3.194 \(\int \frac{\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx\)

Optimal. Leaf size=296 \[ \frac{2 b^{4/3} \tan ^{-1}\left (\frac{\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt{a^{2/3}-b^{2/3}}}\right )}{3 a^2 d \sqrt{a^{2/3}-b^{2/3}}}-\frac{2 b^{4/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 a^2 d \sqrt{b^{2/3}-(-1)^{2/3} a^{2/3}}}-\frac{2 b^{4/3} \tanh ^{-1}\left (\frac{(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 a^2 d \sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}}}+\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cot ^3(c+d x)}{3 a d}-\frac{\cot (c+d x)}{a d} \]

[Out]

(2*b^(4/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*a^2*Sqrt[a^(2/3) - b^(2/3)
]*d) + (b*ArcTanh[Cos[c + d*x]])/(a^2*d) - (2*b^(4/3)*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^(1/3)*Tan[(c + d*x)/2])/
Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]])/(3*a^2*Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]*d) - (2*b^(4/3)*ArcTanh[(
b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x)/2])/Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]])/(3*a^2*Sqrt[(-1)^(1/3)*a^(
2/3) + b^(2/3)]*d) - Cot[c + d*x]/(a*d) - Cot[c + d*x]^3/(3*a*d)

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Rubi [A]  time = 0.393332, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3220, 3770, 3767, 2660, 618, 204, 206} \[ \frac{2 b^{4/3} \tan ^{-1}\left (\frac{\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt{a^{2/3}-b^{2/3}}}\right )}{3 a^2 d \sqrt{a^{2/3}-b^{2/3}}}-\frac{2 b^{4/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 a^2 d \sqrt{b^{2/3}-(-1)^{2/3} a^{2/3}}}-\frac{2 b^{4/3} \tanh ^{-1}\left (\frac{(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 a^2 d \sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}}}+\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cot ^3(c+d x)}{3 a d}-\frac{\cot (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^4/(a + b*Sin[c + d*x]^3),x]

[Out]

(2*b^(4/3)*ArcTan[(b^(1/3) + a^(1/3)*Tan[(c + d*x)/2])/Sqrt[a^(2/3) - b^(2/3)]])/(3*a^2*Sqrt[a^(2/3) - b^(2/3)
]*d) + (b*ArcTanh[Cos[c + d*x]])/(a^2*d) - (2*b^(4/3)*ArcTanh[(b^(1/3) - (-1)^(1/3)*a^(1/3)*Tan[(c + d*x)/2])/
Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]])/(3*a^2*Sqrt[-((-1)^(2/3)*a^(2/3)) + b^(2/3)]*d) - (2*b^(4/3)*ArcTanh[(
b^(1/3) + (-1)^(2/3)*a^(1/3)*Tan[(c + d*x)/2])/Sqrt[(-1)^(1/3)*a^(2/3) + b^(2/3)]])/(3*a^2*Sqrt[(-1)^(1/3)*a^(
2/3) + b^(2/3)]*d) - Cot[c + d*x]/(a*d) - Cot[c + d*x]^3/(3*a*d)

Rule 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

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 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\csc ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx &=\int \left (-\frac{b \csc (c+d x)}{a^2}+\frac{\csc ^4(c+d x)}{a}+\frac{b^2 \sin ^2(c+d x)}{a^2 \left (a+b \sin ^3(c+d x)\right )}\right ) \, dx\\ &=\frac{\int \csc ^4(c+d x) \, dx}{a}-\frac{b \int \csc (c+d x) \, dx}{a^2}+\frac{b^2 \int \frac{\sin ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx}{a^2}\\ &=\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac{b^2 \int \left (\frac{1}{3 b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}+\frac{1}{3 b^{2/3} \left (-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}+\frac{1}{3 b^{2/3} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)\right )}\right ) \, dx}{a^2}-\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a d}\\ &=\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cot (c+d x)}{a d}-\frac{\cot ^3(c+d x)}{3 a d}+\frac{b^{4/3} \int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)} \, dx}{3 a^2}+\frac{b^{4/3} \int \frac{1}{-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)} \, dx}{3 a^2}+\frac{b^{4/3} \int \frac{1}{(-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} \sin (c+d x)} \, dx}{3 a^2}\\ &=\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cot (c+d x)}{a d}-\frac{\cot ^3(c+d x)}{3 a d}+\frac{\left (2 b^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+2 \sqrt [3]{b} x+\sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^2 d}+\frac{\left (2 b^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-\sqrt [3]{-1} \sqrt [3]{a}+2 \sqrt [3]{b} x-\sqrt [3]{-1} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^2 d}+\frac{\left (2 b^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{(-1)^{2/3} \sqrt [3]{a}+2 \sqrt [3]{b} x+(-1)^{2/3} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^2 d}\\ &=\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{\cot (c+d x)}{a d}-\frac{\cot ^3(c+d x)}{3 a d}-\frac{\left (4 b^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^{2/3}-b^{2/3}\right )-x^2} \, dx,x,2 \sqrt [3]{b}+2 \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^2 d}-\frac{\left (4 b^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left ((-1)^{2/3} a^{2/3}-b^{2/3}\right )-x^2} \, dx,x,2 \sqrt [3]{b}-2 \sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^2 d}-\frac{\left (4 b^{4/3}\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (\sqrt [3]{-1} a^{2/3}+b^{2/3}\right )-x^2} \, dx,x,2 \sqrt [3]{b}+2 (-1)^{2/3} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )\right )}{3 a^2 d}\\ &=\frac{2 b^{4/3} \tan ^{-1}\left (\frac{\sqrt [3]{b}+\sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^{2/3}-b^{2/3}}}\right )}{3 a^2 \sqrt{a^{2/3}-b^{2/3}} d}+\frac{b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac{2 b^{4/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-(-1)^{2/3} a^{2/3}+b^{2/3}}}\right )}{3 a^2 \sqrt{-(-1)^{2/3} a^{2/3}+b^{2/3}} d}-\frac{2 b^{4/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 a^2 \sqrt{\sqrt [3]{-1} a^{2/3}+b^{2/3}} d}-\frac{\cot (c+d x)}{a d}-\frac{\cot ^3(c+d x)}{3 a d}\\ \end{align*}

