3.5.33 \(\int \cot ^5(c+d x) (a+b \tan (c+d x))^2 \, dx\) [433]

Optimal. Leaf size=98 \[ 2 a b x+\frac {2 a b \cot (c+d x)}{d}+\frac {\left (a^2-b^2\right ) \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\frac {\left (a^2-b^2\right ) \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3623, 3610, 3612, 3556} \begin {gather*} \frac {\left (a^2-b^2\right ) \cot ^2(c+d x)}{2 d}+\frac {\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}-\frac {a^2 \cot ^4(c+d x)}{4 d}-\frac {2 a b \cot ^3(c+d x)}{3 d}+\frac {2 a b \cot (c+d x)}{d}+2 a b x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]

[Out]

2*a*b*x + (2*a*b*Cot[c + d*x])/d + ((a^2 - b^2)*Cot[c + d*x]^2)/(2*d) - (2*a*b*Cot[c + d*x]^3)/(3*d) - (a^2*Co
t[c + d*x]^4)/(4*d) + ((a^2 - b^2)*Log[Sin[c + d*x]])/d

Rule 3556

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

Rule 3610

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b
*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3612

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c +
b*d)*(x/(a^2 + b^2)), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3623

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(b*c - a*d)^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Ta
n[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Tan[e + f*x], x], x], x] /; FreeQ
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \cot ^5(c+d x) (a+b \tan (c+d x))^2 \, dx &=-\frac {a^2 \cot ^4(c+d x)}{4 d}+\int \cot ^4(c+d x) \left (2 a b-\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=-\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\int \cot ^3(c+d x) \left (-a^2+b^2-2 a b \tan (c+d x)\right ) \, dx\\ &=\frac {\left (a^2-b^2\right ) \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\int \cot ^2(c+d x) \left (-2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=\frac {2 a b \cot (c+d x)}{d}+\frac {\left (a^2-b^2\right ) \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\int \cot (c+d x) \left (a^2-b^2+2 a b \tan (c+d x)\right ) \, dx\\ &=2 a b x+\frac {2 a b \cot (c+d x)}{d}+\frac {\left (a^2-b^2\right ) \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\left (a^2-b^2\right ) \int \cot (c+d x) \, dx\\ &=2 a b x+\frac {2 a b \cot (c+d x)}{d}+\frac {\left (a^2-b^2\right ) \cot ^2(c+d x)}{2 d}-\frac {2 a b \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot ^4(c+d x)}{4 d}+\frac {\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.57, size = 122, normalized size = 1.24 \begin {gather*} -\frac {2 a b \cot ^3(c+d x) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\tan ^2(c+d x)\right )}{3 d}-\frac {b^2 \left (\cot ^2(c+d x)+2 \log (\cos (c+d x))+2 \log (\tan (c+d x))\right )}{2 d}+\frac {a^2 \left (2 \cot ^2(c+d x)-\cot ^4(c+d x)+4 \log (\cos (c+d x))+4 \log (\tan (c+d x))\right )}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]

[Out]

(-2*a*b*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2])/(3*d) - (b^2*(Cot[c + d*x]^2 + 2*Log
[Cos[c + d*x]] + 2*Log[Tan[c + d*x]]))/(2*d) + (a^2*(2*Cot[c + d*x]^2 - Cot[c + d*x]^4 + 4*Log[Cos[c + d*x]] +
4*Log[Tan[c + d*x]]))/(4*d)

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Maple [A]
time = 0.17, size = 87, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+2 a b \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+b^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(87\)
default	\(\frac {a^{2} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+2 a b \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+b^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(87\)
norman	\(\frac {-\frac {a^{2}}{4 d}+\frac {\left (a^{2}-b^{2}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+2 a b x \left (\tan ^{4}\left (d x +c \right )\right )-\frac {2 a b \tan \left (d x +c \right )}{3 d}+\frac {2 a b \left (\tan ^{3}\left (d x +c \right )\right )}{d}}{\tan \left (d x +c \right )^{4}}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(128\)
risch	\(2 a b x -i a^{2} x +i b^{2} x -\frac {2 i a^{2} c}{d}+\frac {2 i b^{2} c}{d}-\frac {2 \left (-12 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}+6 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+24 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-20 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+8 i a b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{d}\)	\(230\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.53, size = 111, normalized size = 1.13 \begin {gather*} \frac {24 \, {\left (d x + c\right )} a b - 6 \, {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {24 \, a b \tan \left (d x + c\right )^{3} - 8 \, a b \tan \left (d x + c\right ) + 6 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} - 3 \, a^{2}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(24*(d*x + c)*a*b - 6*(a^2 - b^2)*log(tan(d*x + c)^2 + 1) + 12*(a^2 - b^2)*log(tan(d*x + c)) + (24*a*b*ta
n(d*x + c)^3 - 8*a*b*tan(d*x + c) + 6*(a^2 - b^2)*tan(d*x + c)^2 - 3*a^2)/tan(d*x + c)^4)/d

