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

Optimal. Leaf size=109 \[ \frac{b^{3/2} (5 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^3}-\frac{b (a-b) \sin (x)}{2 a (a+b)^2 \left (a+b \sin ^2(x)\right )}+\frac{(a+5 b) \tanh ^{-1}(\sin (x))}{2 (a+b)^3}+\frac{\tan (x) \sec (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )} \]

[Out]

(b^(3/2)*(5*a + b)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^3) + ((a + 5*b)*ArcTanh[Sin[x]])/(2*(a
+ b)^3) - ((a - b)*b*Sin[x])/(2*a*(a + b)^2*(a + b*Sin[x]^2)) + (Sec[x]*Tan[x])/(2*(a + b)*(a + b*Sin[x]^2))

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Rubi [A]  time = 0.163165, antiderivative size = 109, 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} \[ \frac{b^{3/2} (5 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^3}-\frac{b (a-b) \sin (x)}{2 a (a+b)^2 \left (a+b \sin ^2(x)\right )}+\frac{(a+5 b) \tanh ^{-1}(\sin (x))}{2 (a+b)^3}+\frac{\tan (x) \sec (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^3/(a + b*Sin[x]^2)^2,x]

[Out]

(b^(3/2)*(5*a + b)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(2*a^(3/2)*(a + b)^3) + ((a + 5*b)*ArcTanh[Sin[x]])/(2*(a
+ b)^3) - ((a - b)*b*Sin[x])/(2*a*(a + b)^2*(a + b*Sin[x]^2)) + (Sec[x]*Tan[x])/(2*(a + b)*(a + b*Sin[x]^2))

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 ^3(x)}{\left (a+b \sin ^2(x)\right )^2} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2 \left (a+b x^2\right )^2} \, dx,x,\sin (x)\right )\\ &=\frac{\sec (x) \tan (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{a+2 b+3 b x^2}{\left (1-x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\sin (x)\right )}{2 (a+b)}\\ &=-\frac{(a-b) b \sin (x)}{2 a (a+b)^2 \left (a+b \sin ^2(x)\right )}+\frac{\sec (x) \tan (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{-2 \left (a^2+4 a b+b^2\right )-2 (a-b) b x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\sin (x)\right )}{4 a (a+b)^2}\\ &=-\frac{(a-b) b \sin (x)}{2 a (a+b)^2 \left (a+b \sin ^2(x)\right )}+\frac{\sec (x) \tan (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )}+\frac{\left (b^2 (5 a+b)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sin (x)\right )}{2 a (a+b)^3}+\frac{(a+5 b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (x)\right )}{2 (a+b)^3}\\ &=\frac{b^{3/2} (5 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{2 a^{3/2} (a+b)^3}+\frac{(a+5 b) \tanh ^{-1}(\sin (x))}{2 (a+b)^3}-\frac{(a-b) b \sin (x)}{2 a (a+b)^2 \left (a+b \sin ^2(x)\right )}+\frac{\sec (x) \tan (x)}{2 (a+b) \left (a+b \sin ^2(x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.01182, size = 183, normalized size = 1.68 \[ \frac{\frac{b^{3/2} (5 a+b) \tan ^{-1}\left (\frac{\sqrt{b} \sin (x)}{\sqrt{a}}\right )}{a^{3/2}}-\frac{b^{3/2} (5 a+b) \tan ^{-1}\left (\frac{\sqrt{a} \csc (x)}{\sqrt{b}}\right )}{a^{3/2}}+\frac{4 b^2 (a+b) \sin (x)}{a (2 a-b \cos (2 x)+b)}+\frac{a+b}{\left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )^2}-\frac{a+b}{\left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^2}-2 (a+5 b) \log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )+2 (a+5 b) \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )}{4 (a+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((b^(3/2)*(5*a + b)*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]])/a^(3/2)) + (b^(3/2)*(5*a + b)*ArcTan[(Sqrt[b]*Sin[x])/
Sqrt[a]])/a^(3/2) - 2*(a + 5*b)*Log[Cos[x/2] - Sin[x/2]] + 2*(a + 5*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*b^2*(a + b)*Sin[x])/(a*(2*a + b - b*Cos[2*x])))/(4*
(a + b)^3)