Mathematica [C]  time = 2.05013, size = 333, normalized size = 1.12 \[ \frac{4 i b^2 \text{RootSum}\left [-8 i \text{$\#$1}^3 a+\text{$\#$1}^6 b-3 \text{$\#$1}^4 b+3 \text{$\#$1}^2 b-b\& ,\frac{-i \text{$\#$1}^4 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 i \text{$\#$1}^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )-i \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (c+d x)+1\right )+2 \text{$\#$1}^4 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )-4 \text{$\#$1}^2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )+2 \tan ^{-1}\left (\frac{\sin (c+d x)}{\cos (c+d x)-\text{$\#$1}}\right )}{-4 i \text{$\#$1}^2 a+\text{$\#$1}^5 b-2 \text{$\#$1}^3 b+\text{$\#$1} b}\& \right ]+8 a \tan \left (\frac{1}{2} (c+d x)\right )-8 a \cot \left (\frac{1}{2} (c+d x)\right )+8 a \sin ^4\left (\frac{1}{2} (c+d x)\right ) \csc ^3(c+d x)-\frac{1}{2} a \sin (c+d x) \csc ^4\left (\frac{1}{2} (c+d x)\right )-24 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+24 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{24 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^4/(a + b*Sin[c + d*x]^3),x]

[Out]

(-8*a*Cot[(c + d*x)/2] + 24*b*Log[Cos[(c + d*x)/2]] - 24*b*Log[Sin[(c + d*x)/2]] + (4*I)*b^2*RootSum[-b + 3*b*
#1^2 - (8*I)*a*#1^3 - 3*b*#1^4 + b*#1^6 & , (2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] - I*Log[1 - 2*Cos[c +
d*x]*#1 + #1^2] - 4*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + (2*I)*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1
^2 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4)/(b*#1 - (4*I)
*a*#1^2 - 2*b*#1^3 + b*#1^5) & ] + 8*a*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - (a*Csc[(c + d*x)/2]^4*Sin[c + d*x])
/2 + 8*a*Tan[(c + d*x)/2])/(24*a^2*d)

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Maple [C]  time = 0.207, size = 176, normalized size = 0.6 \begin{align*}{\frac{1}{24\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{3}{8\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{4\,{b}^{2}}{3\,{a}^{2}d}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{6}+3\,a{{\it \_Z}}^{4}+8\,b{{\it \_Z}}^{3}+3\,a{{\it \_Z}}^{2}+a \right ) }{\frac{{{\it \_R}}^{2}}{{{\it \_R}}^{5}a+2\,{{\it \_R}}^{3}a+4\,{{\it \_R}}^{2}b+{\it \_R}\,a}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -{\it \_R} \right ) }}-{\frac{1}{24\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}}-{\frac{3}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{b}{{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^4/(a+b*sin(d*x+c)^3),x)

[Out]

1/24/d/a*tan(1/2*d*x+1/2*c)^3+3/8/d/a*tan(1/2*d*x+1/2*c)+4/3/d/a^2*b^2*sum(_R^2/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a
)*ln(tan(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a))-1/24/d/a/tan(1/2*d*x+1/2*c)^3-3/8/
d/a/tan(1/2*d*x+1/2*c)-1/d/a^2*b*ln(tan(1/2*d*x+1/2*c))

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Maxima [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*sin(d*x+c)^3),x, algorithm="maxima")

[Out]

Timed out

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Fricas [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*sin(d*x+c)^3),x, algorithm="fricas")

[Out]

Timed out

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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)**4/(a+b*sin(d*x+c)**3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (d x + c\right )^{4}}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^4/(a+b*sin(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(csc(d*x + c)^4/(b*sin(d*x + c)^3 + a), x)