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Fricas [A]
time = 1.14, size = 128, normalized size = 1.31 \begin {gather*} \frac {6 \, {\left (a^{2} - b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{4} + 24 \, a b \tan \left (d x + c\right )^{3} + 3 \, {\left (8 \, a b d x + 3 \, a^{2} - 2 \, b^{2}\right )} \tan \left (d x + c\right )^{4} - 8 \, a b \tan \left (d x + c\right ) + 6 \, {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{2} - 3 \, a^{2}}{12 \, d \tan \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(6*(a^2 - b^2)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^4 + 24*a*b*tan(d*x + c)^3 + 3*(8*a*b
*d*x + 3*a^2 - 2*b^2)*tan(d*x + c)^4 - 8*a*b*tan(d*x + c) + 6*(a^2 - b^2)*tan(d*x + c)^2 - 3*a^2)/(d*tan(d*x +
c)^4)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (88) = 176\).
time = 1.43, size = 178, normalized size = 1.82 \begin {gather*} \begin {cases} \tilde {\infty } a^{2} x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (a + b \tan {\left (c \right )}\right )^{2} \cot ^{5}{\left (c \right )} & \text {for}\: d = 0 \\- \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {a^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {a^{2}}{4 d \tan ^{4}{\left (c + d x \right )}} + 2 a b x + \frac {2 a b}{d \tan {\left (c + d x \right )}} - \frac {2 a b}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {b^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+b*tan(d*x+c))**2,x)

[Out]

Piecewise((zoo*a**2*x, (Eq(c, 0) | Eq(c, -d*x)) & (Eq(d, 0) | Eq(c, -d*x))), (x*(a + b*tan(c))**2*cot(c)**5, E
q(d, 0)), (-a**2*log(tan(c + d*x)**2 + 1)/(2*d) + a**2*log(tan(c + d*x))/d + a**2/(2*d*tan(c + d*x)**2) - a**2
/(4*d*tan(c + d*x)**4) + 2*a*b*x + 2*a*b/(d*tan(c + d*x)) - 2*a*b/(3*d*tan(c + d*x)**3) + b**2*log(tan(c + d*x
)**2 + 1)/(2*d) - b**2*log(tan(c + d*x))/d - b**2/(2*d*tan(c + d*x)**2), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (92) = 184\).
time = 1.06, size = 248, normalized size = 2.53 \begin {gather*} -\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 384 \, {\left (d x + c\right )} a b + 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 192 \, {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 192 \, {\left (a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {400 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 400 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 - 16*a*b*tan(1/2*d*x + 1/2*c)^3 - 36*a^2*tan(1/2*d*x + 1/2*c)^2 + 24*b^2*
tan(1/2*d*x + 1/2*c)^2 - 384*(d*x + c)*a*b + 240*a*b*tan(1/2*d*x + 1/2*c) + 192*(a^2 - b^2)*log(tan(1/2*d*x +
1/2*c)^2 + 1) - 192*(a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + (400*a^2*tan(1/2*d*x + 1/2*c)^4 - 400*b^2*tan
(1/2*d*x + 1/2*c)^4 - 240*a*b*tan(1/2*d*x + 1/2*c)^3 - 36*a^2*tan(1/2*d*x + 1/2*c)^2 + 24*b^2*tan(1/2*d*x + 1/
2*c)^2 + 16*a*b*tan(1/2*d*x + 1/2*c) + 3*a^2)/tan(1/2*d*x + 1/2*c)^4)/d

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Mupad [B]
time = 3.99, size = 129, normalized size = 1.32 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^2-b^2\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^4\,\left (\frac {a^2}{4}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )+\frac {2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )}{3}-2\,a\,b\,{\mathrm {tan}\left (c+d\,x\right )}^3\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^5*(a + b*tan(c + d*x))^2,x)

[Out]

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

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