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

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

[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)

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Rubi [A]  time = 0.165703, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, 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 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 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 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 ^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*}

Mathematica [B]  time = 1.09786, 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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Maple [B]  time = 0.042, size = 165, normalized size = 1.8 \begin{align*} -{\frac{1}{16\,a \left ( \sin \left ( x \right ) +1 \right ) ^{2}}}-{\frac{3}{16\,a \left ( \sin \left ( x \right ) +1 \right ) }}+{\frac{b}{4\,{a}^{2} \left ( \sin \left ( x \right ) +1 \right ) }}+{\frac{3\,\ln \left ( \sin \left ( x \right ) +1 \right ) }{16\,a}}-{\frac{\ln \left ( \sin \left ( x \right ) +1 \right ) b}{4\,{a}^{2}}}+{\frac{\ln \left ( \sin \left ( x \right ) +1 \right ){b}^{2}}{2\,{a}^{3}}}-{\frac{{b}^{3}}{{a}^{3}}{\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}{16\,a \left ( \sin \left ( x \right ) -1 \right ) ^{2}}}-{\frac{3}{16\,a \left ( \sin \left ( x \right ) -1 \right ) }}+{\frac{b}{4\,{a}^{2} \left ( \sin \left ( x \right ) -1 \right ) }}-{\frac{3\,\ln \left ( \sin \left ( x \right ) -1 \right ) }{16\,a}}+{\frac{\ln \left ( \sin \left ( x \right ) -1 \right ) b}{4\,{a}^{2}}}-{\frac{\ln \left ( \sin \left ( x \right ) -1 \right ){b}^{2}}{2\,{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.1347, size = 709, normalized size = 7.88 \begin{align*} \left [\frac{8 \, b^{2} \sqrt{\frac{b}{a + b}} \cos \left (x\right )^{4} \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 (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) -{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left ({\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (x\right )^{2} + 2 \, a^{2}\right )} \sin \left (x\right )}{16 \, a^{3} \cos \left (x\right )^{4}}, \frac{16 \, b^{2} \sqrt{-\frac{b}{a + b}} \arctan \left (\sqrt{-\frac{b}{a + b}} \sin \left (x\right )\right ) \cos \left (x\right )^{4} +{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) -{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left ({\left (3 \, a^{2} - 4 \, a b\right )} \cos \left (x\right )^{2} + 2 \, a^{2}\right )} \sin \left (x\right )}{16 \, a^{3} \cos \left (x\right )^{4}}\right ] \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.1724, size = 171, normalized size = 1.9 \begin{align*} \frac{b^{3} \arctan \left (\frac{b \sin \left (x\right )}{\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 \left (x\right ) + 1\right )}{16 \, a^{3}} - \frac{{\left (3 \, a^{2} - 4 \, a b + 8 \, b^{2}\right )} \log \left (-\sin \left (x\right ) + 1\right )}{16 \, a^{3}} - \frac{3 \, a \sin \left (x\right )^{3} - 4 \, b \sin \left (x\right )^{3} - 5 \, a \sin \left (x\right ) + 4 \, b \sin \left (x\right )}{8 \,{\left (\sin \left (x\right )^{2} - 1\right )}^{2} a^{2}} \end{align*}

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)