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

Optimal. Leaf size=90 \[ -\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b}}+\frac {(3 a-4 b) \tan (x) \sec (x)}{8 a^2}+\frac {\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\sin (x))}{8 a^3}+\frac {\tan (x) \sec ^3(x)}{4 a} \]

[Out]

1/8*(3*a^2-4*a*b+8*b^2)*arctanh(sin(x))/a^3-b^(5/2)*arctanh(sin(x)*b^(1/2)/(a+b)^(1/2))/a^3/(a+b)^(1/2)+1/8*(3
*a-4*b)*sec(x)*tan(x)/a^2+1/4*sec(x)^3*tan(x)/a

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Rubi [A]  time = 0.17, antiderivative size = 90, 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 = {3186, 414, 527, 522, 206, 208} \[ \frac {\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\sin (x))}{8 a^3}-\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b}}+\frac {(3 a-4 b) \tan (x) \sec (x)}{8 a^2}+\frac {\tan (x) \sec ^3(x)}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^5/(a + b*Cos[x]^2),x]

[Out]

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

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 208

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

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 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[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 \cos ^2(x)} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right )^3 \left (a+b-b x^2\right )} \, dx,x,\sin (x)\right )\\ &=\frac {\sec ^3(x) \tan (x)}{4 a}+\frac {\operatorname {Subst}\left (\int \frac {3 a-b-3 b x^2}{\left (1-x^2\right )^2 \left (a+b-b x^2\right )} \, dx,x,\sin (x)\right )}{4 a}\\ &=\frac {(3 a-4 b) \sec (x) \tan (x)}{8 a^2}+\frac {\sec ^3(x) \tan (x)}{4 a}+\frac {\operatorname {Subst}\left (\int \frac {3 a^2-a b+4 b^2-(3 a-4 b) b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\sin (x)\right )}{8 a^2}\\ &=\frac {(3 a-4 b) \sec (x) \tan (x)}{8 a^2}+\frac {\sec ^3(x) \tan (x)}{4 a}-\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\sin (x)\right )}{a^3}+\frac {\left (3 a^2-4 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (x)\right )}{8 a^3}\\ &=\frac {\left (3 a^2-4 a b+8 b^2\right ) \tanh ^{-1}(\sin (x))}{8 a^3}-\frac {b^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^3 \sqrt {a+b}}+\frac {(3 a-4 b) \sec (x) \tan (x)}{8 a^2}+\frac {\sec ^3(x) \tan (x)}{4 a}\\ \end {align*}

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Mathematica [B]  time = 1.18, size = 215, normalized size = 2.39 \[ \frac {-2 \left (3 a^2-4 a b+8 b^2\right ) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+2 \left (3 a^2-4 a b+8 b^2\right ) \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )+\frac {a^2}{\left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^4}-\frac {a^2}{\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^4}+\frac {8 b^{5/2} \log \left (\sqrt {a+b}-\sqrt {b} \sin (x)\right )}{\sqrt {a+b}}-\frac {8 b^{5/2} \log \left (\sqrt {a+b}+\sqrt {b} \sin (x)\right )}{\sqrt {a+b}}+\frac {a (4 b-3 a)}{\sin (x)-1}+\frac {a (4 b-3 a)}{\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^2}}{16 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*a^2 - 4*a*b + 8*b^2)*Log[Cos[x/2] - Sin[x/2]] + 2*(3*a^2 - 4*a*b + 8*b^2)*Log[Cos[x/2] + Sin[x/2]] + (8
*b^(5/2)*Log[Sqrt[a + b] - Sqrt[b]*Sin[x]])/Sqrt[a + b] - (8*b^(5/2)*Log[Sqrt[a + b] + Sqrt[b]*Sin[x]])/Sqrt[a
+ b] + a^2/(Cos[x/2] - Sin[x/2])^4 - a^2/(Cos[x/2] + Sin[x/2])^4 + (a*(-3*a + 4*b))/(Cos[x/2] + Sin[x/2])^2 +
(a*(-3*a + 4*b))/(-1 + Sin[x]))/(16*a^3)

