3.4.8 \(\int \frac {\sec (x)}{a+b \sin ^2(x)} \, dx\) [308]

Optimal. Leaf size=40 \[ \frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}+\frac {\tanh ^{-1}(\sin (x))}{a+b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3269, 400, 212, 211} \begin {gather*} \frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}+\frac {\tanh ^{-1}(\sin (x))}{a+b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[x]/(a + b*Sin[x]^2),x]

[Out]

(Sqrt[b]*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*(a + b)) + ArcTanh[Sin[x]]/(a + b)

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 400

Int[1/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^n),
x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0]

Rule 3269

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]

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(96\) vs. \(2(40)=80\).
time = 0.10, size = 96, normalized size = 2.40 \begin {gather*} \frac {-\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \csc (x)}{\sqrt {b}}\right )+\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )+2 \sqrt {a} \left (-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )\right )}{2 \sqrt {a} (a+b)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]/(a + b*Sin[x]^2),x]

[Out]

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

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Maple [A]
time = 0.20, size = 55, normalized size = 1.38

method	result	size
default	\(\frac {b \arctan \left (\frac {b \sin \left (x \right )}{\sqrt {a b}}\right )}{\left (a +b \right ) \sqrt {a b}}+\frac {\ln \left (1+\sin \left (x \right )\right )}{2 a +2 b}-\frac {\ln \left (\sin \left (x \right )-1\right )}{2 a +2 b}\)	\(55\)
risch	\(-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{a +b}+\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{a +b}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 a \left (a +b \right )}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 a \left (a +b \right )}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)/(a+b*sin(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.50, size = 47, normalized size = 1.18 \begin {gather*} \frac {b \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{\sqrt {a b} {\left (a + b\right )}} + \frac {\log \left (\sin \left (x\right ) + 1\right )}{2 \, {\left (a + b\right )}} - \frac {\log \left (\sin \left (x\right ) - 1\right )}{2 \, {\left (a + b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.42, size = 116, normalized size = 2.90 \begin {gather*} \left [\frac {\sqrt {-\frac {b}{a}} \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 ) + \log \left (\sin \left (x\right ) + 1\right ) - \log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a + b\right )}}, \frac {2 \, \sqrt {\frac {b}{a}} \arctan \left (\sqrt {\frac {b}{a}} \sin \left (x\right )\right ) + \log \left (\sin \left (x\right ) + 1\right ) - \log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a + b\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(-b/a)*log(-(b*cos(x)^2 - 2*a*sqrt(-b/a)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) + log(sin(x) + 1) - l
og(-sin(x) + 1))/(a + b), 1/2*(2*sqrt(b/a)*arctan(sqrt(b/a)*sin(x)) + log(sin(x) + 1) - log(-sin(x) + 1))/(a +
b)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec {\left (x \right )}}{a + b \sin ^{2}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(x)/(a + b*sin(x)**2), x)

