3.1 \(\int (a+b \csc ^2(c+d x))^4 \, dx\)

Optimal. Leaf size=112 \[ -\frac{b^2 \left (6 a^2+8 a b+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac{b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \cot (c+d x)}{d}+a^4 x-\frac{b^3 (4 a+3 b) \cot ^5(c+d x)}{5 d}-\frac{b^4 \cot ^7(c+d x)}{7 d} \]

[Out]

a^4*x - (b*(2*a + b)*(2*a^2 + 2*a*b + b^2)*Cot[c + d*x])/d - (b^2*(6*a^2 + 8*a*b + 3*b^2)*Cot[c + d*x]^3)/(3*d
) - (b^3*(4*a + 3*b)*Cot[c + d*x]^5)/(5*d) - (b^4*Cot[c + d*x]^7)/(7*d)

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Rubi [A]  time = 0.0667195, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4128, 390, 203} \[ -\frac{b^2 \left (6 a^2+8 a b+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac{b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \cot (c+d x)}{d}+a^4 x-\frac{b^3 (4 a+3 b) \cot ^5(c+d x)}{5 d}-\frac{b^4 \cot ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x]^2)^4,x]

[Out]

a^4*x - (b*(2*a + b)*(2*a^2 + 2*a*b + b^2)*Cot[c + d*x])/d - (b^2*(6*a^2 + 8*a*b + 3*b^2)*Cot[c + d*x]^3)/(3*d
) - (b^3*(4*a + 3*b)*Cot[c + d*x]^5)/(5*d) - (b^4*Cot[c + d*x]^7)/(7*d)

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \left (a+b \csc ^2(c+d x)\right )^4 \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^4}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (b (2 a+b) \left (2 a^2+2 a b+b^2\right )+b^2 \left (6 a^2+8 a b+3 b^2\right ) x^2+b^3 (4 a+3 b) x^4+b^4 x^6+\frac{a^4}{1+x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \cot (c+d x)}{d}-\frac{b^2 \left (6 a^2+8 a b+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac{b^3 (4 a+3 b) \cot ^5(c+d x)}{5 d}-\frac{b^4 \cot ^7(c+d x)}{7 d}-\frac{a^4 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=a^4 x-\frac{b (2 a+b) \left (2 a^2+2 a b+b^2\right ) \cot (c+d x)}{d}-\frac{b^2 \left (6 a^2+8 a b+3 b^2\right ) \cot ^3(c+d x)}{3 d}-\frac{b^3 (4 a+3 b) \cot ^5(c+d x)}{5 d}-\frac{b^4 \cot ^7(c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 3.49556, size = 148, normalized size = 1.32 \[ -\frac{16 \sin ^8(c+d x) \left (a+b \csc ^2(c+d x)\right )^4 \left (b \cot (c+d x) \left (2 b \left (105 a^2+56 a b+12 b^2\right ) \csc ^2(c+d x)+420 a^2 b+420 a^3+6 b^2 (14 a+3 b) \csc ^4(c+d x)+224 a b^2+15 b^3 \csc ^6(c+d x)+48 b^3\right )-105 a^4 (c+d x)\right )}{105 d (a (-\cos (2 (c+d x)))+a+2 b)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csc[c + d*x]^2)^4,x]

[Out]

(-16*(a + b*Csc[c + d*x]^2)^4*(-105*a^4*(c + d*x) + b*Cot[c + d*x]*(420*a^3 + 420*a^2*b + 224*a*b^2 + 48*b^3 +
2*b*(105*a^2 + 56*a*b + 12*b^2)*Csc[c + d*x]^2 + 6*b^2*(14*a + 3*b)*Csc[c + d*x]^4 + 15*b^3*Csc[c + d*x]^6))*
Sin[c + d*x]^8)/(105*d*(a + 2*b - a*Cos[2*(c + d*x)])^4)

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Maple [A]  time = 0.037, size = 129, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{4} \left ( dx+c \right ) -4\,{a}^{3}b\cot \left ( dx+c \right ) +6\,{a}^{2}{b}^{2} \left ( -2/3-1/3\, \left ( \csc \left ( dx+c \right ) \right ) ^{2} \right ) \cot \left ( dx+c \right ) +4\,a{b}^{3} \left ( -{\frac{8}{15}}-1/5\, \left ( \csc \left ( dx+c \right ) \right ) ^{4}-{\frac{4\, \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \cot \left ( dx+c \right ) +{b}^{4} \left ( -{\frac{16}{35}}-{\frac{ \left ( \csc \left ( dx+c \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \csc \left ( dx+c \right ) \right ) ^{4}}{35}}-{\frac{8\, \left ( \csc \left ( dx+c \right ) \right ) ^{2}}{35}} \right ) \cot \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*csc(d*x+c)^2)^4,x)

[Out]

1/d*(a^4*(d*x+c)-4*a^3*b*cot(d*x+c)+6*a^2*b^2*(-2/3-1/3*csc(d*x+c)^2)*cot(d*x+c)+4*a*b^3*(-8/15-1/5*csc(d*x+c)
^4-4/15*csc(d*x+c)^2)*cot(d*x+c)+b^4*(-16/35-1/7*csc(d*x+c)^6-6/35*csc(d*x+c)^4-8/35*csc(d*x+c)^2)*cot(d*x+c))

