3.31 \(\int \csc ^6(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=122 \[ -\frac{\left (2 a^2+b^2\right ) \cot ^3(c+d x)}{3 d}-\frac{\left (a^2+2 b^2\right ) \cot (c+d x)}{d}-\frac{a^2 \cot ^5(c+d x)}{5 d}-\frac{a b \cot ^4(c+d x)}{2 d}-\frac{2 a b \cot ^2(c+d x)}{d}+\frac{2 a b \log (\tan (c+d x))}{d}+\frac{b^2 \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.101281, antiderivative size = 122, 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 = {3516, 948} \[ -\frac{\left (2 a^2+b^2\right ) \cot ^3(c+d x)}{3 d}-\frac{\left (a^2+2 b^2\right ) \cot (c+d x)}{d}-\frac{a^2 \cot ^5(c+d x)}{5 d}-\frac{a b \cot ^4(c+d x)}{2 d}-\frac{2 a b \cot ^2(c+d x)}{d}+\frac{2 a b \log (\tan (c+d x))}{d}+\frac{b^2 \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^6*(a + b*Tan[c + d*x])^2,x]

[Out]

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

Rule 3516

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]

Rule 948

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

Rubi steps

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

Mathematica [A]  time = 1.5103, size = 114, normalized size = 0.93 \[ -\frac{2 \cot (c+d x) \left (\left (4 a^2+5 b^2\right ) \csc ^2(c+d x)+3 a^2 \csc ^4(c+d x)+8 a^2+25 b^2\right )+15 b \left (a \csc ^4(c+d x)+2 a \csc ^2(c+d x)-4 a \log (\sin (c+d x))+4 a \log (\cos (c+d x))-2 b \tan (c+d x)\right )}{30 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^6*(a + b*Tan[c + d*x])^2,x]

[Out]

-(2*Cot[c + d*x]*(8*a^2 + 25*b^2 + (4*a^2 + 5*b^2)*Csc[c + d*x]^2 + 3*a^2*Csc[c + d*x]^4) + 15*b*(2*a*Csc[c +
d*x]^2 + a*Csc[c + d*x]^4 + 4*a*Log[Cos[c + d*x]] - 4*a*Log[Sin[c + d*x]] - 2*b*Tan[c + d*x]))/(30*d)

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Maple [A]  time = 0.055, size = 166, normalized size = 1.4 \begin{align*} -{\frac{{b}^{2}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }}+{\frac{4\,{b}^{2}}{3\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-{\frac{8\,{b}^{2}\cot \left ( dx+c \right ) }{3\,d}}-{\frac{ab}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-{\frac{ab}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{ab\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}-{\frac{8\,{a}^{2}\cot \left ( dx+c \right ) }{15\,d}}-{\frac{{a}^{2}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{4\,{a}^{2}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^6*(a+b*tan(d*x+c))^2,x)

[Out]

-1/3/d*b^2/sin(d*x+c)^3/cos(d*x+c)+4/3/d*b^2/sin(d*x+c)/cos(d*x+c)-8/3/d*b^2*cot(d*x+c)-1/2/d*a*b/sin(d*x+c)^4
-1/d*a*b/sin(d*x+c)^2+2*a*b*ln(tan(d*x+c))/d-8/15*a^2*cot(d*x+c)/d-1/5/d*a^2*cot(d*x+c)*csc(d*x+c)^4-4/15/d*a^
2*cot(d*x+c)*csc(d*x+c)^2

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Maxima [A]  time = 1.03974, size = 140, normalized size = 1.15 \begin{align*} \frac{60 \, a b \log \left (\tan \left (d x + c\right )\right ) + 30 \, b^{2} \tan \left (d x + c\right ) - \frac{60 \, a b \tan \left (d x + c\right )^{3} + 30 \,{\left (a^{2} + 2 \, b^{2}\right )} \tan \left (d x + c\right )^{4} + 15 \, a b \tan \left (d x + c\right ) + 10 \,{\left (2 \, a^{2} + b^{2}\right )} \tan \left (d x + c\right )^{2} + 6 \, a^{2}}{\tan \left (d x + c\right )^{5}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/30*(60*a*b*log(tan(d*x + c)) + 30*b^2*tan(d*x + c) - (60*a*b*tan(d*x + c)^3 + 30*(a^2 + 2*b^2)*tan(d*x + c)^
4 + 15*a*b*tan(d*x + c) + 10*(2*a^2 + b^2)*tan(d*x + c)^2 + 6*a^2)/tan(d*x + c)^5)/d

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Fricas [B]  time = 2.42819, size = 628, normalized size = 5.15 \begin{align*} -\frac{16 \,{\left (a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 40 \,{\left (a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \,{\left (a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left (a b \cos \left (d x + c\right )^{5} - 2 \, a b \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) \sin \left (d x + c\right ) - 30 \,{\left (a b \cos \left (d x + c\right )^{5} - 2 \, a b \cos \left (d x + c\right )^{3} + a b \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{4} \, \cos \left (d x + c\right )^{2} + \frac{1}{4}\right ) \sin \left (d x + c\right ) - 30 \, b^{2} - 15 \,{\left (2 \, a b \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{30 \,{\left (d \cos \left (d x + c\right )^{5} - 2 \, d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/30*(16*(a^2 + 5*b^2)*cos(d*x + c)^6 - 40*(a^2 + 5*b^2)*cos(d*x + c)^4 + 30*(a^2 + 5*b^2)*cos(d*x + c)^2 + 3
0*(a*b*cos(d*x + c)^5 - 2*a*b*cos(d*x + c)^3 + a*b*cos(d*x + c))*log(cos(d*x + c)^2)*sin(d*x + c) - 30*(a*b*co
s(d*x + c)^5 - 2*a*b*cos(d*x + c)^3 + a*b*cos(d*x + c))*log(-1/4*cos(d*x + c)^2 + 1/4)*sin(d*x + c) - 30*b^2 -
15*(2*a*b*cos(d*x + c)^3 - 3*a*b*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^5 - 2*d*cos(d*x + c)^3 + d*cos(
d*x + c))*sin(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**6*(a+b*tan(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.40844, size = 177, normalized size = 1.45 \begin{align*} \frac{60 \, a b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + 30 \, b^{2} \tan \left (d x + c\right ) - \frac{137 \, a b \tan \left (d x + c\right )^{5} + 30 \, a^{2} \tan \left (d x + c\right )^{4} + 60 \, b^{2} \tan \left (d x + c\right )^{4} + 60 \, a b \tan \left (d x + c\right )^{3} + 20 \, a^{2} \tan \left (d x + c\right )^{2} + 10 \, b^{2} \tan \left (d x + c\right )^{2} + 15 \, a b \tan \left (d x + c\right ) + 6 \, a^{2}}{\tan \left (d x + c\right )^{5}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^6*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/30*(60*a*b*log(abs(tan(d*x + c))) + 30*b^2*tan(d*x + c) - (137*a*b*tan(d*x + c)^5 + 30*a^2*tan(d*x + c)^4 +
60*b^2*tan(d*x + c)^4 + 60*a*b*tan(d*x + c)^3 + 20*a^2*tan(d*x + c)^2 + 10*b^2*tan(d*x + c)^2 + 15*a*b*tan(d*x
+ c) + 6*a^2)/tan(d*x + c)^5)/d