3.1.20 \(\int \cot ^3(c+d x) (a+b \tan (c+d x))^3 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\) [20]

Optimal. Leaf size=119 \[ -\left (\left (a^3 B-3 a b^2 B-3 a^2 b C+b^3 C\right ) x\right )-\frac {b^2 (b B+3 a C) \log (\cos (c+d x))}{d}+\frac {a^2 (3 b B+a C) \log (\sin (c+d x))}{d}+\frac {b^2 (a B+b C) \tan (c+d x)}{d}-\frac {a B \cot (c+d x) (a+b \tan (c+d x))^2}{d} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3713, 3686, 3718, 3705, 3556} \begin {gather*} \frac {a^2 (a C+3 b B) \log (\sin (c+d x))}{d}-x \left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right )+\frac {b^2 (a B+b C) \tan (c+d x)}{d}-\frac {b^2 (3 a C+b B) \log (\cos (c+d x))}{d}-\frac {a B \cot (c+d x) (a+b \tan (c+d x))^2}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*B - 3*a*b^2*B - 3*a^2*b*C + b^3*C)*x) - (b^2*(b*B + 3*a*C)*Log[Cos[c + d*x]])/d + (a^2*(3*b*B + a*C)*Lo
g[Sin[c + d*x]])/d + (b^2*(a*B + b*C)*Tan[c + d*x])/d - (a*B*Cot[c + d*x]*(a + b*Tan[c + d*x])^2)/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 3686

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e
+ f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

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]

Rule 3713

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

Rule 3718

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])
^(n + 1)/(d*f*(n + 2))), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] &&  !LtQ[n, -1]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.33, size = 113, normalized size = 0.95 \begin {gather*} \frac {-2 a^3 B \cot (c+d x)+i (a+i b)^3 (B+i C) \log (i-\tan (c+d x))+2 a^2 (3 b B+a C) \log (\tan (c+d x))+(i a+b)^3 (B-i C) \log (i+\tan (c+d x))+2 b^3 C \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])^3*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

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

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Maple [A]
time = 0.27, size = 123, normalized size = 1.03

method	result	size
derivativedivides	\(\frac {-B \,b^{3} \ln \left (\cos \left (d x +c \right )\right )+C \,b^{3} \left (\tan \left (d x +c \right )-d x -c \right )+3 B a \,b^{2} \left (d x +c \right )-3 C a \,b^{2} \ln \left (\cos \left (d x +c \right )\right )+3 B \,a^{2} b \ln \left (\sin \left (d x +c \right )\right )+3 C \,a^{2} b \left (d x +c \right )+B \,a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+C \,a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}\)	\(123\)
default	\(\frac {-B \,b^{3} \ln \left (\cos \left (d x +c \right )\right )+C \,b^{3} \left (\tan \left (d x +c \right )-d x -c \right )+3 B a \,b^{2} \left (d x +c \right )-3 C a \,b^{2} \ln \left (\cos \left (d x +c \right )\right )+3 B \,a^{2} b \ln \left (\sin \left (d x +c \right )\right )+3 C \,a^{2} b \left (d x +c \right )+B \,a^{3} \left (-\cot \left (d x +c \right )-d x -c \right )+C \,a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}\)	\(123\)
norman	\(\frac {\left (-B \,a^{3}+3 B a \,b^{2}+3 C \,a^{2} b -C \,b^{3}\right ) x \left (\tan ^{2}\left (d x +c \right )\right )+\frac {C \,b^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{d}-\frac {B \,a^{3} \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}+\frac {a^{2} \left (3 B b +C a \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (3 B \,a^{2} b -B \,b^{3}+C \,a^{3}-3 C a \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(144\)
risch	\(-B \,a^{3} x +3 B a \,b^{2} x +3 C \,a^{2} b x -C \,b^{3} x +3 i C a \,b^{2} x -\frac {6 i B \,a^{2} b c}{d}-i C \,a^{3} x -\frac {2 i C \,a^{3} c}{d}+\frac {2 i B \,b^{3} c}{d}+\frac {6 i C a \,b^{2} c}{d}-\frac {2 i \left (B \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-C \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+B \,a^{3}+C \,b^{3}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-3 i B \,a^{2} b x +i B \,b^{3} x +\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B \,b^{3}}{d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) C a \,b^{2}}{d}\)	\(269\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-B*b^3*ln(cos(d*x+c))+C*b^3*(tan(d*x+c)-d*x-c)+3*B*a*b^2*(d*x+c)-3*C*a*b^2*ln(cos(d*x+c))+3*B*a^2*b*ln(si
n(d*x+c))+3*C*a^2*b*(d*x+c)+B*a^3*(-cot(d*x+c)-d*x-c)+C*a^3*ln(sin(d*x+c)))

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Maxima [A]
time = 0.50, size = 125, normalized size = 1.05 \begin {gather*} \frac {2 \, C b^{3} \tan \left (d x + c\right ) - \frac {2 \, B a^{3}}{\tan \left (d x + c\right )} - 2 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} {\left (d x + c\right )} - {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \log \left (\tan \left (d x + c\right )\right )}{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))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 5.32, size = 145, normalized size = 1.22 \begin {gather*} \frac {2 \, C b^{3} \tan \left (d x + c\right )^{2} - 2 \, B a^{3} - 2 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} d x \tan \left (d x + c\right ) + {\left (C a^{3} + 3 \, B a^{2} b\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 ) - {\left (3 \, C a b^{2} + B b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )}{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)^3*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 2.23, size = 221, normalized size = 1.86 \begin {gather*} \begin {cases} \text {NaN} & \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} \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{3}{\left (c \right )} & \text {for}\: d = 0 \\- B a^{3} x - \frac {B a^{3}}{d \tan {\left (c + d x \right )}} - \frac {3 B a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 B a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 B a b^{2} x + \frac {B b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {C a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 C a^{2} b x + \frac {3 C a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - C b^{3} x + \frac {C b^{3} \tan {\left (c + d x \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))**3*(B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

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

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Giac [A]
time = 1.30, size = 152, normalized size = 1.28 \begin {gather*} \frac {2 \, C b^{3} \tan \left (d x + c\right ) - 2 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} {\left (d x + c\right )} - {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {2 \, {\left (C a^{3} \tan \left (d x + c\right ) + 3 \, B a^{2} b \tan \left (d x + c\right ) + B 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)^3*(a+b*tan(d*x+c))^3*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 8.86, size = 114, normalized size = 0.96 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (C\,a^3+3\,B\,b\,a^2\right )}{d}-\frac {B\,a^3\,\mathrm {cot}\left (c+d\,x\right )}{d}+\frac {C\,b^3\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\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)^3*(B*tan(c + d*x) + C*tan(c + d*x)^2)*(a + b*tan(c + d*x))^3,x)

[Out]

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

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