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

Optimal. Leaf size=107 \[ -\frac{b^2 \left (17 a^2+2 b^2\right ) \cot (c+d x)}{3 d}-\frac{2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (c+d x))}{d}+a^4 x-\frac{4 a b^3 \cot (c+d x) \csc (c+d x)}{3 d}-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))^2}{3 d} \]

[Out]

a^4*x - (2*a*b*(2*a^2 + b^2)*ArcTanh[Cos[c + d*x]])/d - (b^2*(17*a^2 + 2*b^2)*Cot[c + d*x])/(3*d) - (4*a*b^3*C
ot[c + d*x]*Csc[c + d*x])/(3*d) - (b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^2)/(3*d)

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Rubi [A]  time = 0.109771, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {3782, 4048, 3770, 3767, 8} \[ -\frac{b^2 \left (17 a^2+2 b^2\right ) \cot (c+d x)}{3 d}-\frac{2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (c+d x))}{d}+a^4 x-\frac{4 a b^3 \cot (c+d x) \csc (c+d x)}{3 d}-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x])^4,x]

[Out]

a^4*x - (2*a*b*(2*a^2 + b^2)*ArcTanh[Cos[c + d*x]])/d - (b^2*(17*a^2 + 2*b^2)*Cot[c + d*x])/(3*d) - (4*a*b^3*C
ot[c + d*x]*Csc[c + d*x])/(3*d) - (b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^2)/(3*d)

Rule 3782

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) +
3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rule 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (a+b \csc (c+d x))^4 \, dx &=-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))^2}{3 d}+\frac{1}{3} \int (a+b \csc (c+d x)) \left (3 a^3+b \left (9 a^2+2 b^2\right ) \csc (c+d x)+8 a b^2 \csc ^2(c+d x)\right ) \, dx\\ &=-\frac{4 a b^3 \cot (c+d x) \csc (c+d x)}{3 d}-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))^2}{3 d}+\frac{1}{6} \int \left (6 a^4+12 a b \left (2 a^2+b^2\right ) \csc (c+d x)+2 b^2 \left (17 a^2+2 b^2\right ) \csc ^2(c+d x)\right ) \, dx\\ &=a^4 x-\frac{4 a b^3 \cot (c+d x) \csc (c+d x)}{3 d}-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))^2}{3 d}+\left (2 a b \left (2 a^2+b^2\right )\right ) \int \csc (c+d x) \, dx+\frac{1}{3} \left (b^2 \left (17 a^2+2 b^2\right )\right ) \int \csc ^2(c+d x) \, dx\\ &=a^4 x-\frac{2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (c+d x))}{d}-\frac{4 a b^3 \cot (c+d x) \csc (c+d x)}{3 d}-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))^2}{3 d}-\frac{\left (b^2 \left (17 a^2+2 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{3 d}\\ &=a^4 x-\frac{2 a b \left (2 a^2+b^2\right ) \tanh ^{-1}(\cos (c+d x))}{d}-\frac{b^2 \left (17 a^2+2 b^2\right ) \cot (c+d x)}{3 d}-\frac{4 a b^3 \cot (c+d x) \csc (c+d x)}{3 d}-\frac{b^2 \cot (c+d x) (a+b \csc (c+d x))^2}{3 d}\\ \end{align*}

Mathematica [B]  time = 6.23832, size = 568, normalized size = 5.31 \[ \frac{\sin ^4(c+d x) \csc \left (\frac{1}{2} (c+d x)\right ) \left (b^4 \left (-\cos \left (\frac{1}{2} (c+d x)\right )\right )-9 a^2 b^2 \cos \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \csc (c+d x))^4}{3 d (a \sin (c+d x)+b)^4}+\frac{2 \left (2 a^3 b+a b^3\right ) \sin ^4(c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \csc (c+d x))^4}{d (a \sin (c+d x)+b)^4}+\frac{\sin ^4(c+d x) \sec \left (\frac{1}{2} (c+d x)\right ) \left (9 a^2 b^2 \sin \left (\frac{1}{2} (c+d x)\right )+b^4 \sin \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \csc (c+d x))^4}{3 d (a \sin (c+d x)+b)^4}-\frac{2 \left (2 a^3 b+a b^3\right ) \sin ^4(c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a+b \csc (c+d x))^4}{d (a \sin (c+d x)+b)^4}+\frac{a^4 (c+d x) \sin ^4(c+d x) (a+b \csc (c+d x))^4}{d (a \sin (c+d x)+b)^4}-\frac{a b^3 \sin ^4(c+d x) \csc ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \csc (c+d x))^4}{2 d (a \sin (c+d x)+b)^4}-\frac{b^4 \sin ^4(c+d x) \cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \csc (c+d x))^4}{24 d (a \sin (c+d x)+b)^4}+\frac{a b^3 \sin ^4(c+d x) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \csc (c+d x))^4}{2 d (a \sin (c+d x)+b)^4}+\frac{b^4 \sin ^4(c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \csc (c+d x))^4}{24 d (a \sin (c+d x)+b)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csc[c + d*x])^4,x]

[Out]

