\(\int \frac {\sec (c+d x)}{a+b \tan ^2(c+d x)} \, dx\) [452]

Optimal result

Integrand size = 21, antiderivative size = 40 \[ \int \frac {\sec (c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {a-b} d} \]

[Out]

arctanh(sin(d*x+c)*(a-b)^(1/2)/a^(1/2))/d/a^(1/2)/(a-b)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3757, 214} \[ \int \frac {\sec (c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} d \sqrt {a-b}} \]

[In]

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

[Out]

ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]]/(Sqrt[a]*Sqrt[a - b]*d)

Rule 214

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

Rule 3757

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {\sec (c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {a-b} d} \]

[In]

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

[Out]

ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]]/(Sqrt[a]*Sqrt[a - b]*d)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {\operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{d \sqrt {a \left (a -b \right )}}\)	\(36\)
default	\(\frac {\operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{d \sqrt {a \left (a -b \right )}}\)	\(36\)
risch	\(\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{\sqrt {a^{2}-a b}}-1\right )}{2 \sqrt {a^{2}-a b}\, d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{\sqrt {a^{2}-a b}}-1\right )}{2 \sqrt {a^{2}-a b}\, d}\)	\(102\)

[In]

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

[Out]

1/d/(a*(a-b))^(1/2)*arctanh((a-b)*sin(d*x+c)/(a*(a-b))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 3.05 \[ \int \frac {\sec (c+d x)}{a+b \tan ^2(c+d x)} \, dx=\left [\frac {\log \left (-\frac {{\left (a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - a b} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right )}{2 \, \sqrt {a^{2} - a b} d}, -\frac {\sqrt {-a^{2} + a b} \arctan \left (\frac {\sqrt {-a^{2} + a b} \sin \left (d x + c\right )}{a}\right )}{{\left (a^{2} - a b\right )} d}\right ] \]

[In]

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

[Out]

[1/2*log(-((a - b)*cos(d*x + c)^2 - 2*sqrt(a^2 - a*b)*sin(d*x + c) - 2*a + b)/((a - b)*cos(d*x + c)^2 + b))/(s
qrt(a^2 - a*b)*d), -sqrt(-a^2 + a*b)*arctan(sqrt(-a^2 + a*b)*sin(d*x + c)/a)/((a^2 - a*b)*d)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec (c+d x)}{a+b \tan ^2(c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is

Giac [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.18 \[ \int \frac {\sec (c+d x)}{a+b \tan ^2(c+d x)} \, dx=-\frac {\arctan \left (\frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt {-a^{2} + a b}}\right )}{\sqrt {-a^{2} + a b} d} \]

[In]

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

[Out]

-arctan((a*sin(d*x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/(sqrt(-a^2 + a*b)*d)

Mupad [B] (verification not implemented)

Time = 12.44 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int \frac {\sec (c+d x)}{a+b \tan ^2(c+d x)} \, dx=\frac {\mathrm {atanh}\left (\frac {\sin \left (c+d\,x\right )\,\sqrt {a-b}}{\sqrt {a}}\right )}{\sqrt {a}\,d\,\sqrt {a-b}} \]

[In]

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

[Out]

atanh((sin(c + d*x)*(a - b)^(1/2))/a^(1/2))/(a^(1/2)*d*(a - b)^(1/2))