3.4.78 \(\int \cot ^5(e+f x) (b \csc (e+f x))^m \, dx\) [378]

Optimal. Leaf size=69 \[ -\frac {(b \csc (e+f x))^m}{f m}+\frac {2 (b \csc (e+f x))^{2+m}}{b^2 f (2+m)}-\frac {(b \csc (e+f x))^{4+m}}{b^4 f (4+m)} \]

[Out]

-(b*csc(f*x+e))^m/f/m+2*(b*csc(f*x+e))^(2+m)/b^2/f/(2+m)-(b*csc(f*x+e))^(4+m)/b^4/f/(4+m)

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Rubi [A]
time = 0.04, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2686, 276} \begin {gather*} -\frac {(b \csc (e+f x))^{m+4}}{b^4 f (m+4)}+\frac {2 (b \csc (e+f x))^{m+2}}{b^2 f (m+2)}-\frac {(b \csc (e+f x))^m}{f m} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^5*(b*Csc[e + f*x])^m,x]

[Out]

-((b*Csc[e + f*x])^m/(f*m)) + (2*(b*Csc[e + f*x])^(2 + m))/(b^2*f*(2 + m)) - (b*Csc[e + f*x])^(4 + m)/(b^4*f*(
4 + m))

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rubi steps

\begin {align*} \int \cot ^5(e+f x) (b \csc (e+f x))^m \, dx &=-\frac {b \text {Subst}\left (\int (b x)^{-1+m} \left (-1+x^2\right )^2 \, dx,x,\csc (e+f x)\right )}{f}\\ &=-\frac {b \text {Subst}\left (\int \left ((b x)^{-1+m}-\frac {2 (b x)^{1+m}}{b^2}+\frac {(b x)^{3+m}}{b^4}\right ) \, dx,x,\csc (e+f x)\right )}{f}\\ &=-\frac {(b \csc (e+f x))^m}{f m}+\frac {2 (b \csc (e+f x))^{2+m}}{b^2 f (2+m)}-\frac {(b \csc (e+f x))^{4+m}}{b^4 f (4+m)}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 63, normalized size = 0.91 \begin {gather*} -\frac {(b \csc (e+f x))^m \left (8+6 m+m^2-2 m (4+m) \csc ^2(e+f x)+m (2+m) \csc ^4(e+f x)\right )}{f m (2+m) (4+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^5*(b*Csc[e + f*x])^m,x]

[Out]

-(((b*Csc[e + f*x])^m*(8 + 6*m + m^2 - 2*m*(4 + m)*Csc[e + f*x]^2 + m*(2 + m)*Csc[e + f*x]^4))/(f*m*(2 + m)*(4
+ m)))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.36, size = 9618, normalized size = 139.39

method	result	size
risch	\(\text {Expression too large to display}\)	\(9618\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5*(b*csc(f*x+e))^m,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [A]
time = 0.28, size = 83, normalized size = 1.20 \begin {gather*} -\frac {\frac {b^{m} \sin \left (f x + e\right )^{-m}}{m} - \frac {2 \, b^{m} \sin \left (f x + e\right )^{-m}}{{\left (m + 2\right )} \sin \left (f x + e\right )^{2}} + \frac {b^{m} \sin \left (f x + e\right )^{-m}}{{\left (m + 4\right )} \sin \left (f x + e\right )^{4}}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(b*csc(f*x+e))^m,x, algorithm="maxima")

[Out]

-(b^m*sin(f*x + e)^(-m)/m - 2*b^m*sin(f*x + e)^(-m)/((m + 2)*sin(f*x + e)^2) + b^m*sin(f*x + e)^(-m)/((m + 4)*
sin(f*x + e)^4))/f

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Fricas [A]
time = 0.39, size = 120, normalized size = 1.74 \begin {gather*} -\frac {{\left ({\left (m^{2} + 6 \, m + 8\right )} \cos \left (f x + e\right )^{4} - 4 \, {\left (m + 4\right )} \cos \left (f x + e\right )^{2} + 8\right )} \left (\frac {b}{\sin \left (f x + e\right )}\right )^{m}}{{\left (f m^{3} + 6 \, f m^{2} + 8 \, f m\right )} \cos \left (f x + e\right )^{4} + f m^{3} + 6 \, f m^{2} - 2 \, {\left (f m^{3} + 6 \, f m^{2} + 8 \, f m\right )} \cos \left (f x + e\right )^{2} + 8 \, f m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(b*csc(f*x+e))^m,x, algorithm="fricas")

[Out]

