3.568 \(\int \frac{\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=69 \[ -\frac{a^2+b^2}{2 b^3 d (a+b \tan (c+d x))^2}+\frac{2 a}{b^3 d (a+b \tan (c+d x))}+\frac{\log (a+b \tan (c+d x))}{b^3 d} \]

[Out]

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

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Rubi [A]  time = 0.0727754, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3506, 697} \[ -\frac{a^2+b^2}{2 b^3 d (a+b \tan (c+d x))^2}+\frac{2 a}{b^3 d (a+b \tan (c+d x))}+\frac{\log (a+b \tan (c+d x))}{b^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]

[Out]

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

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 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1+\frac{x^2}{b^2}}{(a+x)^3} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2+b^2}{b^2 (a+x)^3}-\frac{2 a}{b^2 (a+x)^2}+\frac{1}{b^2 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\log (a+b \tan (c+d x))}{b^3 d}-\frac{a^2+b^2}{2 b^3 d (a+b \tan (c+d x))^2}+\frac{2 a}{b^3 d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.176622, size = 57, normalized size = 0.83 \[ \frac{-\frac{a^2+b^2}{2 (a+b \tan (c+d x))^2}+\frac{2 a}{a+b \tan (c+d x)}+\log (a+b \tan (c+d x))}{b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + b*Tan[c + d*x])^3,x]

[Out]

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

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Maple [A]  time = 0.116, size = 84, normalized size = 1.2 \begin{align*} -{\frac{{a}^{2}}{2\,d{b}^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1}{2\,bd \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{a}{d{b}^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) }}+{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d{b}^{3}}} \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))^3,x)

[Out]

-1/2/d/b^3/(a+b*tan(d*x+c))^2*a^2-1/2/b/d/(a+b*tan(d*x+c))^2+2*a/b^3/d/(a+b*tan(d*x+c))+ln(a+b*tan(d*x+c))/b^3
/d

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Maxima [A]  time = 1.14823, size = 105, normalized size = 1.52 \begin{align*} \frac{\frac{4 \, a b \tan \left (d x + c\right ) + 3 \, a^{2} - b^{2}}{b^{5} \tan \left (d x + c\right )^{2} + 2 \, a b^{4} \tan \left (d x + c\right ) + a^{2} b^{3}} + \frac{2 \, \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{3}}}{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))^3,x, algorithm="maxima")

[Out]

1/2*((4*a*b*tan(d*x + c) + 3*a^2 - b^2)/(b^5*tan(d*x + c)^2 + 2*a*b^4*tan(d*x + c) + a^2*b^3) + 2*log(b*tan(d*
x + c) + a)/b^3)/d

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Fricas [B]  time = 2.26396, size = 647, normalized size = 9.38 \begin{align*} \frac{4 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b^{2} - b^{4} - 2 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} b^{2} + b^{4} +{\left (a^{4} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) -{\left (a^{2} b^{2} + b^{4} +{\left (a^{4} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right )}{2 \,{\left ({\left (a^{4} b^{3} - b^{7}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} b^{5} + b^{7}\right )} d\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))^3,x, algorithm="fricas")

[Out]

1/2*(4*a^2*b^2*cos(d*x + c)^2 - 3*a^2*b^2 - b^4 - 2*(a^3*b - a*b^3)*cos(d*x + c)*sin(d*x + c) + (a^2*b^2 + b^4
+ (a^4 - b^4)*cos(d*x + c)^2 + 2*(a^3*b + a*b^3)*cos(d*x + c)*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x +
c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - (a^2*b^2 + b^4 + (a^4 - b^4)*cos(d*x + c)^2 + 2*(a^3*b + a*b^3)*cos(d
*x + c)*sin(d*x + c))*log(cos(d*x + c)^2))/((a^4*b^3 - b^7)*d*cos(d*x + c)^2 + 2*(a^3*b^4 + a*b^6)*d*cos(d*x +
c)*sin(d*x + c) + (a^2*b^5 + b^7)*d)

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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{\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)**4/(a+b*tan(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**4/(a + b*tan(c + d*x))**3, x)

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Giac [A]  time = 1.40997, size = 84, normalized size = 1.22 \begin{align*} \frac{\frac{2 \, \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{3}} - \frac{3 \, b \tan \left (d x + c\right )^{2} + 2 \, a \tan \left (d x + c\right ) + b}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} b^{2}}}{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))^3,x, algorithm="giac")

[Out]

1/2*(2*log(abs(b*tan(d*x + c) + a))/b^3 - (3*b*tan(d*x + c)^2 + 2*a*tan(d*x + c) + b)/((b*tan(d*x + c) + a)^2*
b^2))/d