3.260 \(\int \cot (c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=172 \[ \frac{b^2 \left (3 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x)}{d}-\frac{b \left (6 a^2 A b+4 a^3 B-4 a b^2 B-A b^3\right ) \log (\cos (c+d x))}{d}+x \left (4 a^3 A b-6 a^2 b^2 B+a^4 B-4 a A b^3+b^4 B\right )+\frac{a^4 A \log (\sin (c+d x))}{d}+\frac{b (2 a B+A b) (a+b \tan (c+d x))^2}{2 d}+\frac{b B (a+b \tan (c+d x))^3}{3 d} \]

[Out]

(4*a^3*A*b - 4*a*A*b^3 + a^4*B - 6*a^2*b^2*B + b^4*B)*x - (b*(6*a^2*A*b - A*b^3 + 4*a^3*B - 4*a*b^2*B)*Log[Cos
[c + d*x]])/d + (a^4*A*Log[Sin[c + d*x]])/d + (b^2*(3*a*A*b + 3*a^2*B - b^2*B)*Tan[c + d*x])/d + (b*(A*b + 2*a
*B)*(a + b*Tan[c + d*x])^2)/(2*d) + (b*B*(a + b*Tan[c + d*x])^3)/(3*d)

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Rubi [A]  time = 0.47066, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {3607, 3647, 3637, 3624, 3475} \[ \frac{b^2 \left (3 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x)}{d}-\frac{b \left (6 a^2 A b+4 a^3 B-4 a b^2 B-A b^3\right ) \log (\cos (c+d x))}{d}+x \left (4 a^3 A b-6 a^2 b^2 B+a^4 B-4 a A b^3+b^4 B\right )+\frac{a^4 A \log (\sin (c+d x))}{d}+\frac{b (2 a B+A b) (a+b \tan (c+d x))^2}{2 d}+\frac{b B (a+b \tan (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^3*A*b - 4*a*A*b^3 + a^4*B - 6*a^2*b^2*B + b^4*B)*x - (b*(6*a^2*A*b - A*b^3 + 4*a^3*B - 4*a*b^2*B)*Log[Cos
[c + d*x]])/d + (a^4*A*Log[Sin[c + d*x]])/d + (b^2*(3*a*A*b + 3*a^2*B - b^2*B)*Tan[c + d*x])/d + (b*(A*b + 2*a
*B)*(a + b*Tan[c + d*x])^2)/(2*d) + (b*B*(a + b*Tan[c + d*x])^3)/(3*d)

Rule 3607

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

Rule 3647

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] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3637

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]

Rule 3624

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 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 \cot (c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx &=\frac{b B (a+b \tan (c+d x))^3}{3 d}+\frac{1}{3} \int \cot (c+d x) (a+b \tan (c+d x))^2 \left (3 a^2 A+3 \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+3 b (A b+2 a B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{b (A b+2 a B) (a+b \tan (c+d x))^2}{2 d}+\frac{b B (a+b \tan (c+d x))^3}{3 d}+\frac{1}{6} \int \cot (c+d x) (a+b \tan (c+d x)) \left (6 a^3 A+6 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)+6 b \left (3 a A b+3 a^2 B-b^2 B\right ) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{b^2 \left (3 a A b+3 a^2 B-b^2 B\right ) \tan (c+d x)}{d}+\frac{b (A b+2 a B) (a+b \tan (c+d x))^2}{2 d}+\frac{b B (a+b \tan (c+d x))^3}{3 d}-\frac{1}{6} \int \cot (c+d x) \left (-6 a^4 A-6 \left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) \tan (c+d x)-6 b \left (6 a^2 A b-A b^3+4 a^3 B-4 a b^2 B\right ) \tan ^2(c+d x)\right ) \, dx\\ &=\left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) x+\frac{b^2 \left (3 a A b+3 a^2 B-b^2 B\right ) \tan (c+d x)}{d}+\frac{b (A b+2 a B) (a+b \tan (c+d x))^2}{2 d}+\frac{b B (a+b \tan (c+d x))^3}{3 d}+\left (a^4 A\right ) \int \cot (c+d x) \, dx+\left (b \left (6 a^2 A b-A b^3+4 a^3 B-4 a b^2 B\right )\right ) \int \tan (c+d x) \, dx\\ &=\left (4 a^3 A b-4 a A b^3+a^4 B-6 a^2 b^2 B+b^4 B\right ) x-\frac{b \left (6 a^2 A b-A b^3+4 a^3 B-4 a b^2 B\right ) \log (\cos (c+d x))}{d}+\frac{a^4 A \log (\sin (c+d x))}{d}+\frac{b^2 \left (3 a A b+3 a^2 B-b^2 B\right ) \tan (c+d x)}{d}+\frac{b (A b+2 a B) (a+b \tan (c+d x))^2}{2 d}+\frac{b B (a+b \tan (c+d x))^3}{3 d}\\ \end{align*}