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Giac [A]
time = 0.45, size = 49, normalized size = 1.22 \begin {gather*} \frac {b \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{\sqrt {a b} {\left (a + b\right )}} + \frac {\log \left (\sin \left (x\right ) + 1\right )}{2 \, {\left (a + b\right )}} - \frac {\log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a + b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 14.69, size = 856, normalized size = 21.40 \begin {gather*} -\frac {\mathrm {atan}\left (\frac {\frac {\left (4\,b^3\,\sin \left (x\right )+\frac {8\,a\,b^3+4\,b^4+4\,a^2\,b^2-\frac {\sin \left (x\right )\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}\right )\,1{}\mathrm {i}}{2\,\left (a+b\right )}+\frac {\left (4\,b^3\,\sin \left (x\right )-\frac {8\,a\,b^3+4\,b^4+4\,a^2\,b^2+\frac {\sin \left (x\right )\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}\right )\,1{}\mathrm {i}}{2\,\left (a+b\right )}}{\frac {4\,b^3\,\sin \left (x\right )+\frac {8\,a\,b^3+4\,b^4+4\,a^2\,b^2-\frac {\sin \left (x\right )\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}-\frac {4\,b^3\,\sin \left (x\right )-\frac {8\,a\,b^3+4\,b^4+4\,a^2\,b^2+\frac {\sin \left (x\right )\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}}\right )\,1{}\mathrm {i}}{a+b}-\frac {\mathrm {atan}\left (\frac {\frac {\left (2\,b^3\,\sin \left (x\right )+\frac {\sqrt {-a\,b}\,\left (4\,a\,b^3+2\,b^4+2\,a^2\,b^2-\frac {\sin \left (x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{4\,\left (a^2+b\,a\right )}\right )}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-a\,b}\,1{}\mathrm {i}}{a^2+b\,a}+\frac {\left (2\,b^3\,\sin \left (x\right )-\frac {\sqrt {-a\,b}\,\left (4\,a\,b^3+2\,b^4+2\,a^2\,b^2+\frac {\sin \left (x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{4\,\left (a^2+b\,a\right )}\right )}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-a\,b}\,1{}\mathrm {i}}{a^2+b\,a}}{\frac {\left (2\,b^3\,\sin \left (x\right )+\frac {\sqrt {-a\,b}\,\left (4\,a\,b^3+2\,b^4+2\,a^2\,b^2-\frac {\sin \left (x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{4\,\left (a^2+b\,a\right )}\right )}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-a\,b}}{a^2+b\,a}-\frac {\left (2\,b^3\,\sin \left (x\right )-\frac {\sqrt {-a\,b}\,\left (4\,a\,b^3+2\,b^4+2\,a^2\,b^2+\frac {\sin \left (x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{4\,\left (a^2+b\,a\right )}\right )}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-a\,b}}{a^2+b\,a}}\right )\,\sqrt {-a\,b}\,1{}\mathrm {i}}{a\,\left (a+b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)*(a + b*sin(x)^2)),x)

[Out]

- (atan((((4*b^3*sin(x) + (8*a*b^3 + 4*b^4 + 4*a^2*b^2 - (sin(x)*(8*a*b^4 + 8*b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(2
*(a + b)))/(2*(a + b)))*1i)/(2*(a + b)) + ((4*b^3*sin(x) - (8*a*b^3 + 4*b^4 + 4*a^2*b^2 + (sin(x)*(8*a*b^4 + 8
*b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(2*(a + b)))/(2*(a + b)))*1i)/(2*(a + b)))/((4*b^3*sin(x) + (8*a*b^3 + 4*b^4 +
4*a^2*b^2 - (sin(x)*(8*a*b^4 + 8*b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(2*(a + b)))/(2*(a + b)))/(2*(a + b)) - (4*b^3*
sin(x) - (8*a*b^3 + 4*b^4 + 4*a^2*b^2 + (sin(x)*(8*a*b^4 + 8*b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(2*(a + b)))/(2*(a
+ b)))/(2*(a + b))))*1i)/(a + b) - (atan((((2*b^3*sin(x) + ((-a*b)^(1/2)*(4*a*b^3 + 2*b^4 + 2*a^2*b^2 - (sin(x
)*(-a*b)^(1/2)*(8*a*b^4 + 8*b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(4*(a*b + a^2))))/(2*(a*b + a^2)))*(-a*b)^(1/2)*1i)/
(a*b + a^2) + ((2*b^3*sin(x) - ((-a*b)^(1/2)*(4*a*b^3 + 2*b^4 + 2*a^2*b^2 + (sin(x)*(-a*b)^(1/2)*(8*a*b^4 + 8*
b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(4*(a*b + a^2))))/(2*(a*b + a^2)))*(-a*b)^(1/2)*1i)/(a*b + a^2))/(((2*b^3*sin(x)
+ ((-a*b)^(1/2)*(4*a*b^3 + 2*b^4 + 2*a^2*b^2 - (sin(x)*(-a*b)^(1/2)*(8*a*b^4 + 8*b^5 - 8*a^2*b^3 - 8*a^3*b^2)
)/(4*(a*b + a^2))))/(2*(a*b + a^2)))*(-a*b)^(1/2))/(a*b + a^2) - ((2*b^3*sin(x) - ((-a*b)^(1/2)*(4*a*b^3 + 2*b
^4 + 2*a^2*b^2 + (sin(x)*(-a*b)^(1/2)*(8*a*b^4 + 8*b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(4*(a*b + a^2))))/(2*(a*b + a
^2)))*(-a*b)^(1/2))/(a*b + a^2)))*(-a*b)^(1/2)*1i)/(a*(a + b))

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