3.31 \(\int \frac{\cot ^4(c+d x) (B \tan (c+d x)+C \tan ^2(c+d x))}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=137 \[ -\frac{\left (a^2 B+a b C-b^2 B\right ) \log (\sin (c+d x))}{a^3 d}-\frac{b^3 (b B-a C) \log (a \cos (c+d x)+b \sin (c+d x))}{a^3 d \left (a^2+b^2\right )}+\frac{x (b B-a C)}{a^2+b^2}+\frac{(b B-a C) \cot (c+d x)}{a^2 d}-\frac{B \cot ^2(c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.68161, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {3632, 3609, 3649, 3651, 3530, 3475} \[ -\frac{\left (a^2 B+a b C-b^2 B\right ) \log (\sin (c+d x))}{a^3 d}-\frac{b^3 (b B-a C) \log (a \cos (c+d x)+b \sin (c+d x))}{a^3 d \left (a^2+b^2\right )}+\frac{x (b B-a C)}{a^2+b^2}+\frac{(b B-a C) \cot (c+d x)}{a^2 d}-\frac{B \cot ^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[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]

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

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]

Rule 3609

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*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n
+ 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e +
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ
[a, 0])))

Rule 3649

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

Rule 3651

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

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[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 \frac{\cot ^4(c+d x) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx &=\int \frac{\cot ^3(c+d x) (B+C \tan (c+d x))}{a+b \tan (c+d x)} \, dx\\ &=-\frac{B \cot ^2(c+d x)}{2 a d}-\frac{\int \frac{\cot ^2(c+d x) \left (2 (b B-a C)+2 a B \tan (c+d x)+2 b B \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 a}\\ &=\frac{(b B-a C) \cot (c+d x)}{a^2 d}-\frac{B \cot ^2(c+d x)}{2 a d}+\frac{\int \frac{\cot (c+d x) \left (-2 \left (a^2 B-b^2 B+a b C\right )-2 a^2 C \tan (c+d x)+2 b (b B-a C) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 a^2}\\ &=\frac{(b B-a C) x}{a^2+b^2}+\frac{(b B-a C) \cot (c+d x)}{a^2 d}-\frac{B \cot ^2(c+d x)}{2 a d}-\frac{\left (b^3 (b B-a C)\right ) \int \frac{b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^3 \left (a^2+b^2\right )}-\frac{\left (a^2 B-b^2 B+a b C\right ) \int \cot (c+d x) \, dx}{a^3}\\ &=\frac{(b B-a C) x}{a^2+b^2}+\frac{(b B-a C) \cot (c+d x)}{a^2 d}-\frac{B \cot ^2(c+d x)}{2 a d}-\frac{\left (a^2 B-b^2 B+a b C\right ) \log (\sin (c+d x))}{a^3 d}-\frac{b^3 (b B-a C) \log (a \cos (c+d x)+b \sin (c+d x))}{a^3 \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [C]  time = 1.37473, size = 163, normalized size = 1.19 \[ \frac{\frac{2 b^3 (a C-b B) \log (a+b \tan (c+d x))}{a^3 \left (a^2+b^2\right )}-\frac{2 \left (a^2 B+a b C-b^2 B\right ) \log (\tan (c+d x))}{a^3}+\frac{2 (b B-a C) \cot (c+d x)}{a^2}+\frac{(B+i C) \log (-\tan (c+d x)+i)}{a+i b}+\frac{(B-i C) \log (\tan (c+d x)+i)}{a-i b}-\frac{B \cot ^2(c+d x)}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.13, size = 266, normalized size = 1.9 \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) aB}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Cb}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{C\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{B}{2\,ad \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{Bb}{{a}^{2}d\tan \left ( dx+c \right ) }}-{\frac{C}{ad\tan \left ( dx+c \right ) }}-{\frac{B\ln \left ( \tan \left ( dx+c \right ) \right ) }{ad}}+{\frac{\ln \left ( \tan \left ( dx+c \right ) \right ){b}^{2}B}{{a}^{3}d}}-{\frac{\ln \left ( \tan \left ( dx+c \right ) \right ) Cb}{{a}^{2}d}}-{\frac{{b}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{d \left ({a}^{2}+{b}^{2} \right ){a}^{3}}}+{\frac{{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) C}{d \left ({a}^{2}+{b}^{2} \right ){a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/(a^2+b^2)*ln(1+tan(d*x+c)^2)*a*B+1/2/d/(a^2+b^2)*ln(1+tan(d*x+c)^2)*C*b+1/d/(a^2+b^2)*B*arctan(tan(d*x+c
))*b-1/d/(a^2+b^2)*C*arctan(tan(d*x+c))*a-1/2/d/a/tan(d*x+c)^2*B+1/d/a^2/tan(d*x+c)*B*b-1/d/a/tan(d*x+c)*C-1/d
/a*B*ln(tan(d*x+c))+1/d/a^3*ln(tan(d*x+c))*b^2*B-1/d/a^2*ln(tan(d*x+c))*C*b-1/d*b^4/(a^2+b^2)/a^3*ln(a+b*tan(d
*x+c))*B+1/d*b^3/(a^2+b^2)/a^2*ln(a+b*tan(d*x+c))*C

