+b)^2}+\frac {\sec ^3(x) \tan (x)}{4 (a+b)}+\frac {\operatorname {Subst}\left (\int \frac {3 a^2+7 a b+8 b^2+b (3 a+7 b) x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\sin (x)\right )}{8 (a+b)^2}\\ &=\frac {(3 a+7 b) \sec (x) \tan (x)}{8 (a+b)^2}+\frac {\sec ^3(x) \tan (x)}{4 (a+b)}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sin (x)\right )}{(a+b)^3}+\frac {\left (3 a^2+10 a b+15 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (x)\right )}{8 (a+b)^3}\\ &=\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^3}+\frac {\left (3 a^2+10 a b+15 b^2\right ) \tanh ^{-1}(\sin (x))}{8 (a+b)^3}+\frac {(3 a+7 b) \sec (x) \tan (x)}{8 (a+b)^2}+\frac {\sec ^3(x) \tan (x)}{4 (a+b)}\\ \end {align*}

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Mathematica [B]  time = 1.25, size = 214, normalized size = 2.30 \[ -\frac {2 \left (3 a^2+10 a b+15 b^2\right ) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-2 \left (3 a^2+10 a b+15 b^2\right ) \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )-\frac {8 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {8 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \csc (x)}{\sqrt {b}}\right )}{\sqrt {a}}+\frac {(a+b) (3 a+7 b)}{\sin (x)-1}+\frac {(a+b) (3 a+7 b)}{\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^2}-\frac {(a+b)^2}{\left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^4}+\frac {(a+b)^2}{\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^4}}{16 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^5/(a + b*Sin[x]^2),x]

[Out]

-1/16*((8*b^(5/2)*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]])/Sqrt[a] - (8*b^(5/2)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/Sqr
t[a] + 2*(3*a^2 + 10*a*b + 15*b^2)*Log[Cos[x/2] - Sin[x/2]] - 2*(3*a^2 + 10*a*b + 15*b^2)*Log[Cos[x/2] + Sin[x
/2]] - (a + b)^2/(Cos[x/2] - Sin[x/2])^4 + (a + b)^2/(Cos[x/2] + Sin[x/2])^4 + ((a + b)*(3*a + 7*b))/(Cos[x/2]
+ Sin[x/2])^2 + ((a + b)*(3*a + 7*b))/(-1 + Sin[x]))/(a + b)^3

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fricas [A]  time = 0.56, size = 327, normalized size = 3.52 \[ \left [\frac {8 \, b^{2} \sqrt {-\frac {b}{a}} \cos \relax (x)^{4} \log \left (-\frac {b \cos \relax (x)^{2} - 2 \, a \sqrt {-\frac {b}{a}} \sin \relax (x) + a - b}{b \cos \relax (x)^{2} - a - b}\right ) + {\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \relax (x)^{4} \log \left (\sin \relax (x) + 1\right ) - {\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \relax (x)^{4} \log \left (-\sin \relax (x) + 1\right ) + 2 \, {\left ({\left (3 \, a^{2} + 10 \, a b + 7 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sin \relax (x)}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \relax (x)^{4}}, \frac {16 \, b^{2} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {\frac {b}{a}} \sin \relax (x)\right ) \cos \relax (x)^{4} + {\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \relax (x)^{4} \log \left (\sin \relax (x) + 1\right ) - {\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \relax (x)^{4} \log \left (-\sin \relax (x) + 1\right ) + 2 \, {\left ({\left (3 \, a^{2} + 10 \, a b + 7 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sin \relax (x)}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \relax (x)^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^5/(a+b*sin(x)^2),x, algorithm="fricas")

[Out]

[1/16*(8*b^2*sqrt(-b/a)*cos(x)^4*log(-(b*cos(x)^2 - 2*a*sqrt(-b/a)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) + (3*
a^2 + 10*a*b + 15*b^2)*cos(x)^4*log(sin(x) + 1) - (3*a^2 + 10*a*b + 15*b^2)*cos(x)^4*log(-sin(x) + 1) + 2*((3*
a^2 + 10*a*b + 7*b^2)*cos(x)^2 + 2*a^2 + 4*a*b + 2*b^2)*sin(x))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(x)^4), 1/
16*(16*b^2*sqrt(b/a)*arctan(sqrt(b/a)*sin(x))*cos(x)^4 + (3*a^2 + 10*a*b + 15*b^2)*cos(x)^4*log(sin(x) + 1) -
(3*a^2 + 10*a*b + 15*b^2)*cos(x)^4*log(-sin(x) + 1) + 2*((3*a^2 + 10*a*b + 7*b^2)*cos(x)^2 + 2*a^2 + 4*a*b + 2
*b^2)*sin(x))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(x)^4)]

