3.313 \(\int \frac{\cot ^3(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=112 \[ -\frac{B \left (a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac{b^4 B \log (a \cos (c+d x)+b \sin (c+d x))}{a^3 d \left (a^2+b^2\right )}+\frac{b B x}{a^2+b^2}+\frac{b B \cot (c+d x)}{a^2 d}-\frac{B \cot ^2(c+d x)}{2 a d} \]

[Out]

(b*B*x)/(a^2 + b^2) + (b*B*Cot[c + d*x])/(a^2*d) - (B*Cot[c + d*x]^2)/(2*a*d) - ((a^2 - b^2)*B*Log[Sin[c + d*x
]])/(a^3*d) - (b^4*B*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 = 0.324771, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {21, 3569, 3649, 3652, 3530, 3475} \[ -\frac{B \left (a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac{b^4 B \log (a \cos (c+d x)+b \sin (c+d x))}{a^3 d \left (a^2+b^2\right )}+\frac{b B x}{a^2+b^2}+\frac{b B \cot (c+d x)}{a^2 d}-\frac{B \cot ^2(c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*B*x)/(a^2 + b^2) + (b*B*Cot[c + d*x])/(a^2*d) - (B*Cot[c + d*x]^2)/(2*a*d) - ((a^2 - b^2)*B*Log[Sin[c + d*x
]])/(a^3*d) - (b^4*B*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/(a^3*(a^2 + b^2)*d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(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 3652

Int[((A_.) + (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*(A*d - C*d))*x)/((a^2 + b^2)*(c^2 + d^2)), x] + (Dist[
(A*b^2 + 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
+ A*d^2)/((b*c - a*d)*(c^2 + d^2)), Int[(d - c*Tan[e + f*x])/(c + d*Tan[e + f*x]), x], x]) /; FreeQ[{a, b, c,
d, e, f, A, 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 ^3(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx &=B \int \frac{\cot ^3(c+d x)}{a+b \tan (c+d x)} \, dx\\ &=-\frac{B \cot ^2(c+d x)}{2 a d}-\frac{B \int \frac{\cot ^2(c+d x) \left (2 b+2 a \tan (c+d x)+2 b \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 a}\\ &=\frac{b B \cot (c+d x)}{a^2 d}-\frac{B \cot ^2(c+d x)}{2 a d}+\frac{B \int \frac{\cot (c+d x) \left (-2 \left (a^2-b^2\right )+2 b^2 \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 a^2}\\ &=\frac{b B x}{a^2+b^2}+\frac{b B \cot (c+d x)}{a^2 d}-\frac{B \cot ^2(c+d x)}{2 a d}-\frac{\left (\left (a^2-b^2\right ) B\right ) \int \cot (c+d x) \, dx}{a^3}-\frac{\left (b^4 B\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{b B x}{a^2+b^2}+\frac{b B \cot (c+d x)}{a^2 d}-\frac{B \cot ^2(c+d x)}{2 a d}-\frac{\left (a^2-b^2\right ) B \log (\sin (c+d x))}{a^3 d}-\frac{b^4 B \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 = 0.621862, size = 107, normalized size = 0.96 \[ -\frac{B \left (\frac{2 b^4 \log (a \cot (c+d x)+b)}{a^3 \left (a^2+b^2\right )}-\frac{2 b \cot (c+d x)}{a^2}-\frac{\log (-\cot (c+d x)+i)}{a-i b}-\frac{\log (\cot (c+d x)+i)}{a+i b}+\frac{\cot ^2(c+d x)}{a}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(B*((-2*b*Cot[c + d*x])/a^2 + Cot[c + d*x]^2/a - Log[I - Cot[c + d*x]]/(a - I*b) - Log[I + Cot[c + d*x]]/(a +
I*b) + (2*b^4*Log[b + a*Cot[c + d*x]])/(a^3*(a^2 + b^2))))/(2*d)

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Maple [A]  time = 0.095, size = 151, normalized size = 1.4 \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{B\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{B}{2\,ad \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}-{\frac{B\ln \left ( \tan \left ( dx+c \right ) \right ) }{ad}}+{\frac{B\ln \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{{a}^{3}d}}+{\frac{Bb}{{a}^{2}d\tan \left ( dx+c \right ) }}-{\frac{{b}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{{a}^{3}d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.59219, size = 176, normalized size = 1.57 \begin{align*} -\frac{\frac{2 \, B b^{4} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{5} + a^{3} b^{2}} - \frac{2 \,{\left (d x + c\right )} B b}{a^{2} + b^{2}} - \frac{B a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac{2 \,{\left (B a^{2} - B b^{2}\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{3}} - \frac{2 \, B b \tan \left (d x + c\right ) - B a}{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)^3*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.90301, size = 435, normalized size = 3.88 \begin{align*} -\frac{B b^{4} \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} + B a^{4} + B a^{2} b^{2} +{\left (B a^{4} - 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 (2 \, B a^{3} b d x - B a^{4} - B a^{2} b^{2}\right )} \tan \left (d x + c\right )^{2} - 2 \,{\left (B a^{3} b + 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)^3*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.32287, size = 223, normalized size = 1.99 \begin{align*} -\frac{\frac{2 \, B b^{5} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{5} b + a^{3} b^{3}} - \frac{2 \,{\left (d x + c\right )} B b}{a^{2} + b^{2}} - \frac{B a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac{2 \,{\left (B a^{2} - 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 \, B b^{2} \tan \left (d x + c\right )^{2} + 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)^3*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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