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

Optimal. Leaf size=137 \[ \frac{b^2 \left (6 a^2+b^2\right ) \tan (c+d x)}{d}-\frac{a^2 \left (a^2+6 b^2\right ) \cot (c+d x)}{d}+\frac{4 a b \left (a^2+b^2\right ) \log (\tan (c+d x))}{d}-\frac{2 a^3 b \cot ^2(c+d x)}{d}-\frac{a^4 \cot ^3(c+d x)}{3 d}+\frac{2 a b^3 \tan ^2(c+d x)}{d}+\frac{b^4 \tan ^3(c+d x)}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.104227, antiderivative size = 137, 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, 894} \[ \frac{b^2 \left (6 a^2+b^2\right ) \tan (c+d x)}{d}-\frac{a^2 \left (a^2+6 b^2\right ) \cot (c+d x)}{d}+\frac{4 a b \left (a^2+b^2\right ) \log (\tan (c+d x))}{d}-\frac{2 a^3 b \cot ^2(c+d x)}{d}-\frac{a^4 \cot ^3(c+d x)}{3 d}+\frac{2 a b^3 \tan ^2(c+d x)}{d}+\frac{b^4 \tan ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*(a^2 + 6*b^2)*Cot[c + d*x])/d) - (2*a^3*b*Cot[c + d*x]^2)/d - (a^4*Cot[c + d*x]^3)/(3*d) + (4*a*b*(a^2
+ b^2)*Log[Tan[c + d*x]])/d + (b^2*(6*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 894

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] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

\begin{align*} \int \csc ^4(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 )}{x^4} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac{b \operatorname{Subst}\left (\int \left (6 a^2 \left (1+\frac{b^2}{6 a^2}\right )+\frac{a^4 b^2}{x^4}+\frac{4 a^3 b^2}{x^3}+\frac{a^4+6 a^2 b^2}{x^2}+\frac{4 a \left (a^2+b^2\right )}{x}+4 a x+x^2\right ) \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac{a^2 \left (a^2+6 b^2\right ) \cot (c+d x)}{d}-\frac{2 a^3 b \cot ^2(c+d x)}{d}-\frac{a^4 \cot ^3(c+d x)}{3 d}+\frac{4 a b \left (a^2+b^2\right ) \log (\tan (c+d x))}{d}+\frac{b^2 \left (6 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 = 3.67706, size = 188, normalized size = 1.37 \[ -\frac{\sin (c+d x) \tan ^3(c+d x) (a \cot (c+d x)+b)^4 \left (-2 b^2 \left (9 a^2+b^2\right ) \sin (c+d x) \cos ^2(c+d x)+\cos (c+d x) \left (6 a^3 b \cot ^2(c+d x)+a^4 \cot ^3(c+d x)-6 a b^3\right )+2 a \cos ^3(c+d x) \left (a \left (a^2+9 b^2\right ) \cot (c+d x)+6 b \left (a^2+b^2\right ) (\log (\cos (c+d x))-\log (\sin (c+d x)))\right )+b^4 (-\sin (c+d x))\right )}{3 d (a \cos (c+d x)+b \sin (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b + a*Cot[c + d*x])^4*Sin[c + d*x]*(Cos[c + d*x]*(-6*a*b^3 + 6*a^3*b*Cot[c + d*x]^2 + a^4*Cot[c + d*x]^3) +
2*a*Cos[c + d*x]^3*(a*(a^2 + 9*b^2)*Cot[c + d*x] + 6*b*(a^2 + b^2)*(Log[Cos[c + d*x]] - Log[Sin[c + d*x]])) -
b^4*Sin[c + d*x] - 2*b^2*(9*a^2 + b^2)*Cos[c + d*x]^2*Sin[c + d*x])*Tan[c + d*x]^3)/(3*d*(a*Cos[c + d*x] + b*
Sin[c + d*x])^4)

________________________________________________________________________________________

Maple [A]  time = 0.077, size = 184, normalized size = 1.3 \begin{align*}{\frac{2\,{b}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+2\,{\frac{{b}^{3}a}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+4\,{\frac{{b}^{3}a\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2}}{d\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }}-12\,{\frac{{a}^{2}{b}^{2}\cot \left ( dx+c \right ) }{d}}-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{2\,{a}^{4}\cot \left ( dx+c \right ) }{3\,d}}-{\frac{{a}^{4}\cot \left ( dx+c \right ) \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4*(a+b*tan(d*x+c))^4,x)

[Out]