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giac [B]  time = 0.14, size = 177, normalized size = 1.90 \[ \frac {b^{3} \arctan \left (\frac {b \sin \relax (x)}{\sqrt {a b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b}} + \frac {{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (\sin \relax (x) + 1\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (-\sin \relax (x) + 1\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {3 \, a \sin \relax (x)^{3} + 7 \, b \sin \relax (x)^{3} - 5 \, a \sin \relax (x) - 9 \, b \sin \relax (x)}{8 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (\sin \relax (x)^{2} - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^5/(a+b*sin(x)^2),x, algorithm="giac")

[Out]

b^3*arctan(b*sin(x)/sqrt(a*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)) + 1/16*(3*a^2 + 10*a*b + 15*b^2)*lo
g(sin(x) + 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/16*(3*a^2 + 10*a*b + 15*b^2)*log(-sin(x) + 1)/(a^3 + 3*a^2*b
+ 3*a*b^2 + b^3) - 1/8*(3*a*sin(x)^3 + 7*b*sin(x)^3 - 5*a*sin(x) - 9*b*sin(x))/((a^2 + 2*a*b + b^2)*(sin(x)^2
- 1)^2)

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maple [B]  time = 0.32, size = 204, normalized size = 2.19 \[ \frac {1}{2 \left (8 a +8 b \right ) \left (-1+\sin \relax (x )\right )^{2}}-\frac {3 a}{16 \left (a +b \right )^{2} \left (-1+\sin \relax (x )\right )}-\frac {7 b}{16 \left (a +b \right )^{2} \left (-1+\sin \relax (x )\right )}-\frac {3 \ln \left (-1+\sin \relax (x )\right ) a^{2}}{16 \left (a +b \right )^{3}}-\frac {5 \ln \left (-1+\sin \relax (x )\right ) a b}{8 \left (a +b \right )^{3}}-\frac {15 \ln \left (-1+\sin \relax (x )\right ) b^{2}}{16 \left (a +b \right )^{3}}+\frac {b^{3} \arctan \left (\frac {\sin \relax (x ) b}{\sqrt {a b}}\right )}{\left (a +b \right )^{3} \sqrt {a b}}-\frac {1}{2 \left (8 a +8 b \right ) \left (1+\sin \relax (x )\right )^{2}}-\frac {3 a}{16 \left (a +b \right )^{2} \left (1+\sin \relax (x )\right )}-\frac {7 b}{16 \left (a +b \right )^{2} \left (1+\sin \relax (x )\right )}+\frac {3 \ln \left (1+\sin \relax (x )\right ) a^{2}}{16 \left (a +b \right )^{3}}+\frac {5 \ln \left (1+\sin \relax (x )\right ) a b}{8 \left (a +b \right )^{3}}+\frac {15 \ln \left (1+\sin \relax (x )\right ) b^{2}}{16 \left (a +b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^5/(a+b*sin(x)^2),x)

[Out]

1/2/(8*a+8*b)/(-1+sin(x))^2-3/16/(a+b)^2/(-1+sin(x))*a-7/16/(a+b)^2/(-1+sin(x))*b-3/16/(a+b)^3*ln(-1+sin(x))*a
^2-5/8/(a+b)^3*ln(-1+sin(x))*a*b-15/16/(a+b)^3*ln(-1+sin(x))*b^2+b^3/(a+b)^3/(a*b)^(1/2)*arctan(sin(x)*b/(a*b)
^(1/2))-1/2/(8*a+8*b)/(1+sin(x))^2-3/16/(a+b)^2/(1+sin(x))*a-7/16/(a+b)^2/(1+sin(x))*b+3/16/(a+b)^3*ln(1+sin(x
))*a^2+5/8/(a+b)^3*ln(1+sin(x))*a*b+15/16/(a+b)^3*ln(1+sin(x))*b^2

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maxima [B]  time = 0.48, size = 199, normalized size = 2.14 \[ \frac {b^{3} \arctan \left (\frac {b \sin \relax (x)}{\sqrt {a b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b}} + \frac {{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (\sin \relax (x) + 1\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (\sin \relax (x) - 1\right )}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac {{\left (3 \, a + 7 \, b\right )} \sin \relax (x)^{3} - {\left (5 \, a + 9 \, b\right )} \sin \relax (x)}{8 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \relax (x)^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \relax (x)^{2} + a^{2} + 2 \, a b + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^5/(a+b*sin(x)^2),x, algorithm="maxima")

