3.5.31 \(\int \cot ^3(c+d x) (a+b \tan (c+d x))^2 \, dx\) [431]

Optimal. Leaf size=58 \[ -2 a b x-\frac {2 a b \cot (c+d x)}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d}-\frac {\left (a^2-b^2\right ) \log (\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, 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 ) \log (\sin (c+d x))}{d}-\frac {a^2 \cot ^2(c+d x)}{2 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]^3*(a + b*Tan[c + d*x])^2,x]

[Out]

-2*a*b*x - (2*a*b*Cot[c + d*x])/d - (a^2*Cot[c + d*x]^2)/(2*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 ^3(c+d x) (a+b \tan (c+d x))^2 \, dx &=-\frac {a^2 \cot ^2(c+d x)}{2 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 {a^2 \cot ^2(c+d x)}{2 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 {a^2 \cot ^2(c+d x)}{2 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 {a^2 \cot ^2(c+d x)}{2 d}-\frac {\left (a^2-b^2\right ) \log (\sin (c+d x))}{d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.30, size = 92, normalized size = 1.59 \begin {gather*} \frac {-4 a b \cot (c+d x)-a^2 \cot ^2(c+d x)+(a+i b)^2 \log (i-\tan (c+d x))-2 (a-b) (a+b) \log (\tan (c+d x))+(a-i b)^2 \log (i+\tan (c+d x))}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^2,x]

[Out]

(-4*a*b*Cot[c + d*x] - a^2*Cot[c + d*x]^2 + (a + I*b)^2*Log[I - Tan[c + d*x]] - 2*(a - b)*(a + b)*Log[Tan[c +
d*x]] + (a - I*b)^2*Log[I + Tan[c + d*x]])/(2*d)

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Maple [A]
time = 0.20, size = 61, normalized size = 1.05

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+2 a b \left (-\cot \left (d x +c \right )-d x -c \right )+b^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}\)	\(61\)
default	\(\frac {a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+2 a b \left (-\cot \left (d x +c \right )-d x -c \right )+b^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}\)	\(61\)
norman	\(\frac {-\frac {a^{2}}{2 d}-2 a b x \left (\tan ^{2}\left (d x +c \right )\right )-\frac {2 a b \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(92\)
risch	\(-2 a b x +i a^{2} x -i b^{2} x +\frac {2 i a^{2} c}{d}-\frac {2 i b^{2} c}{d}+\frac {2 a \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{d}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 78, normalized size = 1.34 \begin {gather*} -\frac {4 \, {\left (d x + c\right )} a b - {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {4 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.19, size = 86, normalized size = 1.48 \begin {gather*} -\frac {{\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 )^{2} + 4 \, a b \tan \left (d x + c\right ) + {\left (4 \, a b d x + a^{2}\right )} \tan \left (d x + c\right )^{2} + a^{2}}{2 \, d \tan \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (53) = 106\).
time = 0.68, size = 131, normalized size = 2.26 \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 ^{3}{\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 )}} - 2 a b x - \frac {2 a b}{d \tan {\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} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(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)**3, 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) - 2*a*b
*x - 2*a*b/(d*tan(c + d*x)) - b**2*log(tan(c + d*x)**2 + 1)/(2*d) + b**2*log(tan(c + d*x))/d, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (56) = 112\).
time = 0.84, size = 154, normalized size = 2.66 \begin {gather*} -\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 \, {\left (d x + c\right )} a b - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) + 8 \, {\left (a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 + 16*(d*x + c)*a*b - 8*a*b*tan(1/2*d*x + 1/2*c) - 8*(a^2 - b^2)*log(tan(1/2*d
*x + 1/2*c)^2 + 1) + 8*(a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c))) - (12*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*ta
n(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^2)/d

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Mupad [B]
time = 4.01, size = 97, normalized size = 1.67 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^2}{2\,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 )\right )\,\left (a^2-b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\frac {a^2}{2}+2\,b\,\mathrm {tan}\left (c+d\,x\right )\,a\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^3*(a + b*tan(c + d*x))^2,x)

[Out]

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

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