3.328 \(\int \cot ^3(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=57 \[ -\frac {\left (a^2-b^2\right ) \log (\sin (e+f x))}{f}-\frac {(a+b)^2 \csc ^2(e+f x)}{2 f}-\frac {b^2 \log (\cos (e+f x))}{f} \]

[Out]

-1/2*(a+b)^2*csc(f*x+e)^2/f-b^2*ln(cos(f*x+e))/f-(a^2-b^2)*ln(sin(f*x+e))/f

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4138, 446, 88} \[ -\frac {\left (a^2-b^2\right ) \log (\sin (e+f x))}{f}-\frac {(a+b)^2 \csc ^2(e+f x)}{2 f}-\frac {b^2 \log (\cos (e+f x))}{f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^3*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

-((a + b)^2*Csc[e + f*x]^2)/(2*f) - (b^2*Log[Cos[e + f*x]])/f - ((a^2 - b^2)*Log[Sin[e + f*x]])/f

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

Rubi steps

\begin {align*} \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (b+a x^2\right )^2}{x \left (1-x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {(b+a x)^2}{(1-x)^2 x} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {(a+b)^2}{(-1+x)^2}+\frac {a^2-b^2}{-1+x}+\frac {b^2}{x}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {(a+b)^2 \csc ^2(e+f x)}{2 f}-\frac {b^2 \log (\cos (e+f x))}{f}-\frac {\left (a^2-b^2\right ) \log (\sin (e+f x))}{f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 81, normalized size = 1.42 \[ -\frac {2 \left (a \cos ^2(e+f x)+b\right )^2 \left (2 \left (a^2-b^2\right ) \log (\sin (e+f x))+(a+b)^2 \csc ^2(e+f x)+2 b^2 \log (\cos (e+f x))\right )}{f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^3*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

(-2*(b + a*Cos[e + f*x]^2)^2*((a + b)^2*Csc[e + f*x]^2 + 2*b^2*Log[Cos[e + f*x]] + 2*(a^2 - b^2)*Log[Sin[e + f
*x]]))/(f*(a + 2*b + a*Cos[2*(e + f*x)])^2)

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 100, normalized size = 1.75 \[ \frac {a^{2} + 2 \, a b + b^{2} - {\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} \log \left (\cos \left (f x + e\right )^{2}\right ) - {\left ({\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} + b^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (f x + e\right )^{2} + \frac {1}{4}\right )}{2 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/2*(a^2 + 2*a*b + b^2 - (b^2*cos(f*x + e)^2 - b^2)*log(cos(f*x + e)^2) - ((a^2 - b^2)*cos(f*x + e)^2 - a^2 +
b^2)*log(-1/4*cos(f*x + e)^2 + 1/4))/(f*cos(f*x + e)^2 - f)

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*(-b^2/4*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(
1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1)))-2))+a^2/4*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+e
xp(1)))*(1+cos(f*x+exp(1)))+2))+(-((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+ex
p(1))))*b^2-2*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1))))*b*a-((1-cos(
f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1))))*a^2)/16)

________________________________________________________________________________________

maple [A]  time = 1.18, size = 78, normalized size = 1.37 \[ -\frac {a^{2} \left (\cot ^{2}\left (f x +e \right )\right )}{2 f}-\frac {a^{2} \ln \left (\sin \left (f x +e \right )\right )}{f}-\frac {a b}{f \sin \left (f x +e \right )^{2}}-\frac {b^{2}}{2 f \sin \left (f x +e \right )^{2}}+\frac {b^{2} \ln \left (\tan \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x)

[Out]

-1/2*a^2*cot(f*x+e)^2/f-a^2*ln(sin(f*x+e))/f-1/f*a*b/sin(f*x+e)^2-1/2/f*b^2/sin(f*x+e)^2+1/f*b^2*ln(tan(f*x+e)
)

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 60, normalized size = 1.05 \[ -\frac {b^{2} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) + {\left (a^{2} - b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac {a^{2} + 2 \, a b + b^{2}}{\sin \left (f x + e\right )^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

-1/2*(b^2*log(sin(f*x + e)^2 - 1) + (a^2 - b^2)*log(sin(f*x + e)^2) + (a^2 + 2*a*b + b^2)/sin(f*x + e)^2)/f

________________________________________________________________________________________

mupad [B]  time = 4.58, size = 68, normalized size = 1.19 \[ \frac {a^2\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2-b^2\right )}{f}-\frac {{\mathrm {cot}\left (e+f\,x\right )}^2\,\left (\frac {a^2}{2}+a\,b+\frac {b^2}{2}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^3*(a + b/cos(e + f*x)^2)^2,x)

[Out]

(a^2*log(tan(e + f*x)^2 + 1))/(2*f) - (log(tan(e + f*x))*(a^2 - b^2))/f - (cot(e + f*x)^2*(a*b + a^2/2 + b^2/2
))/f

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \cot ^{3}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**3*(a+b*sec(f*x+e)**2)**2,x)

[Out]

Integral((a + b*sec(e + f*x)**2)**2*cot(e + f*x)**3, x)

________________________________________________________________________________________