[Out]

b^3*arctan(b*sin(x)/sqrt(a*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)) + 1/16*(3*a^2 + 10*a*b + 15*b^2)*lo
g(sin(x) + 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/16*(3*a^2 + 10*a*b + 15*b^2)*log(sin(x) - 1)/(a^3 + 3*a^2*b
+ 3*a*b^2 + b^3) - 1/8*((3*a + 7*b)*sin(x)^3 - (5*a + 9*b)*sin(x))/((a^2 + 2*a*b + b^2)*sin(x)^4 - 2*(a^2 + 2*
a*b + b^2)*sin(x)^2 + a^2 + 2*a*b + b^2)

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mupad [B]  time = 17.45, size = 832, normalized size = 8.95 \[ \frac {5\,a^3\,\sin \relax (x)-3\,a^3\,{\sin \relax (x)}^3+3\,a^3\,\mathrm {atanh}\left (\sin \relax (x)\right )+9\,a\,b^2\,\sin \relax (x)+14\,a^2\,b\,\sin \relax (x)-6\,a^3\,\mathrm {atanh}\left (\sin \relax (x)\right )\,{\sin \relax (x)}^2+3\,a^3\,\mathrm {atanh}\left (\sin \relax (x)\right )\,{\sin \relax (x)}^4-7\,a\,b^2\,{\sin \relax (x)}^3-10\,a^2\,b\,{\sin \relax (x)}^3+15\,a\,b^2\,\mathrm {atanh}\left (\sin \relax (x)\right )+10\,a^2\,b\,\mathrm {atanh}\left (\sin \relax (x)\right )-30\,a\,b^2\,\mathrm {atanh}\left (\sin \relax (x)\right )\,{\sin \relax (x)}^2-20\,a^2\,b\,\mathrm {atanh}\left (\sin \relax (x)\right )\,{\sin \relax (x)}^2+15\,a\,b^2\,\mathrm {atanh}\left (\sin \relax (x)\right )\,{\sin \relax (x)}^4+10\,a^2\,b\,\mathrm {atanh}\left (\sin \relax (x)\right )\,{\sin \relax (x)}^4+\mathrm {atan}\left (\frac {a\,\sin \relax (x)\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}-b\,\sin \relax (x)\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}+a^6\,b\,\sin \relax (x)\,\sqrt {-a\,b^5}\,9{}\mathrm {i}+a^2\,b^5\,\sin \relax (x)\,\sqrt {-a\,b^5}\,289{}\mathrm {i}+a^3\,b^4\,\sin \relax (x)\,\sqrt {-a\,b^5}\,300{}\mathrm {i}+a^4\,b^3\,\sin \relax (x)\,\sqrt {-a\,b^5}\,190{}\mathrm {i}+a^5\,b^2\,\sin \relax (x)\,\sqrt {-a\,b^5}\,60{}\mathrm {i}}{9\,a^7\,b^3+60\,a^6\,b^4+190\,a^5\,b^5+300\,a^4\,b^6+225\,a^3\,b^7+64\,a^2\,b^8}\right )\,\sqrt {-a\,b^5}\,8{}\mathrm {i}-\mathrm {atan}\left (\frac {a\,\sin \relax (x)\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}-b\,\sin \relax (x)\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}+a^6\,b\,\sin \relax (x)\,\sqrt {-a\,b^5}\,9{}\mathrm {i}+a^2\,b^5\,\sin \relax (x)\,\sqrt {-a\,b^5}\,289{}\mathrm {i}+a^3\,b^4\,\sin \relax (x)\,\sqrt {-a\,b^5}\,300{}\mathrm {i}+a^4\,b^3\,\sin \relax (x)\,\sqrt {-a\,b^5}\,190{}\mathrm {i}+a^5\,b^2\,\sin \relax (x)\,\sqrt {-a\,b^5}\,60{}\mathrm {i}}{9\,a^7\,b^3+60\,a^6\,b^4+190\,a^5\,b^5+300\,a^4\,b^6+225\,a^3\,b^7+64\,a^2\,b^8}\right )\,{\sin \relax (x)}^2\,\sqrt {-a\,b^5}\,16{}\mathrm {i}+\mathrm {atan}\left (\frac {a\,\sin \relax (x)\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}-b\,\sin \relax (x)\,{\left (-a\,b^5\right )}^{3/2}\,64{}\mathrm {i}+a^6\,b\,\sin \relax (x)\,\sqrt {-a\,b^5}\,9{}\mathrm {i}+a^2\,b^5\,\sin \relax (x)\,\sqrt {-a\,b^5}\,289{}\mathrm {i}+a^3\,b^4\,\sin \relax (x)\,\sqrt {-a\,b^5}\,300{}\mathrm {i}+a^4\,b^3\,\sin \relax (x)\,\sqrt {-a\,b^5}\,190{}\mathrm {i}+a^5\,b^2\,\sin \relax (x)\,\sqrt {-a\,b^5}\,60{}\mathrm {i}}{9\,a^7\,b^3+60\,a^6\,b^4+190\,a^5\,b^5+300\,a^4\,b^6+225\,a^3\,b^7+64\,a^2\,b^8}\right )\,{\sin \relax (x)}^4\,\sqrt {-a\,b^5}\,8{}\mathrm {i}}{8\,a^4\,{\sin \relax (x)}^4-16\,a^4\,{\sin \relax (x)}^2+8\,a^4+24\,a^3\,b\,{\sin \relax (x)}^4-48\,a^3\,b\,{\sin \relax (x)}^2+24\,a^3\,b+24\,a^2\,b^2\,{\sin \relax (x)}^4-48\,a^2\,b^2\,{\sin \relax (x)}^2+24\,a^2\,b^2+8\,a\,b^3\,{\sin \relax (x)}^4-16\,a\,b^3\,{\sin \relax (x)}^2+8\,a\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^5*(a + b*sin(x)^2)),x)

