3.49 \(\int \csc ^6(c+d x) (a+b \tan (c+d x))^4 \, dx\)

Optimal. Leaf size=194 \[ \frac{2 b^2 \left (3 a^2+b^2\right ) \tan (c+d x)}{d}-\frac{2 a^2 \left (a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac{2 a b \left (2 a^2+b^2\right ) \cot ^2(c+d x)}{d}-\frac{\left (12 a^2 b^2+a^4+b^4\right ) \cot (c+d x)}{d}+\frac{4 a b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{d}-\frac{a^3 b \cot ^4(c+d x)}{d}-\frac{a^4 \cot ^5(c+d x)}{5 d}+\frac{2 a b^3 \tan ^2(c+d x)}{d}+\frac{b^4 \tan ^3(c+d x)}{3 d} \]

[Out]

-(((a^4 + 12*a^2*b^2 + b^4)*Cot[c + d*x])/d) - (2*a*b*(2*a^2 + b^2)*Cot[c + d*x]^2)/d - (2*a^2*(a^2 + 3*b^2)*C
ot[c + d*x]^3)/(3*d) - (a^3*b*Cot[c + d*x]^4)/d - (a^4*Cot[c + d*x]^5)/(5*d) + (4*a*b*(a^2 + 2*b^2)*Log[Tan[c
+ d*x]])/d + (2*b^2*(3*a^2 + b^2)*Tan[c + d*x])/d + (2*a*b^3*Tan[c + d*x]^2)/d + (b^4*Tan[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.157077, antiderivative size = 194, 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} \[ \frac{2 b^2 \left (3 a^2+b^2\right ) \tan (c+d x)}{d}-\frac{2 a^2 \left (a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac{2 a b \left (2 a^2+b^2\right ) \cot ^2(c+d x)}{d}-\frac{\left (12 a^2 b^2+a^4+b^4\right ) \cot (c+d x)}{d}+\frac{4 a b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{d}-\frac{a^3 b \cot ^4(c+d x)}{d}-\frac{a^4 \cot ^5(c+d x)}{5 d}+\frac{2 a b^3 \tan ^2(c+d x)}{d}+\frac{b^4 \tan ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^6*(a + b*Tan[c + d*x])^4,x]

[Out]

-(((a^4 + 12*a^2*b^2 + b^4)*Cot[c + d*x])/d) - (2*a*b*(2*a^2 + b^2)*Cot[c + d*x]^2)/d - (2*a^2*(a^2 + 3*b^2)*C
ot[c + d*x]^3)/(3*d) - (a^3*b*Cot[c + d*x]^4)/d - (a^4*Cot[c + d*x]^5)/(5*d) + (4*a*b*(a^2 + 2*b^2)*Log[Tan[c
+ d*x]])/d + (2*b^2*(3*a^2 + b^2)*Tan[c + d*x])/d + (2*a*b^3*Tan[c + d*x]^2)/d + (b^4*Tan[c + d*x]^3)/(3*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))^4 \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^4 \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 (2 \left (3 a^2+b^2\right )+\frac{a^4 b^4}{x^6}+\frac{4 a^3 b^4}{x^5}+\frac{2 a^2 b^2 \left (a^2+3 b^2\right )}{x^4}+\frac{4 a b^2 \left (2 a^2+b^2\right )}{x^3}+\frac{a^4+12 a^2 b^2+b^4}{x^2}+\frac{4 \left (a^3+2 a b^2\right )}{x}+4 a x+x^2\right ) \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac{\left (a^4+12 a^2 b^2+b^4\right ) \cot (c+d x)}{d}-\frac{2 a b \left (2 a^2+b^2\right ) \cot ^2(c+d x)}{d}-\frac{2 a^2 \left (a^2+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac{a^3 b \cot ^4(c+d x)}{d}-\frac{a^4 \cot ^5(c+d x)}{5 d}+\frac{4 a b \left (a^2+2 b^2\right ) \log (\tan (c+d x))}{d}+\frac{2 b^2 \left (3 a^2+b^2\right ) \tan (c+d x)}{d}+\frac{2 a b^3 \tan ^2(c+d x)}{d}+\frac{b^4 \tan ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 3.91889, size = 233, normalized size = 1.2 \[ -\frac{(a+b \tan (c+d x))^4 \left (-5 b^2 \left (18 a^2+5 b^2\right ) \sin (c+d x) \cos ^3(c+d x)+2 a \cos ^2(c+d x) \left (15 b \left (a^2+b^2\right ) \cot ^2(c+d x)+a \left (2 a^2+15 b^2\right ) \cot ^3(c+d x)-15 b^3\right )+\cos ^4(c+d x) \left (\left (150 a^2 b^2+8 a^4+15 b^4\right ) \cot (c+d x)+60 a b \left (a^2+2 b^2\right ) (\log (\cos (c+d x))-\log (\sin (c+d x)))\right )+15 a^3 b \cot ^4(c+d x)+3 a^4 \cot ^5(c+d x)-\frac{5}{2} b^4 \sin (2 (c+d x))\right )}{15 d (a \cos (c+d x)+b \sin (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^6*(a + b*Tan[c + d*x])^4,x]

