\(\int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx\) [70]

Optimal result

Integrand size = 21, antiderivative size = 95 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\cot (c+d x)}{a^3 d}-\frac {3 b \log (\tan (c+d x))}{a^4 d}+\frac {3 b \log (a+b \tan (c+d x))}{a^4 d}-\frac {b}{2 a^2 d (a+b \tan (c+d x))^2}-\frac {2 b}{a^3 d (a+b \tan (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 95, 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 = {3597, 46} \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {3 b \log (\tan (c+d x))}{a^4 d}+\frac {3 b \log (a+b \tan (c+d x))}{a^4 d}-\frac {2 b}{a^3 d (a+b \tan (c+d x))}-\frac {\cot (c+d x)}{a^3 d}-\frac {b}{2 a^2 d (a+b \tan (c+d x))^2} \]

[In]

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

[Out]

-(Cot[c + d*x]/(a^3*d)) - (3*b*Log[Tan[c + d*x]])/(a^4*d) + (3*b*Log[a + b*Tan[c + d*x]])/(a^4*d) - b/(2*a^2*d
*(a + b*Tan[c + d*x])^2) - (2*b)/(a^3*d*(a + b*Tan[c + d*x]))

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 3597

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(241\) vs. \(2(95)=190\).

Time = 4.19 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.54 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {-2 a^3 \left (a^2+b^2\right ) \cot (c+d x)+b \left (-2 a^2 \left (a^2+b^2\right ) (2+3 \log (\sin (c+d x))-3 \log (a \cos (c+d x)+b \sin (c+d x)))-a^2 b^2 \sec ^2(c+d x)+2 a b \left (2 a^2+b^2-6 \left (a^2+b^2\right ) \log (\sin (c+d x))+6 \left (a^2+b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))\right ) \tan (c+d x)-2 b^2 \left (-3 a^2-2 b^2+3 \left (a^2+b^2\right ) \log (\sin (c+d x))-3 \left (a^2+b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))\right ) \tan ^2(c+d x)\right )}{2 a^4 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2} \]

[In]

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

[Out]

(-2*a^3*(a^2 + b^2)*Cot[c + d*x] + b*(-2*a^2*(a^2 + b^2)*(2 + 3*Log[Sin[c + d*x]] - 3*Log[a*Cos[c + d*x] + b*S
in[c + d*x]]) - a^2*b^2*Sec[c + d*x]^2 + 2*a*b*(2*a^2 + b^2 - 6*(a^2 + b^2)*Log[Sin[c + d*x]] + 6*(a^2 + b^2)*
Log[a*Cos[c + d*x] + b*Sin[c + d*x]])*Tan[c + d*x] - 2*b^2*(-3*a^2 - 2*b^2 + 3*(a^2 + b^2)*Log[Sin[c + d*x]] -
3*(a^2 + b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])*Tan[c + d*x]^2))/(2*a^4*(a^2 + b^2)*d*(a + b*Tan[c + d*x]
)^2)

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {-\frac {1}{a^{3} \tan \left (d x +c \right )}-\frac {3 b \ln \left (\tan \left (d x +c \right )\right )}{a^{4}}-\frac {b}{2 a^{2} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {3 b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{4}}-\frac {2 b}{a^{3} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(85\)
default	\(\frac {-\frac {1}{a^{3} \tan \left (d x +c \right )}-\frac {3 b \ln \left (\tan \left (d x +c \right )\right )}{a^{4}}-\frac {b}{2 a^{2} \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {3 b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{4}}-\frac {2 b}{a^{3} \left (a +b \tan \left (d x +c \right )\right )}}{d}\)	\(85\)
risch	\(-\frac {2 i \left (a^{4} {\mathrm e}^{4 i \left (d x +c \right )}-9 a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-4 i a^{3} b \,{\mathrm e}^{4 i \left (d x +c \right )}+9 i a \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+2 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-6 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-4 i a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}-9 i a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a^{4}+5 a^{2} b^{2}+3 b^{4}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +a \right )^{2} \left (-i b +a \right )^{2} a^{3} d}+\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right )}{a^{4} d}-\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}\)	\(276\)

[In]

int(csc(d*x+c)^2/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 565 vs. \(2 (93) = 186\).