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fricas [A]  time = 0.94, size = 270, normalized size = 3.00 \[ \left [\frac {8 \, b^{2} \sqrt {\frac {b}{a + b}} \cos \relax (x)^{4} \log \left (-\frac {b \cos \relax (x)^{2} + 2 \, {\left (a + b\right )} \sqrt {\frac {b}{a + b}} \sin \relax (x) - a - 2 \, b}{b \cos \relax (x)^{2} + a}\right ) + {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \relax (x)^{4} \log \left (\sin \relax (x) + 1\right ) - {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \relax (x)^{4} \log \left (-\sin \relax (x) + 1\right ) + 2 \, {\left ({\left (3 \, a^{2} - 4 \, a b\right )} \cos \relax (x)^{2} + 2 \, a^{2}\right )} \sin \relax (x)}{16 \, a^{3} \cos \relax (x)^{4}}, \frac {16 \, b^{2} \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \sin \relax (x)\right ) \cos \relax (x)^{4} + {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \relax (x)^{4} \log \left (\sin \relax (x) + 1\right ) - {\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \relax (x)^{4} \log \left (-\sin \relax (x) + 1\right ) + 2 \, {\left ({\left (3 \, a^{2} - 4 \, a b\right )} \cos \relax (x)^{2} + 2 \, a^{2}\right )} \sin \relax (x)}{16 \, a^{3} \cos \relax (x)^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 127, normalized size = 1.41 \[ \frac {b^{3} \arctan \left (\frac {b \sin \relax (x)}{\sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{3}} + \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\sin \relax (x) + 1\right )}{16 \, a^{3}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (-\sin \relax (x) + 1\right )}{16 \, a^{3}} - \frac {3 \, a \sin \relax (x)^{3} - 4 \, b \sin \relax (x)^{3} - 5 \, a \sin \relax (x) + 4 \, b \sin \relax (x)}{8 \, {\left (\sin \relax (x)^{2} - 1\right )}^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*arctan(b*sin(x)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a^3) + 1/16*(3*a^2 - 4*a*b + 8*b^2)*log(sin(x) + 1)/a^
3 - 1/16*(3*a^2 - 4*a*b + 8*b^2)*log(-sin(x) + 1)/a^3 - 1/8*(3*a*sin(x)^3 - 4*b*sin(x)^3 - 5*a*sin(x) + 4*b*si
n(x))/((sin(x)^2 - 1)^2*a^2)

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maple [B]  time = 0.13, size = 165, normalized size = 1.83 \[ -\frac {b^{3} \arctanh \left (\frac {\sin \relax (x ) b}{\sqrt {\left (a +b \right ) b}}\right )}{a^{3} \sqrt {\left (a +b \right ) b}}+\frac {1}{16 a \left (\sin \relax (x )-1\right )^{2}}-\frac {3}{16 a \left (\sin \relax (x )-1\right )}+\frac {b}{4 a^{2} \left (\sin \relax (x )-1\right )}-\frac {3 \ln \left (\sin \relax (x )-1\right )}{16 a}+\frac {\ln \left (\sin \relax (x )-1\right ) b}{4 a^{2}}-\frac {\ln \left (\sin \relax (x )-1\right ) b^{2}}{2 a^{3}}-\frac {1}{16 a \left (\sin \relax (x )+1\right )^{2}}-\frac {3}{16 a \left (\sin \relax (x )+1\right )}+\frac {b}{4 a^{2} \left (\sin \relax (x )+1\right )}+\frac {3 \ln \left (\sin \relax (x )+1\right )}{16 a}-\frac {\ln \left (\sin \relax (x )+1\right ) b}{4 a^{2}}+\frac {\ln \left (\sin \relax (x )+1\right ) b^{2}}{2 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^3/a^3/((a+b)*b)^(1/2)*arctanh(sin(x)*b/((a+b)*b)^(1/2))+1/16/a/(sin(x)-1)^2-3/16/a/(sin(x)-1)+1/4/a^2/(sin(
x)-1)*b-3/16/a*ln(sin(x)-1)+1/4/a^2*ln(sin(x)-1)*b-1/2/a^3*ln(sin(x)-1)*b^2-1/16/a/(sin(x)+1)^2-3/16/a/(sin(x)
+1)+1/4/a^2/(sin(x)+1)*b+3/16/a*ln(sin(x)+1)-1/4/a^2*ln(sin(x)+1)*b+1/2/a^3*ln(sin(x)+1)*b^2

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maxima [A]  time = 1.99, size = 145, normalized size = 1.61 \[ \frac {b^{3} \log \left (\frac {b \sin \relax (x) - \sqrt {{\left (a + b\right )} b}}{b \sin \relax (x) + \sqrt {{\left (a + b\right )} b}}\right )}{2 \, \sqrt {{\left (a + b\right )} b} a^{3}} - \frac {{\left (3 \, a - 4 \, b\right )} \sin \relax (x)^{3} - {\left (5 \, a - 4 \, b\right )} \sin \relax (x)}{8 \, {\left (a^{2} \sin \relax (x)^{4} - 2 \, a^{2} \sin \relax (x)^{2} + a^{2}\right )}} + \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\sin \relax (x) + 1\right )}{16 \, a^{3}} - \frac {{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (\sin \relax (x) - 1\right )}{16 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*log((b*sin(x) - sqrt((a + b)*b))/(b*sin(x) + sqrt((a + b)*b)))/(sqrt((a + b)*b)*a^3) - 1/8*((3*a - 4*b
)*sin(x)^3 - (5*a - 4*b)*sin(x))/(a^2*sin(x)^4 - 2*a^2*sin(x)^2 + a^2) + 1/16*(3*a^2 - 4*a*b + 8*b^2)*log(sin(
x) + 1)/a^3 - 1/16*(3*a^2 - 4*a*b + 8*b^2)*log(sin(x) - 1)/a^3

