3.5.40 \(\int \cot ^2(c+d x) (a+b \tan (c+d x))^3 \, dx\) [440]

Optimal. Leaf size=69 \[ -a \left (a^2-3 b^2\right ) x-\frac {b^3 \log (\cos (c+d x))}{d}+\frac {3 a^2 b \log (\sin (c+d x))}{d}-\frac {a^2 \cot (c+d x) (a+b \tan (c+d x))}{d} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3646, 3705, 3556} \begin {gather*} -a x \left (a^2-3 b^2\right )+\frac {3 a^2 b \log (\sin (c+d x))}{d}-\frac {a^2 \cot (c+d x) (a+b \tan (c+d x))}{d}-\frac {b^3 \log (\cos (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(a^2 - 3*b^2)*x) - (b^3*Log[Cos[c + d*x]])/d + (3*a^2*b*Log[Sin[c + d*x]])/d - (a^2*Cot[c + d*x]*(a + b*Ta
n[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 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3705

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/tan[(e_.) + (f_.)*(x_)], x_Symbol
] :> Simp[B*x, x] + (Dist[A, Int[1/Tan[e + f*x], x], x] + Dist[C, Int[Tan[e + f*x], x], x]) /; FreeQ[{e, f, A,
B, C}, x] && NeQ[A, C]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.15, size = 80, normalized size = 1.16 \begin {gather*} -\frac {a^3 \cot (c+d x)-\frac {1}{2} (i a+b)^3 \log (i-\cot (c+d x))+\frac {1}{2} (i a-b)^3 \log (i+\cot (c+d x))-b^3 \log (\tan (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*Cot[c + d*x] - ((I*a + b)^3*Log[I - Cot[c + d*x]])/2 + ((I*a - b)^3*Log[I + Cot[c + d*x]])/2 - b^3*Log[
Tan[c + d*x]])/d)

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Maple [A]
time = 0.18, size = 62, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+3 a^{2} b \ln \left (\sin \left (d x +c \right )\right )+3 b^{2} a \left (d x +c \right )-b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}\)	\(62\)
default	\(\frac {a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+3 a^{2} b \ln \left (\sin \left (d x +c \right )\right )+3 b^{2} a \left (d x +c \right )-b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}\)	\(62\)
norman	\(\frac {\left (-a^{3}+3 b^{2} a \right ) x \tan \left (d x +c \right )-\frac {a^{3}}{d}}{\tan \left (d x +c \right )}+\frac {3 a^{2} b \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {b \left (3 a^{2}-b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(84\)
risch	\(-3 i x b \,a^{2}+i b^{3} x -a^{3} x +3 a \,b^{2} x -\frac {6 i b \,a^{2} c}{d}+\frac {2 i b^{3} c}{d}-\frac {2 i a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 74, normalized size = 1.07 \begin {gather*} \frac {6 \, a^{2} b \log \left (\tan \left (d x + c\right )\right ) - 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )} - {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - \frac {2 \, a^{3}}{\tan \left (d x + c\right )}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.04, size = 97, normalized size = 1.41 \begin {gather*} \frac {3 \, a^{2} b \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 ) - b^{3} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) - 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} d x \tan \left (d x + c\right ) - 2 \, a^{3}}{2 \, d \tan \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.66, size = 114, normalized size = 1.65 \begin {gather*} \begin {cases} \tilde {\infty } a^{3} 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 )^{3} \cot ^{2}{\left (c \right )} & \text {for}\: d = 0 \\- a^{3} x - \frac {a^{3}}{d \tan {\left (c + d x \right )}} - \frac {3 a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 a b^{2} x + \frac {b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.14, size = 88, normalized size = 1.28 \begin {gather*} \frac {6 \, a^{2} b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )} - {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - \frac {2 \, {\left (3 \, a^{2} b \tan \left (d x + c\right ) + a^{3}\right )}}{\tan \left (d x + c\right )}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*a^2*b*log(abs(tan(d*x + c))) - 2*(a^3 - 3*a*b^2)*(d*x + c) - (3*a^2*b - b^3)*log(tan(d*x + c)^2 + 1) -
2*(3*a^2*b*tan(d*x + c) + a^3)/tan(d*x + c))/d

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Mupad [B]
time = 3.91, size = 78, normalized size = 1.13 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^3}{2\,d}-\frac {a^3\,\mathrm {cot}\left (c+d\,x\right )}{d}+\frac {3\,a^2\,b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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