Time = 0.28 (sec) , antiderivative size = 565, normalized size of antiderivative = 5.95 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {2 \, {\left (a^{7} + 4 \, a^{5} b^{2} - 2 \, a^{3} b^{4} - 3 \, a b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (2 \, a^{5} b^{2} - 3 \, a^{3} b^{4} - 3 \, a b^{6}\right )} \cos \left (d x + c\right ) + 3 \, {\left (2 \, {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right ) - {\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7} + {\left (a^{6} b + a^{4} b^{3} - a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 3 \, {\left (2 \, {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right ) - {\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7} + {\left (a^{6} b + a^{4} b^{3} - a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) - {\left (5 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - 4 \, {\left (a^{6} b + 5 \, a^{4} b^{3} + 3 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (2 \, {\left (a^{9} b + 2 \, a^{7} b^{3} + a^{5} b^{5}\right )} d \cos \left (d x + c\right )^{3} - 2 \, {\left (a^{9} b + 2 \, a^{7} b^{3} + a^{5} b^{5}\right )} d \cos \left (d x + c\right ) - {\left ({\left (a^{10} + a^{8} b^{2} - a^{6} b^{4} - a^{4} b^{6}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{8} b^{2} + 2 \, a^{6} b^{4} + a^{4} b^{6}\right )} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(csc(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(2*(a^7 + 4*a^5*b^2 - 2*a^3*b^4 - 3*a*b^6)*cos(d*x + c)^3 - 2*(2*a^5*b^2 - 3*a^3*b^4 - 3*a*b^6)*cos(d*x +
c) + 3*(2*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c)^3 - 2*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c) - (a^4*b
^3 + 2*a^2*b^5 + b^7 + (a^6*b + a^4*b^3 - a^2*b^5 - b^7)*cos(d*x + c)^2)*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) - 3*(2*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c)^3 - 2*(a^5*
b^2 + 2*a^3*b^4 + a*b^6)*cos(d*x + c) - (a^4*b^3 + 2*a^2*b^5 + b^7 + (a^6*b + a^4*b^3 - a^2*b^5 - b^7)*cos(d*x
+ c)^2)*sin(d*x + c))*log(-1/4*cos(d*x + c)^2 + 1/4) - (5*a^4*b^3 + 3*a^2*b^5 - 4*(a^6*b + 5*a^4*b^3 + 3*a^2*
b^5)*cos(d*x + c)^2)*sin(d*x + c))/(2*(a^9*b + 2*a^7*b^3 + a^5*b^5)*d*cos(d*x + c)^3 - 2*(a^9*b + 2*a^7*b^3 +
a^5*b^5)*d*cos(d*x + c) - ((a^10 + a^8*b^2 - a^6*b^4 - a^4*b^6)*d*cos(d*x + c)^2 + (a^8*b^2 + 2*a^6*b^4 + a^4*
b^6)*d)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\csc ^{2}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]

[In]

integrate(csc(d*x+c)**2/(a+b*tan(d*x+c))**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.57 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.14 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\frac {6 \, b^{2} \tan \left (d x + c\right )^{2} + 9 \, a b \tan \left (d x + c\right ) + 2 \, a^{2}}{a^{3} b^{2} \tan \left (d x + c\right )^{3} + 2 \, a^{4} b \tan \left (d x + c\right )^{2} + a^{5} \tan \left (d x + c\right )} - \frac {6 \, b \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4}} + \frac {6 \, b \log \left (\tan \left (d x + c\right )\right )}{a^{4}}}{2 \, d} \]

[In]

integrate(csc(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*((6*b^2*tan(d*x + c)^2 + 9*a*b*tan(d*x + c) + 2*a^2)/(a^3*b^2*tan(d*x + c)^3 + 2*a^4*b*tan(d*x + c)^2 + a
^5*tan(d*x + c)) - 6*b*log(b*tan(d*x + c) + a)/a^4 + 6*b*log(tan(d*x + c))/a^4)/d

Giac [A] (verification not implemented)

none

Time = 0.57 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.19 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {6 \, b \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4}} - \frac {6 \, b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{4}} + \frac {2 \, {\left (3 \, b \tan \left (d x + c\right ) - a\right )}}{a^{4} \tan \left (d x + c\right )} - \frac {9 \, b^{3} \tan \left (d x + c\right )^{2} + 22 \, a b^{2} \tan \left (d x + c\right ) + 14 \, a^{2} b}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2} a^{4}}}{2 \, d} \]

[In]

integrate(csc(d*x+c)^2/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(6*b*log(abs(b*tan(d*x + c) + a))/a^4 - 6*b*log(abs(tan(d*x + c)))/a^4 + 2*(3*b*tan(d*x + c) - a)/(a^4*tan
(d*x + c)) - (9*b^3*tan(d*x + c)^2 + 22*a*b^2*tan(d*x + c) + 14*a^2*b)/((b*tan(d*x + c) + a)^2*a^4))/d

Mupad [B] (verification not implemented)

Time = 4.87 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.04 \[ \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {6\,b\,\mathrm {atanh}\left (\frac {2\,b\,\mathrm {tan}\left (c+d\,x\right )}{a}+1\right )}{a^4\,d}-\frac {\frac {1}{a}+\frac {3\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{a^3}+\frac {9\,b\,\mathrm {tan}\left (c+d\,x\right )}{2\,a^2}}{d\,\left (a^2\,\mathrm {tan}\left (c+d\,x\right )+2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )} \]

[In]

int(1/(sin(c + d*x)^2*(a + b*tan(c + d*x))^3),x)

[Out]

(6*b*atanh((2*b*tan(c + d*x))/a + 1))/(a^4*d) - (1/a + (3*b^2*tan(c + d*x)^2)/a^3 + (9*b*tan(c + d*x))/(2*a^2)
)/(d*(a^2*tan(c + d*x) + b^2*tan(c + d*x)^3 + 2*a*b*tan(c + d*x)^2))