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Maple [A]  time = 0.079, size = 180, normalized size = 1.7 \begin{align*} -{\frac{1}{4\, \left ( a+b \right ) ^{2} \left ( -1+\sin \left ( x \right ) \right ) }}-{\frac{\ln \left ( -1+\sin \left ( x \right ) \right ) a}{4\, \left ( a+b \right ) ^{3}}}-{\frac{5\,\ln \left ( -1+\sin \left ( x \right ) \right ) b}{4\, \left ( a+b \right ) ^{3}}}-{\frac{1}{4\, \left ( a+b \right ) ^{2} \left ( 1+\sin \left ( x \right ) \right ) }}+{\frac{\ln \left ( 1+\sin \left ( x \right ) \right ) a}{4\, \left ( a+b \right ) ^{3}}}+{\frac{5\,\ln \left ( 1+\sin \left ( x \right ) \right ) b}{4\, \left ( a+b \right ) ^{3}}}+{\frac{{b}^{2}\sin \left ( x \right ) }{2\, \left ( a+b \right ) ^{3} \left ( a+b \left ( \sin \left ( x \right ) \right ) ^{2} \right ) }}+{\frac{{b}^{3}\sin \left ( x \right ) }{2\, \left ( a+b \right ) ^{3}a \left ( a+b \left ( \sin \left ( x \right ) \right ) ^{2} \right ) }}+{\frac{5\,{b}^{2}}{2\, \left ( a+b \right ) ^{3}}\arctan \left ({\sin \left ( x \right ) b{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{{b}^{3}}{2\, \left ( a+b \right ) ^{3}a}\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)^2,x)

[Out]

-1/4/(a+b)^2/(-1+sin(x))-1/4/(a+b)^3*ln(-1+sin(x))*a-5/4/(a+b)^3*ln(-1+sin(x))*b-1/4/(a+b)^2/(1+sin(x))+1/4/(a
+b)^3*ln(1+sin(x))*a+5/4/(a+b)^3*ln(1+sin(x))*b+1/2*b^2/(a+b)^3*sin(x)/(a+b*sin(x)^2)+1/2*b^3/(a+b)^3/a*sin(x)
/(a+b*sin(x)^2)+5/2*b^2/(a+b)^3/(a*b)^(1/2)*arctan(sin(x)*b/(a*b)^(1/2))+1/2*b^3/(a+b)^3/a/(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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.14437, size = 1274, normalized size = 11.69 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(((5*a*b^2 + b^3)*cos(x)^4 - (5*a^2*b + 6*a*b^2 + b^3)*cos(x)^2)*sqrt(-b/a)*log(-(b*cos(x)^2 - 2*a*sqrt(-
b/a)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) + ((a^2*b + 5*a*b^2)*cos(x)^4 - (a^3 + 6*a^2*b + 5*a*b^2)*cos(x)^2)
*log(sin(x) + 1) - ((a^2*b + 5*a*b^2)*cos(x)^4 - (a^3 + 6*a^2*b + 5*a*b^2)*cos(x)^2)*log(-sin(x) + 1) - 2*(a^3
+ 2*a^2*b + a*b^2 - (a^2*b - b^3)*cos(x)^2)*sin(x))/((a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(x)^4 - (a^5
+ 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*cos(x)^2), 1/4*(2*((5*a*b^2 + b^3)*cos(x)^4 - (5*a^2*b + 6*a*b^2 +
b^3)*cos(x)^2)*sqrt(b/a)*arctan(sqrt(b/a)*sin(x)) + ((a^2*b + 5*a*b^2)*cos(x)^4 - (a^3 + 6*a^2*b + 5*a*b^2)*co
s(x)^2)*log(sin(x) + 1) - ((a^2*b + 5*a*b^2)*cos(x)^4 - (a^3 + 6*a^2*b + 5*a*b^2)*cos(x)^2)*log(-sin(x) + 1) -
2*(a^3 + 2*a^2*b + a*b^2 - (a^2*b - b^3)*cos(x)^2)*sin(x))/((a^4*b + 3*a^3*b^2 + 3*a^2*b^3 + a*b^4)*cos(x)^4
- (a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*cos(x)^2)]

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

[Out]

Timed out

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Giac [B]  time = 1.14059, size = 262, normalized size = 2.4 \begin{align*} \frac{{\left (a + 5 \, b\right )} \log \left (\sin \left (x\right ) + 1\right )}{4 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac{{\left (a + 5 \, b\right )} \log \left (-\sin \left (x\right ) + 1\right )}{4 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} + \frac{{\left (5 \, a b^{2} + b^{3}\right )} \arctan \left (\frac{b \sin \left (x\right )}{\sqrt{a b}}\right )}{2 \,{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \sqrt{a b}} - \frac{a b \sin \left (x\right )^{3} - b^{2} \sin \left (x\right )^{3} + a^{2} \sin \left (x\right ) + b^{2} \sin \left (x\right )}{2 \,{\left (b \sin \left (x\right )^{4} + a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} - a\right )}{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a + 5*b)*log(sin(x) + 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 1/4*(a + 5*b)*log(-sin(x) + 1)/(a^3 + 3*a^2*b
+ 3*a*b^2 + b^3) + 1/2*(5*a*b^2 + b^3)*arctan(b*sin(x)/sqrt(a*b))/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*sqrt(a*
b)) - 1/2*(a*b*sin(x)^3 - b^2*sin(x)^3 + a^2*sin(x) + b^2*sin(x))/((b*sin(x)^4 + a*sin(x)^2 - b*sin(x)^2 - a)*
(a^3 + 2*a^2*b + a*b^2))