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

Optimal. Leaf size=93 \[ \frac{\left (3 a^2+10 a b+15 b^2\right ) \tanh ^{-1}(\sin (x))}{8 (a+b)^3}+\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^3}+\frac{\tan (x) \sec ^3(x)}{4 (a+b)}+\frac{(3 a+7 b) \tan (x) \sec (x)}{8 (a+b)^2} \]

[Out]

(b^(5/2)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^3) + ((3*a^2 + 10*a*b + 15*b^2)*ArcTanh[Sin[x]])/(
8*(a + b)^3) + ((3*a + 7*b)*Sec[x]*Tan[x])/(8*(a + b)^2) + (Sec[x]^3*Tan[x])/(4*(a + b))

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Rubi [A]  time = 0.145, antiderivative size = 93, 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 = {3190, 414, 527, 522, 206, 205} \[ \frac{\left (3 a^2+10 a b+15 b^2\right ) \tanh ^{-1}(\sin (x))}{8 (a+b)^3}+\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^3}+\frac{\tan (x) \sec ^3(x)}{4 (a+b)}+\frac{(3 a+7 b) \tan (x) \sec (x)}{8 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^5/(a + b*Sin[x]^2),x]

[Out]

(b^(5/2)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*(a + b)^3) + ((3*a^2 + 10*a*b + 15*b^2)*ArcTanh[Sin[x]])/(
8*(a + b)^3) + ((3*a + 7*b)*Sec[x]*Tan[x])/(8*(a + b)^2) + (Sec[x]^3*Tan[x])/(4*(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 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 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 ^5(x)}{a+b \sin ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\sin (x)\right )\\ &=\frac{\sec ^3(x) \tan (x)}{4 (a+b)}+\frac{\operatorname{Subst}\left (\int \frac{3 a+4 b+3 b x^2}{\left (1-x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\sin (x)\right )}{4 (a+b)}\\ &=\frac{(3 a+7 b) \sec (x) \tan (x)}{8 (a+b)^2}+\frac{\sec ^3(x) \tan (x)}{4 (a+b)}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2+7 a b+8 b^2+b (3 a+7 b) x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\sin (x)\right )}{8 (a+b)^2}\\ &=\frac{(3 a+7 b) \sec (x) \tan (x)}{8 (a+b)^2}+\frac{\sec ^3(x) \tan (x)}{4 (a+b)}+\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sin (x)\right )}{(a+b)^3}+\frac{\left (3 a^2+10 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (x)\right )}{8 (a+b)^3}\\ &=\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a} (a+b)^3}+\frac{\left (3 a^2+10 a b+15 b^2\right ) \tanh ^{-1}(\sin (x))}{8 (a+b)^3}+\frac{(3 a+7 b) \sec (x) \tan (x)}{8 (a+b)^2}+\frac{\sec ^3(x) \tan (x)}{4 (a+b)}\\ \end{align*}

Mathematica [B]  time = 1.2299, size = 214, normalized size = 2.3 \[ -\frac{2 \left (3 a^2+10 a b+15 b^2\right ) \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-2 \left (3 a^2+10 a b+15 b^2\right ) \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )-\frac{8 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{\sqrt{a}}+\frac{8 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{a} \csc (x)}{\sqrt{b}}\right )}{\sqrt{a}}+\frac{(a+b) (3 a+7 b)}{\sin (x)-1}+\frac{(a+b) (3 a+7 b)}{\left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^2}-\frac{(a+b)^2}{\left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )^4}+\frac{(a+b)^2}{\left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^4}}{16 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*b^(5/2)*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]])/Sqrt[a] - (8*b^(5/2)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/Sqrt[a]
+ 2*(3*a^2 + 10*a*b + 15*b^2)*Log[Cos[x/2] - Sin[x/2]] - 2*(3*a^2 + 10*a*b + 15*b^2)*Log[Cos[x/2] + Sin[x/2]]
- (a + b)^2/(Cos[x/2] - Sin[x/2])^4 + (a + b)^2/(Cos[x/2] + Sin[x/2])^4 + ((a + b)*(3*a + 7*b))/(Cos[x/2] + Si
n[x/2])^2 + ((a + b)*(3*a + 7*b))/(-1 + Sin[x]))/(16*(a + b)^3)