(a^4*(c + d*x)*(a + b*Csc[c + d*x])^4*Sin[c + d*x]^4)/(d*(b + a*Sin[c + d*x])^4) + ((-9*a^2*b^2*Cos[(c + d*x)/
2] - b^4*Cos[(c + d*x)/2])*Csc[(c + d*x)/2]*(a + b*Csc[c + d*x])^4*Sin[c + d*x]^4)/(3*d*(b + a*Sin[c + d*x])^4
) - (a*b^3*Csc[(c + d*x)/2]^2*(a + b*Csc[c + d*x])^4*Sin[c + d*x]^4)/(2*d*(b + a*Sin[c + d*x])^4) - (b^4*Cot[(
c + d*x)/2]*Csc[(c + d*x)/2]^2*(a + b*Csc[c + d*x])^4*Sin[c + d*x]^4)/(24*d*(b + a*Sin[c + d*x])^4) - (2*(2*a^
3*b + a*b^3)*(a + b*Csc[c + d*x])^4*Log[Cos[(c + d*x)/2]]*Sin[c + d*x]^4)/(d*(b + a*Sin[c + d*x])^4) + (2*(2*a
^3*b + a*b^3)*(a + b*Csc[c + d*x])^4*Log[Sin[(c + d*x)/2]]*Sin[c + d*x]^4)/(d*(b + a*Sin[c + d*x])^4) + (a*b^3
*(a + b*Csc[c + d*x])^4*Sec[(c + d*x)/2]^2*Sin[c + d*x]^4)/(2*d*(b + a*Sin[c + d*x])^4) + ((a + b*Csc[c + d*x]
)^4*Sec[(c + d*x)/2]*(9*a^2*b^2*Sin[(c + d*x)/2] + b^4*Sin[(c + d*x)/2])*Sin[c + d*x]^4)/(3*d*(b + a*Sin[c + d
*x])^4) + (b^4*(a + b*Csc[c + d*x])^4*Sec[(c + d*x)/2]^2*Sin[c + d*x]^4*Tan[(c + d*x)/2])/(24*d*(b + a*Sin[c +
d*x])^4)

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Maple [A]  time = 0.05, size = 139, normalized size = 1.3 \begin{align*}{a}^{4}x+{\frac{{a}^{4}c}{d}}+4\,{\frac{{a}^{3}b\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-6\,{\frac{{a}^{2}{b}^{2}\cot \left ( dx+c \right ) }{d}}-2\,{\frac{a{b}^{3}\cot \left ( dx+c \right ) \csc \left ( dx+c \right ) }{d}}+2\,{\frac{a{b}^{3}\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-{\frac{2\,{b}^{4}\cot \left ( dx+c \right ) }{3\,d}}-{\frac{{b}^{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((a+b*csc(d*x+c))^4,x)

[Out]

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

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Maxima [A]  time = 1.00559, size = 169, normalized size = 1.58 \begin{align*} a^{4} x + \frac{a b^{3}{\left (\frac{2 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )}}{d} - \frac{4 \, a^{3} b \log \left (\cot \left (d x + c\right ) + \csc \left (d x + c\right )\right )}{d} - \frac{6 \, a^{2} b^{2}}{d \tan \left (d x + c\right )} - \frac{{\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} b^{4}}{3 \, d \tan \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.523606, size = 524, normalized size = 4.9 \begin{align*} -\frac{2 \,{\left (9 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (2 \, a^{3} b + a b^{3} -{\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) + 3 \,{\left (2 \, a^{3} b + a b^{3} -{\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) \sin \left (d x + c\right ) - 3 \,{\left (6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right ) - 3 \,{\left (a^{4} d x \cos \left (d x + c\right )^{2} - a^{4} d x + 2 \, a b^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \,{\left (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))^4,x, algorithm="fricas")

[Out]

-1/3*(2*(9*a^2*b^2 + b^4)*cos(d*x + c)^3 - 3*(2*a^3*b + a*b^3 - (2*a^3*b + a*b^3)*cos(d*x + c)^2)*log(1/2*cos(
d*x + c) + 1/2)*sin(d*x + c) + 3*(2*a^3*b + a*b^3 - (2*a^3*b + a*b^3)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) +
1/2)*sin(d*x + c) - 3*(6*a^2*b^2 + b^4)*cos(d*x + c) - 3*(a^4*d*x*cos(d*x + c)^2 - a^4*d*x + 2*a*b^3*cos(d*x +
c))*sin(d*x + c))/((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{\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))**4,x)

[Out]

Integral((a + b*csc(c + d*x))**4, x)

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Giac [B]  time = 1.59847, size = 277, normalized size = 2.59 \begin{align*} \frac{b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 24 \,{\left (d x + c\right )} a^{4} + 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 48 \,{\left (2 \, a^{3} b + a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{176 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 88 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 9 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{4}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b^4*tan(1/2*d*x + 1/2*c)^3 + 12*a*b^3*tan(1/2*d*x + 1/2*c)^2 + 24*(d*x + c)*a^4 + 72*a^2*b^2*tan(1/2*d*x
+ 1/2*c) + 9*b^4*tan(1/2*d*x + 1/2*c) + 48*(2*a^3*b + a*b^3)*log(abs(tan(1/2*d*x + 1/2*c))) - (176*a^3*b*tan(
1/2*d*x + 1/2*c)^3 + 88*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 + 9*b^4*tan(1/2*d*x +
1/2*c)^2 + 12*a*b^3*tan(1/2*d*x + 1/2*c) + b^4)/tan(1/2*d*x + 1/2*c)^3)/d