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

Optimal. Leaf size=66 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} d}+\frac{\tan (c+d x)}{2 a d \left (a+b \tan ^2(c+d x)\right )} \]

[Out]

ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]]/(2*a^(3/2)*Sqrt[b]*d) + Tan[c + d*x]/(2*a*d*(a + b*Tan[c + d*x]^2))

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Rubi [A]  time = 0.0619239, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3675, 199, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} \sqrt{b} d}+\frac{\tan (c+d x)}{2 a d \left (a+b \tan ^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]]/(2*a^(3/2)*Sqrt[b]*d) + Tan[c + d*x]/(2*a*d*(a + b*Tan[c + d*x]^2))

Rule 3675

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])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

Mathematica [A]  time = 0.266365, size = 63, normalized size = 0.95 \[ \frac{\frac{\tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{\sqrt{b}}+\frac{\sqrt{a} \tan (c+d x)}{a+b \tan ^2(c+d x)}}{2 a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]]/Sqrt[b] + (Sqrt[a]*Tan[c + d*x])/(a + b*Tan[c + d*x]^2))/(2*a^(3/2)*d)

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Maple [A]  time = 0.083, size = 57, normalized size = 0.9 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{2\,ad \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{1}{2\,ad}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*tan(d*x+c)/a/d/(a+b*tan(d*x+c)^2)+1/2/d/a/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(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(d*x+c)^2/(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.69403, size = 771, normalized size = 11.68 \begin{align*} \left [\frac{4 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) -{\left ({\left (a - b\right )} \cos \left (d x + c\right )^{2} + b\right )} \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 )}{8 \,{\left (a^{2} b^{2} d +{\left (a^{3} b - a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2}\right )}}, \frac{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) -{\left ({\left (a - b\right )} \cos \left (d x + c\right )^{2} + b\right )} \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 )}{4 \,{\left (a^{2} b^{2} d +{\left (a^{3} b - a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(4*a*b*cos(d*x + c)*sin(d*x + c) - ((a - b)*cos(d*x + c)^2 + b)*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)*cos(d*x + c)^2 + b^2)))/(a^2*b^2*d + (a^3*b - a^
2*b^2)*d*cos(d*x + c)^2), 1/4*(2*a*b*cos(d*x + c)*sin(d*x + c) - ((a - b)*cos(d*x + c)^2 + b)*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^2*b^2*d + (a^3*b - a^2*b^2)*d
*cos(d*x + c)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (c + d x \right )}}{\left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.73682, size = 95, normalized size = 1.44 \begin{align*} \frac{\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} a} + \frac{\tan \left (d x + c\right )}{{\left (b \tan \left (d x + c\right )^{2} + a\right )} a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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