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Maple [B]  time = 0.068, size = 204, normalized size = 2.2 \begin{align*}{\frac{1}{ \left ( 16\,a+16\,b \right ) \left ( -1+\sin \left ( x \right ) \right ) ^{2}}}-{\frac{3\,a}{16\, \left ( a+b \right ) ^{2} \left ( -1+\sin \left ( x \right ) \right ) }}-{\frac{7\,b}{16\, \left ( a+b \right ) ^{2} \left ( -1+\sin \left ( x \right ) \right ) }}-{\frac{3\,\ln \left ( -1+\sin \left ( x \right ) \right ){a}^{2}}{16\, \left ( a+b \right ) ^{3}}}-{\frac{5\,\ln \left ( -1+\sin \left ( x \right ) \right ) ab}{8\, \left ( a+b \right ) ^{3}}}-{\frac{15\,\ln \left ( -1+\sin \left ( x \right ) \right ){b}^{2}}{16\, \left ( a+b \right ) ^{3}}}-{\frac{1}{ \left ( 16\,a+16\,b \right ) \left ( 1+\sin \left ( x \right ) \right ) ^{2}}}-{\frac{3\,a}{16\, \left ( a+b \right ) ^{2} \left ( 1+\sin \left ( x \right ) \right ) }}-{\frac{7\,b}{16\, \left ( a+b \right ) ^{2} \left ( 1+\sin \left ( x \right ) \right ) }}+{\frac{3\,\ln \left ( 1+\sin \left ( x \right ) \right ){a}^{2}}{16\, \left ( a+b \right ) ^{3}}}+{\frac{5\,\ln \left ( 1+\sin \left ( x \right ) \right ) ab}{8\, \left ( a+b \right ) ^{3}}}+{\frac{15\,\ln \left ( 1+\sin \left ( x \right ) \right ){b}^{2}}{16\, \left ( a+b \right ) ^{3}}}+{\frac{{b}^{3}}{ \left ( a+b \right ) ^{3}}\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)^5/(a+b*sin(x)^2),x)

[Out]

1/2/(8*a+8*b)/(-1+sin(x))^2-3/16/(a+b)^2/(-1+sin(x))*a-7/16/(a+b)^2/(-1+sin(x))*b-3/16/(a+b)^3*ln(-1+sin(x))*a
^2-5/8/(a+b)^3*ln(-1+sin(x))*a*b-15/16/(a+b)^3*ln(-1+sin(x))*b^2-1/2/(8*a+8*b)/(1+sin(x))^2-3/16/(a+b)^2/(1+si
n(x))*a-7/16/(a+b)^2/(1+sin(x))*b+3/16/(a+b)^3*ln(1+sin(x))*a^2+5/8/(a+b)^3*ln(1+sin(x))*a*b+15/16/(a+b)^3*ln(
1+sin(x))*b^2+b^3/(a+b)^3/(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)^5/(a+b*sin(x)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.96914, size = 825, normalized size = 8.87 \begin{align*} \left [\frac{8 \, b^{2} \sqrt{-\frac{b}{a}} \cos \left (x\right )^{4} \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 (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) -{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left ({\left (3 \, a^{2} + 10 \, a b + 7 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sin \left (x\right )}{16 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (x\right )^{4}}, \frac{16 \, b^{2} \sqrt{\frac{b}{a}} \arctan \left (\sqrt{\frac{b}{a}} \sin \left (x\right )\right ) \cos \left (x\right )^{4} +{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) -{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \,{\left ({\left (3 \, a^{2} + 10 \, a b + 7 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sin \left (x\right )}{16 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \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*sin(x)^2),x, algorithm="fricas")

[Out]

[1/16*(8*b^2*sqrt(-b/a)*cos(x)^4*log(-(b*cos(x)^2 - 2*a*sqrt(-b/a)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) + (3*
a^2 + 10*a*b + 15*b^2)*cos(x)^4*log(sin(x) + 1) - (3*a^2 + 10*a*b + 15*b^2)*cos(x)^4*log(-sin(x) + 1) + 2*((3*
a^2 + 10*a*b + 7*b^2)*cos(x)^2 + 2*a^2 + 4*a*b + 2*b^2)*sin(x))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(x)^4), 1/
16*(16*b^2*sqrt(b/a)*arctan(sqrt(b/a)*sin(x))*cos(x)^4 + (3*a^2 + 10*a*b + 15*b^2)*cos(x)^4*log(sin(x) + 1) -
(3*a^2 + 10*a*b + 15*b^2)*cos(x)^4*log(-sin(x) + 1) + 2*((3*a^2 + 10*a*b + 7*b^2)*cos(x)^2 + 2*a^2 + 4*a*b + 2
*b^2)*sin(x))/((a^3 + 3*a^2*b + 3*a*b^2 + b^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*sin(x)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.10426, size = 239, normalized size = 2.57 \begin{align*} \frac{b^{3} \arctan \left (\frac{b \sin \left (x\right )}{\sqrt{a b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt{a b}} + \frac{{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (x\right ) + 1\right )}{16 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac{{\left (3 \, a^{2} + 10 \, a b + 15 \, b^{2}\right )} \log \left (-\sin \left (x\right ) + 1\right )}{16 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac{3 \, a \sin \left (x\right )^{3} + 7 \, b \sin \left (x\right )^{3} - 5 \, a \sin \left (x\right ) - 9 \, b \sin \left (x\right )}{8 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}{\left (\sin \left (x\right )^{2} - 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*arctan(b*sin(x)/sqrt(a*b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)) + 1/16*(3*a^2 + 10*a*b + 15*b^2)*lo
g(sin(x) + 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/16*(3*a^2 + 10*a*b + 15*b^2)*log(-sin(x) + 1)/(a^3 + 3*a^2*b
+ 3*a*b^2 + b^3) - 1/8*(3*a*sin(x)^3 + 7*b*sin(x)^3 - 5*a*sin(x) - 9*b*sin(x))/((a^2 + 2*a*b + b^2)*(sin(x)^2
- 1)^2)