2/3/d*b^4*tan(d*x+c)+1/3/d*b^4*tan(d*x+c)*sec(d*x+c)^2+2/d*b^3*a/cos(d*x+c)^2+4/d*b^3*a*ln(tan(d*x+c))+6/d*a^2
*b^2/sin(d*x+c)/cos(d*x+c)-12/d*a^2*b^2*cot(d*x+c)-2/d*b*a^3/sin(d*x+c)^2+4*a^3*b*ln(tan(d*x+c))/d-2/3*a^4*cot
(d*x+c)/d-1/3/d*a^4*cot(d*x+c)*csc(d*x+c)^2

________________________________________________________________________________________

Maxima [A]  time = 1.07015, size = 162, normalized size = 1.18 \begin{align*} \frac{b^{4} \tan \left (d x + c\right )^{3} + 6 \, a b^{3} \tan \left (d x + c\right )^{2} + 12 \,{\left (a^{3} b + a b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) + 3 \,{\left (6 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right ) - \frac{6 \, a^{3} b \tan \left (d x + c\right ) + a^{4} + 3 \,{\left (a^{4} + 6 \, a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+b*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/3*(b^4*tan(d*x + c)^3 + 6*a*b^3*tan(d*x + c)^2 + 12*(a^3*b + a*b^3)*log(tan(d*x + c)) + 3*(6*a^2*b^2 + b^4)*
tan(d*x + c) - (6*a^3*b*tan(d*x + c) + a^4 + 3*(a^4 + 6*a^2*b^2)*tan(d*x + c)^2)/tan(d*x + c)^3)/d

________________________________________________________________________________________

Fricas [A]  time = 2.19935, size = 633, normalized size = 4.62 \begin{align*} -\frac{2 \,{\left (a^{4} + 18 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} + 18 \, a^{2} b^{2} \cos \left (d x + c\right )^{2} - 3 \,{\left (a^{4} + 18 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + b^{4} + 6 \,{\left ({\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{5} -{\left (a^{3} b + 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 ) - 6 \,{\left ({\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{5} -{\left (a^{3} b + 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 ) + 6 \,{\left (a b^{3} \cos \left (d x + c\right ) -{\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3 \,{\left (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)^4*(a+b*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/3*(2*(a^4 + 18*a^2*b^2 + b^4)*cos(d*x + c)^6 + 18*a^2*b^2*cos(d*x + c)^2 - 3*(a^4 + 18*a^2*b^2 + b^4)*cos(d
*x + c)^4 + b^4 + 6*((a^3*b + a*b^3)*cos(d*x + c)^5 - (a^3*b + a*b^3)*cos(d*x + c)^3)*log(cos(d*x + c)^2)*sin(
d*x + c) - 6*((a^3*b + a*b^3)*cos(d*x + c)^5 - (a^3*b + a*b^3)*cos(d*x + c)^3)*log(-1/4*cos(d*x + c)^2 + 1/4)*
sin(d*x + c) + 6*(a*b^3*cos(d*x + c) - (a^3*b + a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/((d*cos(d*x + c)^5 - d*co
s(d*x + c)^3)*sin(d*x + c))

________________________________________________________________________________________

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)**4*(a+b*tan(d*x+c))**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.64509, size = 217, normalized size = 1.58 \begin{align*} \frac{b^{4} \tan \left (d x + c\right )^{3} + 6 \, a b^{3} \tan \left (d x + c\right )^{2} + 18 \, a^{2} b^{2} \tan \left (d x + c\right ) + 3 \, b^{4} \tan \left (d x + c\right ) + 12 \,{\left (a^{3} b + a b^{3}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac{22 \, a^{3} b \tan \left (d x + c\right )^{3} + 22 \, a b^{3} \tan \left (d x + c\right )^{3} + 3 \, a^{4} \tan \left (d x + c\right )^{2} + 18 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} + 6 \, a^{3} b \tan \left (d x + c\right ) + a^{4}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4*(a+b*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/3*(b^4*tan(d*x + c)^3 + 6*a*b^3*tan(d*x + c)^2 + 18*a^2*b^2*tan(d*x + c) + 3*b^4*tan(d*x + c) + 12*(a^3*b +
a*b^3)*log(abs(tan(d*x + c))) - (22*a^3*b*tan(d*x + c)^3 + 22*a*b^3*tan(d*x + c)^3 + 3*a^4*tan(d*x + c)^2 + 18
*a^2*b^2*tan(d*x + c)^2 + 6*a^3*b*tan(d*x + c) + a^4)/tan(d*x + c)^3)/d