3.14 \(\int \cot ^4(c+d x) (a+b \tan (c+d x))^2 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=88 \[ -\frac {\left (a^2 B-2 a b C-b^2 B\right ) \log (\sin (c+d x))}{d}-\frac {a^2 B \cot ^2(c+d x)}{2 d}+x \left (b^2 C-a (a C+2 b B)\right )-\frac {a (a C+2 b B) \cot (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3632, 3604, 3628, 3531, 3475} \[ -\frac {\left (a^2 B-2 a b C-b^2 B\right ) \log (\sin (c+d x))}{d}-\frac {a^2 B \cot ^2(c+d x)}{2 d}+x \left (b^2 C-a (a C+2 b B)\right )-\frac {a (a C+2 b B) \cot (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^2*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

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

Rule 3475

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

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 3604

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((B*c - A*d)*(b*c - a*d)^2*(c + d*Tan[e + f*x])^(n + 1))/(f*d^2*(n +
1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[B*(b*c - a*d)^2 + A*d*(a^2
*c - b^2*c + 2*a*b*d) + d*(B*(a^2*c - b^2*c + 2*a*b*d) + A*(2*a*b*c - a^2*d + b^2*d))*Tan[e + f*x] + b^2*B*(c^
2 + d^2)*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] && LtQ[n, -1]

Rule 3628

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

Rule 3632

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]

Rubi steps

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

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Mathematica [C]  time = 0.35, size = 123, normalized size = 1.40 \[ \frac {-2 \left (a^2 B-2 a b C-b^2 B\right ) \log (\tan (c+d x))-a^2 B \cot ^2(c+d x)-2 a (a C+2 b B) \cot (c+d x)+(a-i b)^2 (B-i C) \log (\tan (c+d x)+i)+(a+i b)^2 (B+i C) \log (-\tan (c+d x)+i)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*(a + b*Tan[c + d*x])^2*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

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

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fricas [A]  time = 1.55, size = 122, normalized size = 1.39 \[ -\frac {{\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + B a^{2} + {\left (B a^{2} + 2 \, {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} d x\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (C a^{2} + 2 \, B a b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 8.12, size = 237, normalized size = 2.69 \[ -\frac {B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} {\left (d x + c\right )} - 8 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) + 8 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.62, size = 141, normalized size = 1.60 \[ -\frac {a^{2} B \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a^{2} B \ln \left (\sin \left (d x +c \right )\right )}{d}-a^{2} C x -\frac {C \cot \left (d x +c \right ) a^{2}}{d}-\frac {C \,a^{2} c}{d}-2 B x a b -\frac {2 B \cot \left (d x +c \right ) a b}{d}-\frac {2 B a b c}{d}+\frac {2 C a b \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {b^{2} B \ln \left (\sin \left (d x +c \right )\right )}{d}+b^{2} C x +\frac {C \,b^{2} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+b*tan(d*x+c))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.70, size = 120, normalized size = 1.36 \[ -\frac {2 \, {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} {\left (d x + c\right )} - {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {B a^{2} + 2 \, {\left (C a^{2} + 2 \, B a b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+b*tan(d*x+c))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 8.98, size = 127, normalized size = 1.44 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-B\,a^2+2\,C\,a\,b+B\,b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\frac {B\,a^2}{2}+\mathrm {tan}\left (c+d\,x\right )\,\left (C\,a^2+2\,B\,b\,a\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^4*(B*tan(c + d*x) + C*tan(c + d*x)^2)*(a + b*tan(c + d*x))^2,x)

[Out]

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

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sympy [A]  time = 4.31, size = 212, normalized size = 2.41 \[ \begin {cases} \text {NaN} & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (a + b \tan {\relax (c )}\right )^{2} \left (B \tan {\relax (c )} + C \tan ^{2}{\relax (c )}\right ) \cot ^{4}{\relax (c )} & \text {for}\: d = 0 \\\frac {B a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B a^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - 2 B a b x - \frac {2 B a b}{d \tan {\left (c + d x \right )}} - \frac {B b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - C a^{2} x - \frac {C a^{2}}{d \tan {\left (c + d x \right )}} - \frac {C a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {2 C a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + C b^{2} x & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(a+b*tan(d*x+c))**2*(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))**2*(B*tan(c) + C*tan(c
)**2)*cot(c)**4, Eq(d, 0)), (B*a**2*log(tan(c + d*x)**2 + 1)/(2*d) - B*a**2*log(tan(c + d*x))/d - B*a**2/(2*d*
tan(c + d*x)**2) - 2*B*a*b*x - 2*B*a*b/(d*tan(c + d*x)) - B*b**2*log(tan(c + d*x)**2 + 1)/(2*d) + B*b**2*log(t
an(c + d*x))/d - C*a**2*x - C*a**2/(d*tan(c + d*x)) - C*a*b*log(tan(c + d*x)**2 + 1)/d + 2*C*a*b*log(tan(c + d
*x))/d + C*b**2*x, True))

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