\(\int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [568]

Optimal result

Integrand size = 21, antiderivative size = 69 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\log (a+b \tan (c+d x))}{b^3 d}-\frac {a^2+b^2}{2 b^3 d (a+b \tan (c+d x))^2}+\frac {2 a}{b^3 d (a+b \tan (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3587, 711} \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {a^2+b^2}{2 b^3 d (a+b \tan (c+d x))^2}+\frac {2 a}{b^3 d (a+b \tan (c+d x))}+\frac {\log (a+b \tan (c+d x))}{b^3 d} \]

[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 711

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]

Rule 3587

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1+\frac {x^2}{b^2}}{(a+x)^3} \, dx,x,b \tan (c+d x)\right )}{b d} \\ & = \frac {\text {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] (verified)

Time = 0.22 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.83 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\log (a+b \tan (c+d x))-\frac {a^2+b^2}{2 (a+b \tan (c+d x))^2}+\frac {2 a}{a+b \tan (c+d x)}}{b^3 d} \]

[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)

Maple [A] (verified)

Time = 37.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3}}-\frac {a^{2}+b^{2}}{2 b^{3} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {2 a}{b^{3} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(63\)
default	\(\frac {\frac {\ln \left (a +b \tan \left (d x +c \right )\right )}{b^{3}}-\frac {a^{2}+b^{2}}{2 b^{3} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {2 a}{b^{3} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(63\)
risch	\(\frac {-2 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a^{2}-2 i a b}{b^{2} \left (i a +b \right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+i a \,{\mathrm e}^{2 i \left (d x +c \right )}-b +i a \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{b^{3} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{b^{3} d}\)	\(160\)

[In]

int(sec(d*x+c)^4/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (67) = 134\).

Time = 0.27 (sec) , antiderivative size = 284, normalized size of antiderivative = 4.12 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {4 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b^{2} - b^{4} - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} b^{2} + b^{4} + {\left (a^{4} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - {\left (a^{2} b^{2} + b^{4} + {\left (a^{4} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right )}{2 \, {\left ({\left (a^{4} b^{3} - b^{7}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} b^{5} + b^{7}\right )} d\right )}} \]

[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)

Sympy [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {4 \, a b \tan \left (d x + c\right ) + 3 \, a^{2} - b^{2}}{b^{5} \tan \left (d x + c\right )^{2} + 2 \, a b^{4} \tan \left (d x + c\right ) + a^{2} b^{3}} + \frac {2 \, \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{3}}}{2 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.56 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {2 \, \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{3}} - \frac {3 \, b \tan \left (d x + c\right )^{2} + 2 \, a \tan \left (d x + c\right ) + b}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} b^{2}}}{2 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 4.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.16 \[ \int \frac {\sec ^4(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {3\,a^2-b^2}{2\,b^3}+\frac {2\,a\,\mathrm {tan}\left (c+d\,x\right )}{b^2}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}+\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{b^3\,d} \]

[In]

int(1/(cos(c + d*x)^4*(a + b*tan(c + d*x))^3),x)

[Out]

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