Mathematica [C]  time = 1.39383, size = 149, normalized size = 0.87 \[ \frac{6 b^2 \left (3 a^2 B+3 a A b-b^2 B\right ) \tan (c+d x)+6 a^4 A \log (\tan (c+d x))+3 b (2 a B+A b) (a+b \tan (c+d x))^2-3 (a+i b)^4 (A+i B) \log (-\tan (c+d x)+i)-3 (a-i b)^4 (A-i B) \log (\tan (c+d x)+i)+2 b B (a+b \tan (c+d x))^3}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(a + I*b)^4*(A + I*B)*Log[I - Tan[c + d*x]] + 6*a^4*A*Log[Tan[c + d*x]] - 3*(a - I*b)^4*(A - I*B)*Log[I +
Tan[c + d*x]] + 6*b^2*(3*a*A*b + 3*a^2*B - b^2*B)*Tan[c + d*x] + 3*b*(A*b + 2*a*B)*(a + b*Tan[c + d*x])^2 + 2*
b*B*(a + b*Tan[c + d*x])^3)/(6*d)

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Maple [A]  time = 0.089, size = 277, normalized size = 1.6 \begin{align*}{\frac{A{b}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{A{b}^{4}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{B \left ( \tan \left ( dx+c \right ) \right ) ^{3}{b}^{4}}{3\,d}}-{\frac{B{b}^{4}\tan \left ( dx+c \right ) }{d}}+B{b}^{4}x+{\frac{B{b}^{4}c}{d}}-4\,Aa{b}^{3}x+4\,{\frac{Aa{b}^{3}\tan \left ( dx+c \right ) }{d}}-4\,{\frac{Aa{b}^{3}c}{d}}+2\,{\frac{Ba{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}+4\,{\frac{Ba{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-6\,{\frac{A{a}^{2}{b}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-6\,B{a}^{2}{b}^{2}x+6\,{\frac{B{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}-6\,{\frac{B{a}^{2}{b}^{2}c}{d}}+4\,Ax{a}^{3}b+4\,{\frac{A{a}^{3}bc}{d}}-4\,{\frac{B{a}^{3}b\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{A{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+B{a}^{4}x+{\frac{B{a}^{4}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*A*b^4*tan(d*x+c)^2+1/d*A*b^4*ln(cos(d*x+c))+1/3/d*B*tan(d*x+c)^3*b^4-1/d*B*b^4*tan(d*x+c)+B*b^4*x+1/d*B*
b^4*c-4*A*a*b^3*x+4/d*A*a*b^3*tan(d*x+c)-4/d*A*a*b^3*c+2/d*B*a*b^3*tan(d*x+c)^2+4/d*B*a*b^3*ln(cos(d*x+c))-6/d
*A*a^2*b^2*ln(cos(d*x+c))-6*B*a^2*b^2*x+6/d*B*a^2*b^2*tan(d*x+c)-6/d*B*a^2*b^2*c+4*A*x*a^3*b+4/d*A*a^3*b*c-4/d
*B*a^3*b*ln(cos(d*x+c))+a^4*A*ln(sin(d*x+c))/d+B*a^4*x+1/d*B*a^4*c

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Maxima [A]  time = 1.48472, size = 236, normalized size = 1.37 \begin{align*} \frac{2 \, B b^{4} \tan \left (d x + c\right )^{3} + 6 \, A a^{4} \log \left (\tan \left (d x + c\right )\right ) + 3 \,{\left (4 \, B a b^{3} + A b^{4}\right )} \tan \left (d x + c\right )^{2} + 6 \,{\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )}{\left (d x + c\right )} - 3 \,{\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 \,{\left (6 \, B a^{2} b^{2} + 4 \, A a b^{3} - B b^{4}\right )} \tan \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*b^4*tan(d*x + c)^3 + 6*A*a^4*log(tan(d*x + c)) + 3*(4*B*a*b^3 + A*b^4)*tan(d*x + c)^2 + 6*(B*a^4 + 4*
A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3 + B*b^4)*(d*x + c) - 3*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)
*log(tan(d*x + c)^2 + 1) + 6*(6*B*a^2*b^2 + 4*A*a*b^3 - B*b^4)*tan(d*x + c))/d