-((m^2 + 6*m + 8)*cos(f*x + e)^4 - 4*(m + 4)*cos(f*x + e)^2 + 8)*(b/sin(f*x + e))^m/((f*m^3 + 6*f*m^2 + 8*f*m)
*cos(f*x + e)^4 + f*m^3 + 6*f*m^2 - 2*(f*m^3 + 6*f*m^2 + 8*f*m)*cos(f*x + e)^2 + 8*f*m)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} x \left (b \csc {\left (e \right )}\right )^{m} \cot ^{5}{\left (e \right )} & \text {for}\: f = 0 \\\frac {\int \frac {\cot ^{5}{\left (e + f x \right )}}{\csc ^{4}{\left (e + f x \right )}}\, dx}{b^{4}} & \text {for}\: m = -4 \\\frac {\int \frac {\cot ^{5}{\left (e + f x \right )}}{\csc ^{2}{\left (e + f x \right )}}\, dx}{b^{2}} & \text {for}\: m = -2 \\- \frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {\log {\left (\tan {\left (e + f x \right )} \right )}}{f} + \frac {1}{2 f \tan ^{2}{\left (e + f x \right )}} - \frac {1}{4 f \tan ^{4}{\left (e + f x \right )}} & \text {for}\: m = 0 \\- \frac {m^{2} \left (b \csc {\left (e + f x \right )}\right )^{m} \cot ^{4}{\left (e + f x \right )}}{f m^{3} + 6 f m^{2} + 8 f m} - \frac {2 m \left (b \csc {\left (e + f x \right )}\right )^{m} \cot ^{4}{\left (e + f x \right )}}{f m^{3} + 6 f m^{2} + 8 f m} + \frac {4 m \left (b \csc {\left (e + f x \right )}\right )^{m} \cot ^{2}{\left (e + f x \right )}}{f m^{3} + 6 f m^{2} + 8 f m} - \frac {8 \left (b \csc {\left (e + f x \right )}\right )^{m}}{f m^{3} + 6 f m^{2} + 8 f m} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**5*(b*csc(f*x+e))**m,x)

[Out]

Piecewise((x*(b*csc(e))**m*cot(e)**5, Eq(f, 0)), (Integral(cot(e + f*x)**5/csc(e + f*x)**4, x)/b**4, Eq(m, -4)
), (Integral(cot(e + f*x)**5/csc(e + f*x)**2, x)/b**2, Eq(m, -2)), (-log(tan(e + f*x)**2 + 1)/(2*f) + log(tan(
e + f*x))/f + 1/(2*f*tan(e + f*x)**2) - 1/(4*f*tan(e + f*x)**4), Eq(m, 0)), (-m**2*(b*csc(e + f*x))**m*cot(e +
f*x)**4/(f*m**3 + 6*f*m**2 + 8*f*m) - 2*m*(b*csc(e + f*x))**m*cot(e + f*x)**4/(f*m**3 + 6*f*m**2 + 8*f*m) + 4
*m*(b*csc(e + f*x))**m*cot(e + f*x)**2/(f*m**3 + 6*f*m**2 + 8*f*m) - 8*(b*csc(e + f*x))**m/(f*m**3 + 6*f*m**2
+ 8*f*m), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(b*csc(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((b*csc(f*x + e))^m*cot(f*x + e)^5, x)

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Mupad [B]
time = 7.63, size = 222, normalized size = 3.22 \begin {gather*} -\frac {{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^m\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {2\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )}{f\,m}-\frac {\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )\,\left (6\,m^2+4\,m+48\right )}{f\,m\,\left (m^2+6\,m+8\right )}+\frac {2\,\left (2\,{\sin \left (e+f\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )\,\left (4\,m^2+8\,m-32\right )}{f\,m\,\left (m^2+6\,m+8\right )}\right )}{16\,{\sin \left (e+f\,x\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^5*(b/sin(e + f*x))^m,x)

[Out]

-((b/sin(e + f*x))^m*(sin(4*e + 4*f*x)*1i + 2*sin(2*e + 2*f*x)^2 - 1)*((2*(2*sin(2*e + 2*f*x)^2 - 1)*(sin(4*e
+ 4*f*x)*1i - 2*sin(2*e + 2*f*x)^2 + 1))/(f*m) - ((sin(4*e + 4*f*x)*1i - 2*sin(2*e + 2*f*x)^2 + 1)*(4*m + 6*m^
2 + 48))/(f*m*(6*m + m^2 + 8)) + (2*(2*sin(e + f*x)^2 - 1)*(sin(4*e + 4*f*x)*1i - 2*sin(2*e + 2*f*x)^2 + 1)*(8
*m + 4*m^2 - 32))/(f*m*(6*m + m^2 + 8))))/(16*sin(e + f*x)^4)

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