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]

((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)

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Rubi [A]  time = 0.0981543, antiderivative size = 59, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [B]  time = 0.36146, 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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Maple [A]  time = 0.039, size = 92, normalized size = 1.6 \begin{align*} -{\frac{1}{4\,a \left ( \sin \left ( x \right ) +1 \right ) }}+{\frac{\ln \left ( \sin \left ( x \right ) +1 \right ) }{4\,a}}-{\frac{\ln \left ( \sin \left ( x \right ) +1 \right ) b}{2\,{a}^{2}}}+{\frac{{b}^{2}}{{a}^{2}}{\it Artanh} \left ({\sin \left ( x \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}-{\frac{1}{4\,a \left ( \sin \left ( x \right ) -1 \right ) }}-{\frac{\ln \left ( \sin \left ( x \right ) -1 \right ) }{4\,a}}+{\frac{\ln \left ( \sin \left ( x \right ) -1 \right ) b}{2\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.99673, size = 536, normalized size = 9.08 \begin{align*} \left [\frac{2 \, b \sqrt{\frac{b}{a + b}} \cos \left (x\right )^{2} \log \left (-\frac{b \cos \left (x\right )^{2} - 2 \,{\left (a + b\right )} \sqrt{\frac{b}{a + b}} \sin \left (x\right ) - a - 2 \, b}{b \cos \left (x\right )^{2} + a}\right ) +{\left (a - 2 \, b\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) -{\left (a - 2 \, b\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, a \sin \left (x\right )}{4 \, a^{2} \cos \left (x\right )^{2}}, -\frac{4 \, b \sqrt{-\frac{b}{a + b}} \arctan \left (\sqrt{-\frac{b}{a + b}} \sin \left (x\right )\right ) \cos \left (x\right )^{2} -{\left (a - 2 \, b\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) +{\left (a - 2 \, b\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \, a \sin \left (x\right )}{4 \, a^{2} \cos \left (x\right )^{2}}\right ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (x \right )}}{a + b \cos ^{2}{\left (x \right )}}\, dx \end{align*}

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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Giac [A]  time = 1.22048, size = 115, normalized size = 1.95 \begin{align*} -\frac{b^{2} \arctan \left (\frac{b \sin \left (x\right )}{\sqrt{-a b - b^{2}}}\right )}{\sqrt{-a b - b^{2}} a^{2}} + \frac{{\left (a - 2 \, b\right )} \log \left (\sin \left (x\right ) + 1\right )}{4 \, a^{2}} - \frac{{\left (a - 2 \, b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{4 \, a^{2}} - \frac{\sin \left (x\right )}{2 \,{\left (\sin \left (x\right )^{2} - 1\right )} a} \end{align*}

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)