\(\int \frac {\sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx\) [159]

Optimal result

Integrand size = 23, antiderivative size = 32 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} d} \]

[Out]

arctan(b^(1/2)*tan(d*x+c)/a^(1/2))/d/a^(1/2)/b^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3756, 211} \[ \int \frac {\sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} d} \]

[In]

Int[Sec[c + d*x]^2/(a + b*Tan[c + d*x]^2),x]

[Out]

ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]]/(Sqrt[a]*Sqrt[b]*d)

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 3756

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b} d} \]

[In]

Integrate[Sec[c + d*x]^2/(a + b*Tan[c + d*x]^2),x]

[Out]

ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]]/(Sqrt[a]*Sqrt[b]*d)

Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {\arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{d \sqrt {a b}}\)	\(24\)
default	\(\frac {\arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{d \sqrt {a b}}\)	\(24\)
risch	\(-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i b a +a \sqrt {-a b}+b \sqrt {-a b}}{\sqrt {-a b}\, \left (a -b \right )}\right )}{2 \sqrt {-a b}\, d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i b a -a \sqrt {-a b}-b \sqrt {-a b}}{\sqrt {-a b}\, \left (a -b \right )}\right )}{2 \sqrt {-a b}\, d}\)	\(121\)

[In]

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

[Out]

1/d/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (24) = 48\).

Time = 0.26 (sec) , antiderivative size = 205, normalized size of antiderivative = 6.41 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\left [-\frac {\sqrt {-a b} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (d x + c\right )^{3} - b \cos \left (d x + c\right )\right )} \sqrt {-a b} \sin \left (d x + c\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right )}{4 \, a b d}, -\frac {\sqrt {a b} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (d x + c\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right )}{2 \, a b d}\right ] \]

[In]

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

[Out]

[-1/4*sqrt(-a*b)*log(((a^2 + 6*a*b + b^2)*cos(d*x + c)^4 - 2*(3*a*b + b^2)*cos(d*x + c)^2 + 4*((a + b)*cos(d*x
+ c)^3 - b*cos(d*x + c))*sqrt(-a*b)*sin(d*x + c) + b^2)/((a^2 - 2*a*b + b^2)*cos(d*x + c)^4 + 2*(a*b - b^2)*c
os(d*x + c)^2 + b^2))/(a*b*d), -1/2*sqrt(a*b)*arctan(1/2*((a + b)*cos(d*x + c)^2 - b)*sqrt(a*b)/(a*b*cos(d*x +
c)*sin(d*x + c)))/(a*b*d)]

Sympy [F]

\[ \int \frac {\sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{a + b \tan ^{2}{\left (c + d x \right )}}\, dx \]

[In]

integrate(sec(d*x+c)**2/(a+b*tan(d*x+c)**2),x)

[Out]

Integral(sec(c + d*x)**2/(a + b*tan(c + d*x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} d} \]

[In]

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

[Out]

arctan(b*tan(d*x + c)/sqrt(a*b))/(sqrt(a*b)*d)

Giac [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{\sqrt {a b} d} \]

[In]

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

[Out]

(pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a*b)))/(sqrt(a*b)*d)

Mupad [B] (verification not implemented)

Time = 26.64 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {\sec ^2(c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (c+d\,x\right )}{\sqrt {a}}\right )}{\sqrt {a}\,\sqrt {b}\,d} \]

[In]

int(1/(cos(c + d*x)^2*(a + b*tan(c + d*x)^2)),x)

[Out]

atan((b^(1/2)*tan(c + d*x))/a^(1/2))/(a^(1/2)*b^(1/2)*d)