[Out]

-((15*a^3*b*Cot[c + d*x]^4 + 3*a^4*Cot[c + d*x]^5 + 2*a*Cos[c + d*x]^2*(-15*b^3 + 15*b*(a^2 + b^2)*Cot[c + d*x
]^2 + a*(2*a^2 + 15*b^2)*Cot[c + d*x]^3) + Cos[c + d*x]^4*((8*a^4 + 150*a^2*b^2 + 15*b^4)*Cot[c + d*x] + 60*a*
b*(a^2 + 2*b^2)*(Log[Cos[c + d*x]] - Log[Sin[c + d*x]])) - 5*b^2*(18*a^2 + 5*b^2)*Cos[c + d*x]^3*Sin[c + d*x]
- (5*b^4*Sin[2*(c + d*x)])/2)*(a + b*Tan[c + d*x])^4)/(15*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^4)

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Maple [A]  time = 0.082, size = 301, normalized size = 1.6 \begin{align*}{\frac{{b}^{4}}{3\,d\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{4\,{b}^{4}}{3\,d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-{\frac{8\,{b}^{4}\cot \left ( dx+c \right ) }{3\,d}}+2\,{\frac{{b}^{3}a}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-4\,{\frac{{b}^{3}a}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+8\,{\frac{{b}^{3}a\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{{a}^{2}{b}^{2}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) }}+8\,{\frac{{a}^{2}{b}^{2}}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-16\,{\frac{{a}^{2}{b}^{2}\cot \left ( dx+c \right ) }{d}}-{\frac{b{a}^{3}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}-2\,{\frac{b{a}^{3}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+4\,{\frac{b{a}^{3}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}-{\frac{8\,{a}^{4}\cot \left ( dx+c \right ) }{15\,d}}-{\frac{{a}^{4}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{4\,{a}^{4}\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))^4,x)

[Out]

1/3/d*b^4/sin(d*x+c)/cos(d*x+c)^3+4/3/d*b^4/sin(d*x+c)/cos(d*x+c)-8/3/d*b^4*cot(d*x+c)+2/d*b^3*a/sin(d*x+c)^2/
cos(d*x+c)^2-4/d*b^3*a/sin(d*x+c)^2+8/d*b^3*a*ln(tan(d*x+c))-2/d*a^2*b^2/sin(d*x+c)^3/cos(d*x+c)+8/d*a^2*b^2/s
in(d*x+c)/cos(d*x+c)-16/d*a^2*b^2*cot(d*x+c)-1/d*b*a^3/sin(d*x+c)^4-2/d*b*a^3/sin(d*x+c)^2+4*a^3*b*ln(tan(d*x+
c))/d-8/15*a^4*cot(d*x+c)/d-1/5/d*a^4*cot(d*x+c)*csc(d*x+c)^4-4/15/d*a^4*cot(d*x+c)*csc(d*x+c)^2

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Maxima [A]  time = 1.06929, size = 231, normalized size = 1.19 \begin{align*} \frac{5 \, b^{4} \tan \left (d x + c\right )^{3} + 30 \, a b^{3} \tan \left (d x + c\right )^{2} + 60 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) + 30 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right ) - \frac{15 \, a^{3} b \tan \left (d x + c\right ) + 15 \,{\left (a^{4} + 12 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{4} + 3 \, a^{4} + 30 \,{\left (2 \, a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )^{3} + 10 \,{\left (a^{4} + 3 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{5}}}{15 \, 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))^4,x, algorithm="maxima")

[Out]

1/15*(5*b^4*tan(d*x + c)^3 + 30*a*b^3*tan(d*x + c)^2 + 60*(a^3*b + 2*a*b^3)*log(tan(d*x + c)) + 30*(3*a^2*b^2
+ b^4)*tan(d*x + c) - (15*a^3*b*tan(d*x + c) + 15*(a^4 + 12*a^2*b^2 + b^4)*tan(d*x + c)^4 + 3*a^4 + 30*(2*a^3*
b + a*b^3)*tan(d*x + c)^3 + 10*(a^4 + 3*a^2*b^2)*tan(d*x + c)^2)/tan(d*x + c)^5)/d

