3.33 \(\int \frac {\sec ^3(x)}{a+b \cos ^2(x)} \, dx\)

Optimal. Leaf size=59 \[ \frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^2 \sqrt {a+b}}+\frac {(a-2 b) \tanh ^{-1}(\sin (x))}{2 a^2}+\frac {\tan (x) \sec (x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3186, 414, 522, 206, 208} \[ \frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^2 \sqrt {a+b}}+\frac {(a-2 b) \tanh ^{-1}(\sin (x))}{2 a^2}+\frac {\tan (x) \sec (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^3/(a + b*Cos[x]^2),x]

[Out]

((a - 2*b)*ArcTanh[Sin[x]])/(2*a^2) + (b^(3/2)*ArcTanh[(Sqrt[b]*Sin[x])/Sqrt[a + b]])/(a^2*Sqrt[a + b]) + (Sec
[x]*Tan[x])/(2*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 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 ^3(x)}{a+b \cos ^2(x)} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2 \left (a+b-b x^2\right )} \, dx,x,\sin (x)\right )\\ &=\frac {\sec (x) \tan (x)}{2 a}+\frac {\operatorname {Subst}\left (\int \frac {a-b-b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\sin (x)\right )}{2 a}\\ &=\frac {\sec (x) \tan (x)}{2 a}+\frac {(a-2 b) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (x)\right )}{2 a^2}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\sin (x)\right )}{a^2}\\ &=\frac {(a-2 b) \tanh ^{-1}(\sin (x))}{2 a^2}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a^2 \sqrt {a+b}}+\frac {\sec (x) \tan (x)}{2 a}\\ \end {align*}

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Mathematica [B]  time = 0.39, size = 152, normalized size = 2.58 \[ \frac {-\frac {2 b^{3/2} \log \left (\sqrt {a+b}-\sqrt {b} \sin (x)\right )}{\sqrt {a+b}}+\frac {2 b^{3/2} \log \left (\sqrt {a+b}+\sqrt {b} \sin (x)\right )}{\sqrt {a+b}}-2 (a-2 b) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+2 (a-2 b) \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )+\frac {a}{\left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^2}-\frac {a}{\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^2}}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^3/(a + b*Cos[x]^2),x]

[Out]

(-2*(a - 2*b)*Log[Cos[x/2] - Sin[x/2]] + 2*(a - 2*b)*Log[Cos[x/2] + Sin[x/2]] - (2*b^(3/2)*Log[Sqrt[a + b] - S
qrt[b]*Sin[x]])/Sqrt[a + b] + (2*b^(3/2)*Log[Sqrt[a + b] + Sqrt[b]*Sin[x]])/Sqrt[a + b] + a/(Cos[x/2] - Sin[x/
2])^2 - a/(Cos[x/2] + Sin[x/2])^2)/(4*a^2)

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fricas [A]  time = 0.67, size = 186, normalized size = 3.15 \[ \left [\frac {2 \, b \sqrt {\frac {b}{a + b}} \cos \relax (x)^{2} \log \left (-\frac {b \cos \relax (x)^{2} - 2 \, {\left (a + b\right )} \sqrt {\frac {b}{a + b}} \sin \relax (x) - a - 2 \, b}{b \cos \relax (x)^{2} + a}\right ) + {\left (a - 2 \, b\right )} \cos \relax (x)^{2} \log \left (\sin \relax (x) + 1\right ) - {\left (a - 2 \, b\right )} \cos \relax (x)^{2} \log \left (-\sin \relax (x) + 1\right ) + 2 \, a \sin \relax (x)}{4 \, a^{2} \cos \relax (x)^{2}}, -\frac {4 \, b \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \sin \relax (x)\right ) \cos \relax (x)^{2} - {\left (a - 2 \, b\right )} \cos \relax (x)^{2} \log \left (\sin \relax (x) + 1\right ) + {\left (a - 2 \, b\right )} \cos \relax (x)^{2} \log \left (-\sin \relax (x) + 1\right ) - 2 \, a \sin \relax (x)}{4 \, a^{2} \cos \relax (x)^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+b*cos(x)^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 85, normalized size = 1.44 \[ -\frac {b^{2} \arctan \left (\frac {b \sin \relax (x)}{\sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a^{2}} + \frac {{\left (a - 2 \, b\right )} \log \left (\sin \relax (x) + 1\right )}{4 \, a^{2}} - \frac {{\left (a - 2 \, b\right )} \log \left (-\sin \relax (x) + 1\right )}{4 \, a^{2}} - \frac {\sin \relax (x)}{2 \, {\left (\sin \relax (x)^{2} - 1\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+b*cos(x)^2),x, algorithm="giac")

[Out]

-b^2*arctan(b*sin(x)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a^2) + 1/4*(a - 2*b)*log(sin(x) + 1)/a^2 - 1/4*(a - 2
*b)*log(-sin(x) + 1)/a^2 - 1/2*sin(x)/((sin(x)^2 - 1)*a)

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maple [A]  time = 0.12, size = 92, normalized size = 1.56 \[ \frac {b^{2} \arctanh \left (\frac {\sin \relax (x ) b}{\sqrt {\left (a +b \right ) b}}\right )}{a^{2} \sqrt {\left (a +b \right ) b}}-\frac {1}{4 a \left (\sin \relax (x )-1\right )}-\frac {\ln \left (\sin \relax (x )-1\right )}{4 a}+\frac {\ln \left (\sin \relax (x )-1\right ) b}{2 a^{2}}-\frac {1}{4 a \left (\sin \relax (x )+1\right )}+\frac {\ln \left (\sin \relax (x )+1\right )}{4 a}-\frac {\ln \left (\sin \relax (x )+1\right ) b}{2 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^3/(a+b*cos(x)^2),x)

