3.420 \(\int \cot ^3(c+d x) (a+b \tan (c+d x)) \, dx\)

Optimal. Leaf size=46 \[ -\frac{a \cot ^2(c+d x)}{2 d}-\frac{a \log (\sin (c+d x))}{d}-\frac{b \cot (c+d x)}{d}-b x \]

[Out]

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

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Rubi [A]  time = 0.064159, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3529, 3531, 3475} \[ -\frac{a \cot ^2(c+d x)}{2 d}-\frac{a \log (\sin (c+d x))}{d}-\frac{b \cot (c+d x)}{d}-b x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3529

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 3531

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 3475

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

Rubi steps

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

Mathematica [C]  time = 0.265902, size = 66, normalized size = 1.43 \[ -\frac{a \left (\cot ^2(c+d x)+2 \log (\tan (c+d x))+2 \log (\cos (c+d x))\right )}{2 d}-\frac{b \cot (c+d x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\tan ^2(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.04, size = 52, normalized size = 1.1 \begin{align*} -bx-{\frac{\cot \left ( dx+c \right ) b}{d}}-{\frac{bc}{d}}-{\frac{ \left ( \cot \left ( dx+c \right ) \right ) ^{2}a}{2\,d}}-{\frac{a\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.72042, size = 78, normalized size = 1.7 \begin{align*} -\frac{2 \,{\left (d x + c\right )} b - a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, a \log \left (\tan \left (d x + c\right )\right ) + \frac{2 \, b \tan \left (d x + c\right ) + a}{\tan \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.69637, size = 188, normalized size = 4.09 \begin{align*} -\frac{a \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} +{\left (2 \, b d x + a\right )} \tan \left (d x + c\right )^{2} + 2 \, b \tan \left (d x + c\right ) + a}{2 \, d \tan \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.70375, size = 83, normalized size = 1.8 \begin{align*} \begin{cases} \tilde{\infty } a 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 ) \cot ^{3}{\left (c \right )} & \text{for}\: d = 0 \\\frac{a \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac{a \log{\left (\tan{\left (c + d x \right )} \right )}}{d} - \frac{a}{2 d \tan ^{2}{\left (c + d x \right )}} - b x - \frac{b}{d \tan{\left (c + d x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(a+b*tan(d*x+c)),x)

[Out]

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

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Giac [B]  time = 1.34567, size = 153, normalized size = 3.33 \begin{align*} -\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 \,{\left (d x + c\right )} b - 8 \, a \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) + 8 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{12 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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