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Fricas [B]  time = 2.63441, size = 922, normalized size = 4.75 \begin{align*} -\frac{8 \,{\left (a^{4} + 30 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{8} - 20 \,{\left (a^{4} + 30 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{6} + 15 \,{\left (a^{4} + 30 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{4} - 5 \, b^{4} - 10 \,{\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 30 \,{\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{7} - 2 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} +{\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) \sin \left (d x + c\right ) - 30 \,{\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{7} - 2 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} +{\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac{1}{4} \, \cos \left (d x + c\right )^{2} + \frac{1}{4}\right ) \sin \left (d x + c\right ) - 15 \,{\left (2 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} + 2 \, a b^{3} \cos \left (d x + c\right ) - 3 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{15 \,{\left (d \cos \left (d x + c\right )^{7} - 2 \, d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{3}\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))^4,x, algorithm="fricas")

[Out]

-1/15*(8*(a^4 + 30*a^2*b^2 + 5*b^4)*cos(d*x + c)^8 - 20*(a^4 + 30*a^2*b^2 + 5*b^4)*cos(d*x + c)^6 + 15*(a^4 +
30*a^2*b^2 + 5*b^4)*cos(d*x + c)^4 - 5*b^4 - 10*(9*a^2*b^2 + b^4)*cos(d*x + c)^2 + 30*((a^3*b + 2*a*b^3)*cos(d
*x + c)^7 - 2*(a^3*b + 2*a*b^3)*cos(d*x + c)^5 + (a^3*b + 2*a*b^3)*cos(d*x + c)^3)*log(cos(d*x + c)^2)*sin(d*x
+ c) - 30*((a^3*b + 2*a*b^3)*cos(d*x + c)^7 - 2*(a^3*b + 2*a*b^3)*cos(d*x + c)^5 + (a^3*b + 2*a*b^3)*cos(d*x
+ c)^3)*log(-1/4*cos(d*x + c)^2 + 1/4)*sin(d*x + c) - 15*(2*(a^3*b + 2*a*b^3)*cos(d*x + c)^5 + 2*a*b^3*cos(d*x
+ c) - 3*(a^3*b + 2*a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/((d*cos(d*x + c)^7 - 2*d*cos(d*x + c)^5 + d*cos(d*x
+ c)^3)*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))**4,x)

[Out]

Timed out

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Giac [A]  time = 2.65297, size = 317, normalized size = 1.63 \begin{align*} \frac{5 \, b^{4} \tan \left (d x + c\right )^{3} + 30 \, a b^{3} \tan \left (d x + c\right )^{2} + 90 \, a^{2} b^{2} \tan \left (d x + c\right ) + 30 \, b^{4} \tan \left (d x + c\right ) + 60 \,{\left (a^{3} b + 2 \, a b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac{137 \, a^{3} b \tan \left (d x + c\right )^{5} + 274 \, a b^{3} \tan \left (d x + c\right )^{5} + 15 \, a^{4} \tan \left (d x + c\right )^{4} + 180 \, a^{2} b^{2} \tan \left (d x + c\right )^{4} + 15 \, b^{4} \tan \left (d x + c\right )^{4} + 60 \, a^{3} b \tan \left (d x + c\right )^{3} + 30 \, a b^{3} \tan \left (d x + c\right )^{3} + 10 \, a^{4} \tan \left (d x + c\right )^{2} + 30 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 15 \, a^{3} b \tan \left (d x + c\right ) + 3 \, a^{4}}{\tan \left (d x + c\right )^{5}}}{15 \, 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))^4,x, algorithm="giac")

[Out]

1/15*(5*b^4*tan(d*x + c)^3 + 30*a*b^3*tan(d*x + c)^2 + 90*a^2*b^2*tan(d*x + c) + 30*b^4*tan(d*x + c) + 60*(a^3
*b + 2*a*b^3)*log(abs(tan(d*x + c))) - (137*a^3*b*tan(d*x + c)^5 + 274*a*b^3*tan(d*x + c)^5 + 15*a^4*tan(d*x +
c)^4 + 180*a^2*b^2*tan(d*x + c)^4 + 15*b^4*tan(d*x + c)^4 + 60*a^3*b*tan(d*x + c)^3 + 30*a*b^3*tan(d*x + c)^3
+ 10*a^4*tan(d*x + c)^2 + 30*a^2*b^2*tan(d*x + c)^2 + 15*a^3*b*tan(d*x + c) + 3*a^4)/tan(d*x + c)^5)/d