[Out]

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

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maxima [A]  time = 1.26, size = 92, normalized size = 1.56 \[ -\frac {b^{2} \log \left (\frac {b \sin \relax (x) - \sqrt {{\left (a + b\right )} b}}{b \sin \relax (x) + \sqrt {{\left (a + b\right )} b}}\right )}{2 \, \sqrt {{\left (a + b\right )} b} a^{2}} + \frac {{\left (a - 2 \, b\right )} \log \left (\sin \relax (x) + 1\right )}{4 \, a^{2}} - \frac {{\left (a - 2 \, b\right )} \log \left (\sin \relax (x) - 1\right )}{4 \, a^{2}} - \frac {\sin \relax (x)}{2 \, {\left (a \sin \relax (x)^{2} - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^3/(a+b*cos(x)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.53, size = 483, normalized size = 8.19 \[ -\frac {a^2\,\sin \relax (x)+a^2\,\mathrm {atanh}\left (\sin \relax (x)\right )-2\,b^2\,\mathrm {atanh}\left (\sin \relax (x)\right )+a\,b\,\sin \relax (x)-a\,b\,\mathrm {atanh}\left (\sin \relax (x)\right )-a^2\,\mathrm {atanh}\left (\sin \relax (x)\right )\,{\sin \relax (x)}^2+2\,b^2\,\mathrm {atanh}\left (\sin \relax (x)\right )\,{\sin \relax (x)}^2+a\,b\,\mathrm {atanh}\left (\sin \relax (x)\right )\,{\sin \relax (x)}^2+\mathrm {atan}\left (\frac {b^5\,\sin \relax (x)\,\sqrt {b^4+a\,b^3}\,8{}\mathrm {i}-a\,\sin \relax (x)\,{\left (b^4+a\,b^3\right )}^{3/2}\,4{}\mathrm {i}-b\,\sin \relax (x)\,{\left (b^4+a\,b^3\right )}^{3/2}\,8{}\mathrm {i}+a\,b^4\,\sin \relax (x)\,\sqrt {b^4+a\,b^3}\,12{}\mathrm {i}+a^4\,b\,\sin \relax (x)\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}+a^2\,b^3\,\sin \relax (x)\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}-a^3\,b^2\,\sin \relax (x)\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}}{-a^5\,b^2+a^4\,b^3+5\,a^3\,b^4+3\,a^2\,b^5}\right )\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {b^5\,\sin \relax (x)\,\sqrt {b^4+a\,b^3}\,8{}\mathrm {i}-a\,\sin \relax (x)\,{\left (b^4+a\,b^3\right )}^{3/2}\,4{}\mathrm {i}-b\,\sin \relax (x)\,{\left (b^4+a\,b^3\right )}^{3/2}\,8{}\mathrm {i}+a\,b^4\,\sin \relax (x)\,\sqrt {b^4+a\,b^3}\,12{}\mathrm {i}+a^4\,b\,\sin \relax (x)\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}+a^2\,b^3\,\sin \relax (x)\,\sqrt {b^4+a\,b^3}\,1{}\mathrm {i}-a^3\,b^2\,\sin \relax (x)\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}}{-a^5\,b^2+a^4\,b^3+5\,a^3\,b^4+3\,a^2\,b^5}\right )\,{\sin \relax (x)}^2\,\sqrt {b^4+a\,b^3}\,2{}\mathrm {i}}{2\,a^3\,{\sin \relax (x)}^2-2\,a^3+2\,b\,a^2\,{\sin \relax (x)}^2-2\,b\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^3*(a + b*cos(x)^2)),x)

[Out]

-(a^2*sin(x) + a^2*atanh(sin(x)) - 2*b^2*atanh(sin(x)) + atan((b^5*sin(x)*(a*b^3 + b^4)^(1/2)*8i - a*sin(x)*(a
*b^3 + b^4)^(3/2)*4i - b*sin(x)*(a*b^3 + b^4)^(3/2)*8i + a*b^4*sin(x)*(a*b^3 + b^4)^(1/2)*12i + a^4*b*sin(x)*(
a*b^3 + b^4)^(1/2)*1i + a^2*b^3*sin(x)*(a*b^3 + b^4)^(1/2)*1i - a^3*b^2*sin(x)*(a*b^3 + b^4)^(1/2)*2i)/(3*a^2*
b^5 + 5*a^3*b^4 + a^4*b^3 - a^5*b^2))*(a*b^3 + b^4)^(1/2)*2i + a*b*sin(x) - a*b*atanh(sin(x)) - a^2*atanh(sin(
x))*sin(x)^2 + 2*b^2*atanh(sin(x))*sin(x)^2 - atan((b^5*sin(x)*(a*b^3 + b^4)^(1/2)*8i - a*sin(x)*(a*b^3 + b^4)
^(3/2)*4i - b*sin(x)*(a*b^3 + b^4)^(3/2)*8i + a*b^4*sin(x)*(a*b^3 + b^4)^(1/2)*12i + a^4*b*sin(x)*(a*b^3 + b^4
)^(1/2)*1i + a^2*b^3*sin(x)*(a*b^3 + b^4)^(1/2)*1i - a^3*b^2*sin(x)*(a*b^3 + b^4)^(1/2)*2i)/(3*a^2*b^5 + 5*a^3
*b^4 + a^4*b^3 - a^5*b^2))*sin(x)^2*(a*b^3 + b^4)^(1/2)*2i + a*b*atanh(sin(x))*sin(x)^2)/(2*a^3*sin(x)^2 - 2*a
^2*b - 2*a^3 + 2*a^2*b*sin(x)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\relax (x )}}{a + b \cos ^{2}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**3/(a+b*cos(x)**2),x)

[Out]

Integral(sec(x)**3/(a + b*cos(x)**2), x)

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