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Maxima [A]  time = 1.10806, size = 190, normalized size = 1.7 \begin{align*} a^{4} x - \frac{4 \, a^{3} b}{d \tan \left (d x + c\right )} - \frac{2 \,{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a^{2} b^{2}}{d \tan \left (d x + c\right )^{3}} - \frac{4 \,{\left (15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} + 3\right )} a b^{3}}{15 \, d \tan \left (d x + c\right )^{5}} - \frac{{\left (35 \, \tan \left (d x + c\right )^{6} + 35 \, \tan \left (d x + c\right )^{4} + 21 \, \tan \left (d x + c\right )^{2} + 5\right )} b^{4}}{35 \, d \tan \left (d x + c\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csc(d*x+c)^2)^4,x, algorithm="maxima")

[Out]

a^4*x - 4*a^3*b/(d*tan(d*x + c)) - 2*(3*tan(d*x + c)^2 + 1)*a^2*b^2/(d*tan(d*x + c)^3) - 4/15*(15*tan(d*x + c)
^4 + 10*tan(d*x + c)^2 + 3)*a*b^3/(d*tan(d*x + c)^5) - 1/35*(35*tan(d*x + c)^6 + 35*tan(d*x + c)^4 + 21*tan(d*
x + c)^2 + 5)*b^4/(d*tan(d*x + c)^7)

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Fricas [B]  time = 0.504415, size = 602, normalized size = 5.38 \begin{align*} -\frac{4 \,{\left (105 \, a^{3} b + 105 \, a^{2} b^{2} + 56 \, a b^{3} + 12 \, b^{4}\right )} \cos \left (d x + c\right )^{7} - 14 \,{\left (90 \, a^{3} b + 105 \, a^{2} b^{2} + 56 \, a b^{3} + 12 \, b^{4}\right )} \cos \left (d x + c\right )^{5} + 70 \,{\left (18 \, a^{3} b + 24 \, a^{2} b^{2} + 14 \, a b^{3} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{3} - 105 \,{\left (4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right ) - 105 \,{\left (a^{4} d x \cos \left (d x + c\right )^{6} - 3 \, a^{4} d x \cos \left (d x + c\right )^{4} + 3 \, a^{4} d x \cos \left (d x + c\right )^{2} - a^{4} d x\right )} \sin \left (d x + c\right )}{105 \,{\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csc(d*x+c)^2)^4,x, algorithm="fricas")

[Out]

-1/105*(4*(105*a^3*b + 105*a^2*b^2 + 56*a*b^3 + 12*b^4)*cos(d*x + c)^7 - 14*(90*a^3*b + 105*a^2*b^2 + 56*a*b^3
+ 12*b^4)*cos(d*x + c)^5 + 70*(18*a^3*b + 24*a^2*b^2 + 14*a*b^3 + 3*b^4)*cos(d*x + c)^3 - 105*(4*a^3*b + 6*a^
2*b^2 + 4*a*b^3 + b^4)*cos(d*x + c) - 105*(a^4*d*x*cos(d*x + c)^6 - 3*a^4*d*x*cos(d*x + c)^4 + 3*a^4*d*x*cos(d
*x + c)^2 - a^4*d*x)*sin(d*x + c))/((d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)*sin(d*x +
c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \csc ^{2}{\left (c + d x \right )}\right )^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csc(d*x+c)**2)**4,x)

[Out]

Integral((a + b*csc(c + d*x)**2)**4, x)

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Giac [B]  time = 1.25578, size = 474, normalized size = 4.23 \begin{align*} \frac{15 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 336 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 147 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3360 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2800 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 735 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 13440 \,{\left (d x + c\right )} a^{4} + 26880 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 30240 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 16800 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3675 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{26880 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 30240 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 16800 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 3675 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 3360 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2800 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 735 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 336 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 147 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 15 \, b^{4}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7}}}{13440 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*csc(d*x+c)^2)^4,x, algorithm="giac")

[Out]

1/13440*(15*b^4*tan(1/2*d*x + 1/2*c)^7 + 336*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 147*b^4*tan(1/2*d*x + 1/2*c)^5 + 3
360*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 2800*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 735*b^4*tan(1/2*d*x + 1/2*c)^3 + 1344
0*(d*x + c)*a^4 + 26880*a^3*b*tan(1/2*d*x + 1/2*c) + 30240*a^2*b^2*tan(1/2*d*x + 1/2*c) + 16800*a*b^3*tan(1/2*
d*x + 1/2*c) + 3675*b^4*tan(1/2*d*x + 1/2*c) - (26880*a^3*b*tan(1/2*d*x + 1/2*c)^6 + 30240*a^2*b^2*tan(1/2*d*x
+ 1/2*c)^6 + 16800*a*b^3*tan(1/2*d*x + 1/2*c)^6 + 3675*b^4*tan(1/2*d*x + 1/2*c)^6 + 3360*a^2*b^2*tan(1/2*d*x
+ 1/2*c)^4 + 2800*a*b^3*tan(1/2*d*x + 1/2*c)^4 + 735*b^4*tan(1/2*d*x + 1/2*c)^4 + 336*a*b^3*tan(1/2*d*x + 1/2*
c)^2 + 147*b^4*tan(1/2*d*x + 1/2*c)^2 + 15*b^4)/tan(1/2*d*x + 1/2*c)^7)/d