[Out]

(5*a^3*sin(x) - 3*a^3*sin(x)^3 + 3*a^3*atanh(sin(x)) + atan((a*sin(x)*(-a*b^5)^(3/2)*64i - b*sin(x)*(-a*b^5)^(
3/2)*64i + a^6*b*sin(x)*(-a*b^5)^(1/2)*9i + a^2*b^5*sin(x)*(-a*b^5)^(1/2)*289i + a^3*b^4*sin(x)*(-a*b^5)^(1/2)
*300i + a^4*b^3*sin(x)*(-a*b^5)^(1/2)*190i + a^5*b^2*sin(x)*(-a*b^5)^(1/2)*60i)/(64*a^2*b^8 + 225*a^3*b^7 + 30
0*a^4*b^6 + 190*a^5*b^5 + 60*a^6*b^4 + 9*a^7*b^3))*(-a*b^5)^(1/2)*8i + 9*a*b^2*sin(x) + 14*a^2*b*sin(x) - 6*a^
3*atanh(sin(x))*sin(x)^2 + 3*a^3*atanh(sin(x))*sin(x)^4 - 7*a*b^2*sin(x)^3 - 10*a^2*b*sin(x)^3 + 15*a*b^2*atan
h(sin(x)) + 10*a^2*b*atanh(sin(x)) - atan((a*sin(x)*(-a*b^5)^(3/2)*64i - b*sin(x)*(-a*b^5)^(3/2)*64i + a^6*b*s
in(x)*(-a*b^5)^(1/2)*9i + a^2*b^5*sin(x)*(-a*b^5)^(1/2)*289i + a^3*b^4*sin(x)*(-a*b^5)^(1/2)*300i + a^4*b^3*si
n(x)*(-a*b^5)^(1/2)*190i + a^5*b^2*sin(x)*(-a*b^5)^(1/2)*60i)/(64*a^2*b^8 + 225*a^3*b^7 + 300*a^4*b^6 + 190*a^
5*b^5 + 60*a^6*b^4 + 9*a^7*b^3))*sin(x)^2*(-a*b^5)^(1/2)*16i + atan((a*sin(x)*(-a*b^5)^(3/2)*64i - b*sin(x)*(-
a*b^5)^(3/2)*64i + a^6*b*sin(x)*(-a*b^5)^(1/2)*9i + a^2*b^5*sin(x)*(-a*b^5)^(1/2)*289i + a^3*b^4*sin(x)*(-a*b^
5)^(1/2)*300i + a^4*b^3*sin(x)*(-a*b^5)^(1/2)*190i + a^5*b^2*sin(x)*(-a*b^5)^(1/2)*60i)/(64*a^2*b^8 + 225*a^3*
b^7 + 300*a^4*b^6 + 190*a^5*b^5 + 60*a^6*b^4 + 9*a^7*b^3))*sin(x)^4*(-a*b^5)^(1/2)*8i - 30*a*b^2*atanh(sin(x))
*sin(x)^2 - 20*a^2*b*atanh(sin(x))*sin(x)^2 + 15*a*b^2*atanh(sin(x))*sin(x)^4 + 10*a^2*b*atanh(sin(x))*sin(x)^
4)/(8*a^4*sin(x)^4 - 16*a^4*sin(x)^2 + 8*a*b^3 + 24*a^3*b + 8*a^4 + 24*a^2*b^2 - 48*a^2*b^2*sin(x)^2 + 24*a^2*
b^2*sin(x)^4 - 16*a*b^3*sin(x)^2 - 48*a^3*b*sin(x)^2 + 8*a*b^3*sin(x)^4 + 24*a^3*b*sin(x)^4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**5/(a+b*sin(x)**2),x)

[Out]

Timed out

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