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Fricas [A]  time = 2.36052, size = 423, normalized size = 2.46 \begin{align*} \frac{2 \, B b^{4} \tan \left (d x + c\right )^{3} + 3 \, A a^{4} \log \left (\frac{\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + 6 \,{\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} d x + 3 \,{\left (4 \, B a b^{3} + A b^{4}\right )} \tan \left (d x + c\right )^{2} - 3 \,{\left (4 \, B a^{3} b + 6 \, A a^{2} b^{2} - 4 \, B a b^{3} - A b^{4}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 6 \,{\left (6 \, B a^{2} b^{2} + 4 \, A a b^{3} - B b^{4}\right )} \tan \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*b^4*tan(d*x + c)^3 + 3*A*a^4*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1)) + 6*(B*a^4 + 4*A*a^3*b - 6*B*a^
2*b^2 - 4*A*a*b^3 + B*b^4)*d*x + 3*(4*B*a*b^3 + A*b^4)*tan(d*x + c)^2 - 3*(4*B*a^3*b + 6*A*a^2*b^2 - 4*B*a*b^3
- A*b^4)*log(1/(tan(d*x + c)^2 + 1)) + 6*(6*B*a^2*b^2 + 4*A*a*b^3 - B*b^4)*tan(d*x + c))/d

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Sympy [A]  time = 4.89496, size = 291, normalized size = 1.69 \begin{align*} \begin{cases} - \frac{A a^{4} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{A a^{4} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + 4 A a^{3} b x + \frac{3 A a^{2} b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - 4 A a b^{3} x + \frac{4 A a b^{3} \tan{\left (c + d x \right )}}{d} - \frac{A b^{4} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{A b^{4} \tan ^{2}{\left (c + d x \right )}}{2 d} + B a^{4} x + \frac{2 B a^{3} b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - 6 B a^{2} b^{2} x + \frac{6 B a^{2} b^{2} \tan{\left (c + d x \right )}}{d} - \frac{2 B a b^{3} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac{2 B a b^{3} \tan ^{2}{\left (c + d x \right )}}{d} + B b^{4} x + \frac{B b^{4} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac{B b^{4} \tan{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (A + B \tan{\left (c \right )}\right ) \left (a + b \tan{\left (c \right )}\right )^{4} \cot{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+b*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

Piecewise((-A*a**4*log(tan(c + d*x)**2 + 1)/(2*d) + A*a**4*log(tan(c + d*x))/d + 4*A*a**3*b*x + 3*A*a**2*b**2*
log(tan(c + d*x)**2 + 1)/d - 4*A*a*b**3*x + 4*A*a*b**3*tan(c + d*x)/d - A*b**4*log(tan(c + d*x)**2 + 1)/(2*d)
+ A*b**4*tan(c + d*x)**2/(2*d) + B*a**4*x + 2*B*a**3*b*log(tan(c + d*x)**2 + 1)/d - 6*B*a**2*b**2*x + 6*B*a**2
*b**2*tan(c + d*x)/d - 2*B*a*b**3*log(tan(c + d*x)**2 + 1)/d + 2*B*a*b**3*tan(c + d*x)**2/d + B*b**4*x + B*b**
4*tan(c + d*x)**3/(3*d) - B*b**4*tan(c + d*x)/d, Ne(d, 0)), (x*(A + B*tan(c))*(a + b*tan(c))**4*cot(c), True))

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Giac [A]  time = 2.6021, size = 258, normalized size = 1.5 \begin{align*} \frac{2 \, B b^{4} \tan \left (d x + c\right )^{3} + 12 \, B a b^{3} \tan \left (d x + c\right )^{2} + 3 \, A b^{4} \tan \left (d x + c\right )^{2} + 6 \, A a^{4} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + 36 \, B a^{2} b^{2} \tan \left (d x + c\right ) + 24 \, A a b^{3} \tan \left (d x + c\right ) - 6 \, B b^{4} \tan \left (d x + c\right ) + 6 \,{\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )}{\left (d x + c\right )} - 3 \,{\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*b^4*tan(d*x + c)^3 + 12*B*a*b^3*tan(d*x + c)^2 + 3*A*b^4*tan(d*x + c)^2 + 6*A*a^4*log(abs(tan(d*x + c
))) + 36*B*a^2*b^2*tan(d*x + c) + 24*A*a*b^3*tan(d*x + c) - 6*B*b^4*tan(d*x + c) + 6*(B*a^4 + 4*A*a^3*b - 6*B*
a^2*b^2 - 4*A*a*b^3 + B*b^4)*(d*x + c) - 3*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*log(tan(d*x +
c)^2 + 1))/d