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Maxima [A]  time = 1.59209, size = 213, normalized size = 1.55 \begin{align*} -\frac{\frac{2 \,{\left (C a - B b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} - \frac{2 \,{\left (C a b^{3} - B b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{5} + a^{3} b^{2}} - \frac{{\left (B a + C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac{2 \,{\left (B a^{2} + C a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{3}} + \frac{B a + 2 \,{\left (C a - B b\right )} \tan \left (d x + c\right )}{a^{2} \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)^4*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*(2*(C*a - B*b)*(d*x + c)/(a^2 + b^2) - 2*(C*a*b^3 - B*b^4)*log(b*tan(d*x + c) + a)/(a^5 + a^3*b^2) - (B*a
+ C*b)*log(tan(d*x + c)^2 + 1)/(a^2 + b^2) + 2*(B*a^2 + C*a*b - B*b^2)*log(tan(d*x + c))/a^3 + (B*a + 2*(C*a
- B*b)*tan(d*x + c))/(a^2*tan(d*x + c)^2))/d

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Fricas [A]  time = 1.33, size = 518, normalized size = 3.78 \begin{align*} -\frac{B a^{4} + B a^{2} b^{2} +{\left (B a^{4} + C a^{3} b + C a b^{3} - B b^{4}\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 (C a b^{3} - B b^{4}\right )} \log \left (\frac{b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} +{\left (B a^{4} + B a^{2} b^{2} + 2 \,{\left (C a^{4} - B a^{3} b\right )} d x\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (C a^{4} - B a^{3} b + C a^{2} b^{2} - B a b^{3}\right )} \tan \left (d x + c\right )}{2 \,{\left (a^{5} + a^{3} b^{2}\right )} 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)^4*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(B*a^4 + B*a^2*b^2 + (B*a^4 + C*a^3*b + C*a*b^3 - B*b^4)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x
+ c)^2 - (C*a*b^3 - B*b^4)*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1))*tan(d*x
+ c)^2 + (B*a^4 + B*a^2*b^2 + 2*(C*a^4 - B*a^3*b)*d*x)*tan(d*x + c)^2 + 2*(C*a^4 - B*a^3*b + C*a^2*b^2 - B*a*b
^3)*tan(d*x + c))/((a^5 + a^3*b^2)*d*tan(d*x + c)^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.76978, size = 289, normalized size = 2.11 \begin{align*} -\frac{\frac{2 \,{\left (C a - B b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} - \frac{{\left (B a + C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \,{\left (C a b^{4} - B b^{5}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{5} b + a^{3} b^{3}} + \frac{2 \,{\left (B a^{2} + C a b - B b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac{3 \, B a^{2} \tan \left (d x + c\right )^{2} + 3 \, C a b \tan \left (d x + c\right )^{2} - 3 \, B b^{2} \tan \left (d x + c\right )^{2} - 2 \, C a^{2} \tan \left (d x + c\right ) + 2 \, B a b \tan \left (d x + c\right ) - B a^{2}}{a^{3} \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)^4*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(2*(C*a - B*b)*(d*x + c)/(a^2 + b^2) - (B*a + C*b)*log(tan(d*x + c)^2 + 1)/(a^2 + b^2) - 2*(C*a*b^4 - B*b
^5)*log(abs(b*tan(d*x + c) + a))/(a^5*b + a^3*b^3) + 2*(B*a^2 + C*a*b - B*b^2)*log(abs(tan(d*x + c)))/a^3 - (3
*B*a^2*tan(d*x + c)^2 + 3*C*a*b*tan(d*x + c)^2 - 3*B*b^2*tan(d*x + c)^2 - 2*C*a^2*tan(d*x + c) + 2*B*a*b*tan(d
*x + c) - B*a^2)/(a^3*tan(d*x + c)^2))/d