\(\int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx\) [21]

Optimal result

Integrand size = 19, antiderivative size = 87 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {a \cot (c+d x)}{d}-\frac {b \cot ^2(c+d x)}{d}-\frac {2 a \cot ^3(c+d x)}{3 d}-\frac {b \cot ^4(c+d x)}{4 d}-\frac {a \cot ^5(c+d x)}{5 d}+\frac {b \log (\tan (c+d x))}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {780} \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {a \cot ^5(c+d x)}{5 d}-\frac {2 a \cot ^3(c+d x)}{3 d}-\frac {a \cot (c+d x)}{d}-\frac {b \cot ^4(c+d x)}{4 d}-\frac {b \cot ^2(c+d x)}{d}+\frac {b \log (\tan (c+d x))}{d} \]

[In]

Int[Csc[c + d*x]^6*(a + b*Tan[c + d*x]),x]

[Out]

-((a*Cot[c + d*x])/d) - (b*Cot[c + d*x]^2)/d - (2*a*Cot[c + d*x]^3)/(3*d) - (b*Cot[c + d*x]^4)/(4*d) - (a*Cot[
c + d*x]^5)/(5*d) + (b*Log[Tan[c + d*x]])/d

Rule 780

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a+b x) \left (1+x^2\right )^2}{x^6} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a}{x^6}+\frac {b}{x^5}+\frac {2 a}{x^4}+\frac {2 b}{x^3}+\frac {a}{x^2}+\frac {b}{x}\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {a \cot (c+d x)}{d}-\frac {b \cot ^2(c+d x)}{d}-\frac {2 a \cot ^3(c+d x)}{3 d}-\frac {b \cot ^4(c+d x)}{4 d}-\frac {a \cot ^5(c+d x)}{5 d}+\frac {b \log (\tan (c+d x))}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.33 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {8 a \cot (c+d x)}{15 d}-\frac {b \csc ^2(c+d x)}{2 d}-\frac {4 a \cot (c+d x) \csc ^2(c+d x)}{15 d}-\frac {b \csc ^4(c+d x)}{4 d}-\frac {a \cot (c+d x) \csc ^4(c+d x)}{5 d}-\frac {b \log (\cos (c+d x))}{d}+\frac {b \log (\sin (c+d x))}{d} \]

[In]

Integrate[Csc[c + d*x]^6*(a + b*Tan[c + d*x]),x]

[Out]

(-8*a*Cot[c + d*x])/(15*d) - (b*Csc[c + d*x]^2)/(2*d) - (4*a*Cot[c + d*x]*Csc[c + d*x]^2)/(15*d) - (b*Csc[c +
d*x]^4)/(4*d) - (a*Cot[c + d*x]*Csc[c + d*x]^4)/(5*d) - (b*Log[Cos[c + d*x]])/d + (b*Log[Sin[c + d*x]])/d

Maple [A] (verified)

Time = 5.39 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a \left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (d x +c \right )\right )}{15}\right ) \cot \left (d x +c \right )}{d}\)	\(66\)
default	\(\frac {b \left (-\frac {1}{4 \sin \left (d x +c \right )^{4}}-\frac {1}{2 \sin \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a \left (-\frac {8}{15}-\frac {\left (\csc ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\csc ^{2}\left (d x +c \right )\right )}{15}\right ) \cot \left (d x +c \right )}{d}\)	\(66\)
risch	\(\frac {2 b \,{\mathrm e}^{8 i \left (d x +c \right )}-10 b \,{\mathrm e}^{6 i \left (d x +c \right )}-\frac {32 i a \,{\mathrm e}^{4 i \left (d x +c \right )}}{3}+10 b \,{\mathrm e}^{4 i \left (d x +c \right )}+\frac {16 i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{3}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-\frac {16 i a}{15}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(134\)

[In]

int(csc(d*x+c)^6*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(b*(-1/4/sin(d*x+c)^4-1/2/sin(d*x+c)^2+ln(tan(d*x+c)))+a*(-8/15-1/5*csc(d*x+c)^4-4/15*csc(d*x+c)^2)*cot(d*
x+c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 174 vs. \(2 (81) = 162\).

Time = 0.27 (sec) , antiderivative size = 174, normalized size of antiderivative = 2.00 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {32 \, a \cos \left (d x + c\right )^{5} - 80 \, a \cos \left (d x + c\right )^{3} + 30 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) \sin \left (d x + c\right ) - 30 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + b\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) + 60 \, a \cos \left (d x + c\right ) - 15 \, {\left (2 \, b \cos \left (d x + c\right )^{2} - 3 \, b\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right )} \]

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(32*a*cos(d*x + c)^5 - 80*a*cos(d*x + c)^3 + 30*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(cos(d*x
+ c)^2)*sin(d*x + c) - 30*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + b)*log(-1/4*cos(d*x + c)^2 + 1/4)*sin(d*x +
c) + 60*a*cos(d*x + c) - 15*(2*b*cos(d*x + c)^2 - 3*b)*sin(d*x + c))/((d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2
+ d)*sin(d*x + c))

Sympy [F]

\[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right ) \csc ^{6}{\left (c + d x \right )}\, dx \]

[In]

integrate(csc(d*x+c)**6*(a+b*tan(d*x+c)),x)

[Out]

Integral((a + b*tan(c + d*x))*csc(c + d*x)**6, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {60 \, b \log \left (\tan \left (d x + c\right )\right ) - \frac {60 \, a \tan \left (d x + c\right )^{4} + 60 \, b \tan \left (d x + c\right )^{3} + 40 \, a \tan \left (d x + c\right )^{2} + 15 \, b \tan \left (d x + c\right ) + 12 \, a}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/60*(60*b*log(tan(d*x + c)) - (60*a*tan(d*x + c)^4 + 60*b*tan(d*x + c)^3 + 40*a*tan(d*x + c)^2 + 15*b*tan(d*x
+ c) + 12*a)/tan(d*x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.97 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {60 \, b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {137 \, b \tan \left (d x + c\right )^{5} + 60 \, a \tan \left (d x + c\right )^{4} + 60 \, b \tan \left (d x + c\right )^{3} + 40 \, a \tan \left (d x + c\right )^{2} + 15 \, b \tan \left (d x + c\right ) + 12 \, a}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

1/60*(60*b*log(abs(tan(d*x + c))) - (137*b*tan(d*x + c)^5 + 60*a*tan(d*x + c)^4 + 60*b*tan(d*x + c)^3 + 40*a*t
an(d*x + c)^2 + 15*b*tan(d*x + c) + 12*a)/tan(d*x + c)^5)/d

Mupad [B] (verification not implemented)

Time = 4.57 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \csc ^6(c+d x) (a+b \tan (c+d x)) \, dx=\frac {b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^4+b\,{\mathrm {tan}\left (c+d\,x\right )}^3+\frac {2\,a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{3}+\frac {b\,\mathrm {tan}\left (c+d\,x\right )}{4}+\frac {a}{5}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5} \]

[In]

int((a + b*tan(c + d*x))/sin(c + d*x)^6,x)

[Out]

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