3.569 \(\int \frac{\sec ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=22 \[ -\frac{1}{2 b d (a+b \tan (c+d x))^2} \]

[Out]

-1/(2*b*d*(a + b*Tan[c + d*x])^2)

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Rubi [A]  time = 0.0429776, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3506, 32} \[ -\frac{1}{2 b d (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]

[Out]

-1/(2*b*d*(a + b*Tan[c + d*x])^2)

Rule 3506

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [B]  time = 0.174164, size = 58, normalized size = 2.64 \[ \frac{2 \tan (c+d x) (a+b \tan (c+d x))-b \sec ^2(c+d x)}{2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]

[Out]

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

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Maple [A]  time = 0.051, size = 21, normalized size = 1. \begin{align*} -{\frac{1}{2\,bd \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}} \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))^3,x)

[Out]

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

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Maxima [A]  time = 1.19315, size = 27, normalized size = 1.23 \begin{align*} -\frac{1}{2 \,{\left (b \tan \left (d x + c\right ) + a\right )}^{2} b 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))^3,x, algorithm="maxima")

[Out]

-1/2/((b*tan(d*x + c) + a)^2*b*d)

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Fricas [B]  time = 2.16357, size = 313, normalized size = 14.23 \begin{align*} -\frac{4 \, a^{2} b \cos \left (d x + c\right )^{2} - a^{2} b + b^{3} - 2 \,{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{6} + a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} d\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))^3,x, algorithm="fricas")

[Out]

-1/2*(4*a^2*b*cos(d*x + c)^2 - a^2*b + b^3 - 2*(a^3 - a*b^2)*cos(d*x + c)*sin(d*x + c))/((a^6 + a^4*b^2 - a^2*
b^4 - b^6)*d*cos(d*x + c)^2 + 2*(a^5*b + 2*a^3*b^3 + a*b^5)*d*cos(d*x + c)*sin(d*x + c) + (a^4*b^2 + 2*a^2*b^4
+ b^6)*d)

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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{\left (c + d x \right )}\right )^{3}}\, 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))**3,x)

[Out]

Integral(sec(c + d*x)**2/(a + b*tan(c + d*x))**3, x)

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Giac [A]  time = 1.48829, size = 27, normalized size = 1.23 \begin{align*} -\frac{1}{2 \,{\left (b \tan \left (d x + c\right ) + a\right )}^{2} b 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))^3,x, algorithm="giac")

[Out]

-1/2/((b*tan(d*x + c) + a)^2*b*d)