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

Optimal. Leaf size=61 \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^2}+\frac{(a+3 b) \tanh ^{-1}(\sin (x))}{2 (a+b)^2}+\frac{\tan (x) \sec (x)}{2 (a+b)} \]

[Out]

(b^(3/2)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^2) + ((a + 3*b)*ArcTanh[Sin[x]])/(2*(a + b)^2) + (
Sec[x]*Tan[x])/(2*(a + b))

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Rubi [A]  time = 0.0867202, antiderivative size = 61, 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 = {3190, 414, 522, 206, 205} \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^2}+\frac{(a+3 b) \tanh ^{-1}(\sin (x))}{2 (a+b)^2}+\frac{\tan (x) \sec (x)}{2 (a+b)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^3/(a + b*Sin[x]^2),x]

[Out]

(b^(3/2)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^2) + ((a + 3*b)*ArcTanh[Sin[x]])/(2*(a + b)^2) + (
Sec[x]*Tan[x])/(2*(a + b))

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]

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 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]

Rubi steps

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

Mathematica [B]  time = 0.315132, size = 147, normalized size = 2.41 \[ \frac{\frac{2 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a}}-\frac{2 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \csc (x)}{\sqrt{b}}\right )}{\sqrt{a}}+\frac{a+b}{\left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )^2}-\frac{a+b}{\left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^2}-2 (a+3 b) \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+2 (a+3 b) \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )}{4 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.062, size = 112, normalized size = 1.8 \begin{align*} -{\frac{1}{ \left ( 4\,a+4\,b \right ) \left ( -1+\sin \left ( x \right ) \right ) }}-{\frac{\ln \left ( -1+\sin \left ( x \right ) \right ) a}{4\, \left ( a+b \right ) ^{2}}}-{\frac{3\,\ln \left ( -1+\sin \left ( x \right ) \right ) b}{4\, \left ( a+b \right ) ^{2}}}-{\frac{1}{ \left ( 4\,a+4\,b \right ) \left ( 1+\sin \left ( x \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( x \right ) \right ) a}{4\, \left ( a+b \right ) ^{2}}}+{\frac{3\,\ln \left ( 1+\sin \left ( x \right ) \right ) b}{4\, \left ( a+b \right ) ^{2}}}+{\frac{{b}^{2}}{ \left ( a+b \right ) ^{2}}\arctan \left ({\sin \left ( x \right ) b{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.80953, size = 556, normalized size = 9.11 \begin{align*} \left [\frac{2 \, b \sqrt{-\frac{b}{a}} \cos \left (x\right )^{2} \log \left (-\frac{b \cos \left (x\right )^{2} - 2 \, a \sqrt{-\frac{b}{a}} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right ) +{\left (a + 3 \, b\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) -{\left (a + 3 \, b\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left (a + b\right )} \sin \left (x\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (x\right )^{2}}, \frac{4 \, b \sqrt{\frac{b}{a}} \arctan \left (\sqrt{\frac{b}{a}} \sin \left (x\right )\right ) \cos \left (x\right )^{2} +{\left (a + 3 \, b\right )} \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) -{\left (a + 3 \, b\right )} \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left (a + b\right )} \sin \left (x\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \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*sin(x)^2),x, algorithm="fricas")

[Out]

[1/4*(2*b*sqrt(-b/a)*cos(x)^2*log(-(b*cos(x)^2 - 2*a*sqrt(-b/a)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) + (a + 3
*b)*cos(x)^2*log(sin(x) + 1) - (a + 3*b)*cos(x)^2*log(-sin(x) + 1) + 2*(a + b)*sin(x))/((a^2 + 2*a*b + b^2)*co
s(x)^2), 1/4*(4*b*sqrt(b/a)*arctan(sqrt(b/a)*sin(x))*cos(x)^2 + (a + 3*b)*cos(x)^2*log(sin(x) + 1) - (a + 3*b)
*cos(x)^2*log(-sin(x) + 1) + 2*(a + b)*sin(x))/((a^2 + 2*a*b + b^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 \sin ^{2}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.14683, size = 138, normalized size = 2.26 \begin{align*} \frac{b^{2} \arctan \left (\frac{b \sin \left (x\right )}{\sqrt{a b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt{a b}} + \frac{{\left (a + 3 \, b\right )} \log \left (\sin \left (x\right ) + 1\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{{\left (a + 3 \, b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{\sin \left (x\right )}{2 \,{\left (\sin \left (x\right )^{2} - 1\right )}{\left (a + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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