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mupad [B]  time = 2.64, size = 969, normalized size = 10.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^5*(a + b*cos(x)^2)),x)

[Out]

(5*a^3*sin(x) + atan((b^7*sin(x)*(a*b^5 + b^6)^(1/2)*128i - a*sin(x)*(a*b^5 + b^6)^(3/2)*64i - b*sin(x)*(a*b^5
+ b^6)^(3/2)*128i + a*b^6*sin(x)*(a*b^5 + b^6)^(1/2)*192i + a^6*b*sin(x)*(a*b^5 + b^6)^(1/2)*9i + a^2*b^5*sin
(x)*(a*b^5 + b^6)^(1/2)*64i + a^3*b^4*sin(x)*(a*b^5 + b^6)^(1/2)*40i + a^4*b^3*sin(x)*(a*b^5 + b^6)^(1/2)*25i
- a^5*b^2*sin(x)*(a*b^5 + b^6)^(1/2)*6i)/(40*a^3*b^7 + 65*a^4*b^6 + 19*a^5*b^5 + 3*a^6*b^4 + 9*a^7*b^3))*(a*b^
5 + b^6)^(1/2)*8i - 3*a^3*sin(x)^3 + 3*a^3*atanh(sin(x)) + 8*b^3*atanh(sin(x)) - 4*a*b^2*sin(x) + a^2*b*sin(x)
- atan((b^7*sin(x)*(a*b^5 + b^6)^(1/2)*128i - a*sin(x)*(a*b^5 + b^6)^(3/2)*64i - b*sin(x)*(a*b^5 + b^6)^(3/2)
*128i + a*b^6*sin(x)*(a*b^5 + b^6)^(1/2)*192i + a^6*b*sin(x)*(a*b^5 + b^6)^(1/2)*9i + a^2*b^5*sin(x)*(a*b^5 +
b^6)^(1/2)*64i + a^3*b^4*sin(x)*(a*b^5 + b^6)^(1/2)*40i + a^4*b^3*sin(x)*(a*b^5 + b^6)^(1/2)*25i - a^5*b^2*sin
(x)*(a*b^5 + b^6)^(1/2)*6i)/(40*a^3*b^7 + 65*a^4*b^6 + 19*a^5*b^5 + 3*a^6*b^4 + 9*a^7*b^3))*sin(x)^2*(a*b^5 +
b^6)^(1/2)*16i + atan((b^7*sin(x)*(a*b^5 + b^6)^(1/2)*128i - a*sin(x)*(a*b^5 + b^6)^(3/2)*64i - b*sin(x)*(a*b^
5 + b^6)^(3/2)*128i + a*b^6*sin(x)*(a*b^5 + b^6)^(1/2)*192i + a^6*b*sin(x)*(a*b^5 + b^6)^(1/2)*9i + a^2*b^5*si
n(x)*(a*b^5 + b^6)^(1/2)*64i + a^3*b^4*sin(x)*(a*b^5 + b^6)^(1/2)*40i + a^4*b^3*sin(x)*(a*b^5 + b^6)^(1/2)*25i
- a^5*b^2*sin(x)*(a*b^5 + b^6)^(1/2)*6i)/(40*a^3*b^7 + 65*a^4*b^6 + 19*a^5*b^5 + 3*a^6*b^4 + 9*a^7*b^3))*sin(
x)^4*(a*b^5 + b^6)^(1/2)*8i - 6*a^3*atanh(sin(x))*sin(x)^2 + 3*a^3*atanh(sin(x))*sin(x)^4 - 16*b^3*atanh(sin(x
))*sin(x)^2 + 8*b^3*atanh(sin(x))*sin(x)^4 + 4*a*b^2*sin(x)^3 + a^2*b*sin(x)^3 + 4*a*b^2*atanh(sin(x)) - a^2*b
*atanh(sin(x)) - 8*a*b^2*atanh(sin(x))*sin(x)^2 + 2*a^2*b*atanh(sin(x))*sin(x)^2 + 4*a*b^2*atanh(sin(x))*sin(x
)^4 - a^2*b*atanh(sin(x))*sin(x)^4)/(8*a^4*sin(x)^4 - 16*a^4*sin(x)^2 + 8*a^3*b + 8*a^4 - 16*a^3*b*sin(x)^2 +
8*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*cos(x)**2),x)

[Out]

Timed out

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