3.470 \(\int \frac{\sec ^4(c+d x)}{(a+b \tan ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=77 \[ \frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{3/2} d}-\frac{(a-b) \tan (c+d x)}{2 a b d \left (a+b \tan ^2(c+d x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.0759769, antiderivative size = 77, 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, 385, 205} \[ \frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{3/2} d}-\frac{(a-b) \tan (c+d x)}{2 a b d \left (a+b \tan ^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

((a + b)*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/(2*a^(3/2)*b^(3/2)*d) - ((a - b)*Tan[c + d*x])/(2*a*b*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 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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 ^4(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{\left (a+b x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{(a-b) \tan (c+d x)}{2 a b d \left (a+b \tan ^2(c+d x)\right )}+\frac{(a+b) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (c+d x)\right )}{2 a b d}\\ &=\frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )}{2 a^{3/2} b^{3/2} d}-\frac{(a-b) \tan (c+d x)}{2 a b d \left (a+b \tan ^2(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.290043, size = 83, normalized size = 1.08 \[ \frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (c+d x)}{\sqrt{a}}\right )+\frac{\sqrt{a} \sqrt{b} (b-a) \sin (2 (c+d x))}{(a-b) \cos (2 (c+d x))+a+b}}{2 a^{3/2} b^{3/2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

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Maple [A]  time = 0.088, size = 112, normalized size = 1.5 \begin{align*} -{\frac{\tan \left ( dx+c \right ) }{2\,db \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{\tan \left ( dx+c \right ) }{2\,ad \left ( a+b \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{1}{2\,db}\arctan \left ({b\tan \left ( dx+c \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\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)^4/(a+b*tan(d*x+c)^2)^2,x)

[Out]

-1/2/d/b*tan(d*x+c)/(a+b*tan(d*x+c)^2)+1/2*tan(d*x+c)/a/d/(a+b*tan(d*x+c)^2)+1/2/d/b/(a*b)^(1/2)*arctan(b*tan(
d*x+c)/(a*b)^(1/2))+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)^4/(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.73774, size = 844, normalized size = 10.96 \begin{align*} \left [-\frac{4 \,{\left (a^{2} b - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + a b + b^{2}\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^{3} d +{\left (a^{3} b^{2} - a^{2} b^{3}\right )} d \cos \left (d x + c\right )^{2}\right )}}, -\frac{2 \,{\left (a^{2} b - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + a b + b^{2}\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^{3} d +{\left (a^{3} b^{2} - a^{2} b^{3}\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)^4/(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\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)**4/(a+b*tan(d*x+c)**2)**2,x)

[Out]

Integral(sec(c + d*x)**4/(a + b*tan(c + d*x)**2)**2, x)

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Giac [A]  time = 1.70943, size = 124, normalized size = 1.61 \begin{align*} \frac{\frac{{\left (\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 )\right )}{\left (a + b\right )}}{\sqrt{a b} a b} - \frac{a \tan \left (d x + c\right ) - b \tan \left (d x + c\right )}{{\left (b \tan \left (d x + c\right )^{2} + a\right )} a b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(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)))*(a + b)/(sqrt(a*b)*a*b) - (a*tan
(d*x + c) - b*tan(d*x + c))/((b*tan(d*x + c)^2 + a)*a*b))/d