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

Optimal. Leaf size=66 \[ -\frac {(a C+b B) \cot (c+d x)}{d}-\frac {(a B-b C) \log (\sin (c+d x))}{d}-x (a C+b B)-\frac {a B \cot ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3632, 3591, 3529, 3531, 3475} \[ -\frac {(a C+b B) \cot (c+d x)}{d}-\frac {(a B-b C) \log (\sin (c+d x))}{d}-x (a C+b B)-\frac {a B \cot ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*B + a*C)*x) - ((b*B + a*C)*Cot[c + d*x])/d - (a*B*Cot[c + d*x]^2)/(2*d) - ((a*B - 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 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 3591

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(A*b - a*B)*(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[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 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)) \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)) (B+C \tan (c+d x)) \, dx\\ &=-\frac {a B \cot ^2(c+d x)}{2 d}+\int \cot ^2(c+d x) (b B+a C-(a B-b C) \tan (c+d x)) \, dx\\ &=-\frac {(b B+a C) \cot (c+d x)}{d}-\frac {a B \cot ^2(c+d x)}{2 d}+\int \cot (c+d x) (-a B+b C-(b B+a C) \tan (c+d x)) \, dx\\ &=-(b B+a C) x-\frac {(b B+a C) \cot (c+d x)}{d}-\frac {a B \cot ^2(c+d x)}{2 d}+(-a B+b C) \int \cot (c+d x) \, dx\\ &=-(b B+a C) x-\frac {(b B+a C) \cot (c+d x)}{d}-\frac {a B \cot ^2(c+d x)}{2 d}-\frac {(a B-b C) \log (\sin (c+d x))}{d}\\ \end {align*}

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Mathematica [C]  time = 0.47, size = 77, normalized size = 1.17 \[ -\frac {2 (a C+b B) \cot (c+d x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\tan ^2(c+d x)\right )+2 (a B-b C) (\log (\tan (c+d x))+\log (\cos (c+d x)))+a B \cot ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 95, normalized size = 1.44 \[ -\frac {{\left (B a - C b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + {\left (2 \, {\left (C a + B b\right )} d x + B a\right )} \tan \left (d x + c\right )^{2} + B a + 2 \, {\left (C a + B b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 5.64, size = 179, normalized size = 2.71 \[ -\frac {B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, {\left (C a + B b\right )} {\left (d x + c\right )} - 8 \, {\left (B a - C b\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) + 8 \, {\left (B a - C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.52, size = 96, normalized size = 1.45 \[ -\frac {a B \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a B \ln \left (\sin \left (d x +c \right )\right )}{d}-a C x -\frac {C \cot \left (d x +c \right ) a}{d}-\frac {C a c}{d}-B x b -\frac {B \cot \left (d x +c \right ) b}{d}-\frac {B b c}{d}+\frac {C b \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.66, size = 86, normalized size = 1.30 \[ -\frac {2 \, {\left (C a + B b\right )} {\left (d x + c\right )} - {\left (B a - C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (B a - C b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {B a + 2 \, {\left (C a + B b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.94, size = 108, normalized size = 1.64 \[ -\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a-C\,b\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\frac {B\,a}{2}+\mathrm {tan}\left (c+d\,x\right )\,\left (B\,b+C\,a\right )\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,\left (b+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d} \]

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)),x)

[Out]

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

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sympy [A]  time = 2.33, size = 150, normalized size = 2.27 \[ \begin {cases} \text {NaN} & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (a + b \tan {\relax (c )}\right ) \left (B \tan {\relax (c )} + C \tan ^{2}{\relax (c )}\right ) \cot ^{4}{\relax (c )} & \text {for}\: d = 0 \\\frac {B a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B a \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B a}{2 d \tan ^{2}{\left (c + d x \right )}} - B b x - \frac {B b}{d \tan {\left (c + d x \right )}} - C a x - \frac {C a}{d \tan {\left (c + d x \right )}} - \frac {C b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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