3.256 \(\int \cot ^6(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=233 \[ \frac{a \left (5 a^2 A-15 a b B-12 A b^2\right ) \cot ^3(c+d x)}{15 d}+\frac{\left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \cot ^2(c+d x)}{2 d}-\frac{\left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \cot (c+d x)}{d}+\frac{\left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \log (\sin (c+d x))}{d}-x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )-\frac{a^2 (5 a B+7 A b) \cot ^4(c+d x)}{20 d}-\frac{a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d} \]

[Out]

-((a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*x) - ((a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*Cot[c + d*x])/d + ((3*
a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Cot[c + d*x]^2)/(2*d) + (a*(5*a^2*A - 12*A*b^2 - 15*a*b*B)*Cot[c + d*x]^3
)/(15*d) - (a^2*(7*A*b + 5*a*B)*Cot[c + d*x]^4)/(20*d) + ((3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Log[Sin[c +
d*x]])/d - (a*A*Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2)/(5*d)

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Rubi [A]  time = 0.495875, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3605, 3635, 3628, 3529, 3531, 3475} \[ \frac{a \left (5 a^2 A-15 a b B-12 A b^2\right ) \cot ^3(c+d x)}{15 d}+\frac{\left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \cot ^2(c+d x)}{2 d}-\frac{\left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \cot (c+d x)}{d}+\frac{\left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \log (\sin (c+d x))}{d}-x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )-\frac{a^2 (5 a B+7 A b) \cot ^4(c+d x)}{20 d}-\frac{a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

-((a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*x) - ((a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*Cot[c + d*x])/d + ((3*
a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Cot[c + d*x]^2)/(2*d) + (a*(5*a^2*A - 12*A*b^2 - 15*a*b*B)*Cot[c + d*x]^3
)/(15*d) - (a^2*(7*A*b + 5*a*B)*Cot[c + d*x]^4)/(20*d) + ((3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Log[Sin[c +
d*x]])/d - (a*A*Cot[c + d*x]^5*(a + b*Tan[c + d*x])^2)/(5*d)

Rule 3605

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*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e
+ f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*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] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

Rule 3635

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 - a*d)*(c^2*C - B*c*d + A*d^2)*
(c + d*Tan[e + f*x])^(n + 1))/(d^2*f*(n + 1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x
])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b*(c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d +
a*C*d)*Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] &&
NeQ[b*c - a*d, 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 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 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 ^6(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac{1}{5} \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (a (7 A b+5 a B)-5 \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)-b (3 a A-5 b B) \tan ^2(c+d x)\right ) \, dx\\ &=-\frac{a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}-\frac{a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac{1}{5} \int \cot ^4(c+d x) \left (-a \left (5 a^2 A-12 A b^2-15 a b B\right )-5 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-b^2 (3 a A-5 b B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (5 a^2 A-12 A b^2-15 a b B\right ) \cot ^3(c+d x)}{15 d}-\frac{a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}-\frac{a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac{1}{5} \int \cot ^3(c+d x) \left (-5 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )+5 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)\right ) \, dx\\ &=\frac{\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot ^2(c+d x)}{2 d}+\frac{a \left (5 a^2 A-12 A b^2-15 a b B\right ) \cot ^3(c+d x)}{15 d}-\frac{a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}-\frac{a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac{1}{5} \int \cot ^2(c+d x) \left (5 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )+5 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)\right ) \, dx\\ &=-\frac{\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \cot (c+d x)}{d}+\frac{\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot ^2(c+d x)}{2 d}+\frac{a \left (5 a^2 A-12 A b^2-15 a b B\right ) \cot ^3(c+d x)}{15 d}-\frac{a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}-\frac{a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\frac{1}{5} \int \cot (c+d x) \left (5 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right )-5 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)\right ) \, dx\\ &=-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x-\frac{\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \cot (c+d x)}{d}+\frac{\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot ^2(c+d x)}{2 d}+\frac{a \left (5 a^2 A-12 A b^2-15 a b B\right ) \cot ^3(c+d x)}{15 d}-\frac{a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}-\frac{a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}+\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \int \cot (c+d x) \, dx\\ &=-\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x-\frac{\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \cot (c+d x)}{d}+\frac{\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cot ^2(c+d x)}{2 d}+\frac{a \left (5 a^2 A-12 A b^2-15 a b B\right ) \cot ^3(c+d x)}{15 d}-\frac{a^2 (7 A b+5 a B) \cot ^4(c+d x)}{20 d}+\frac{\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \log (\sin (c+d x))}{d}-\frac{a A \cot ^5(c+d x) (a+b \tan (c+d x))^2}{5 d}\\ \end{align*}

Mathematica [C]  time = 1.15635, size = 237, normalized size = 1.02 \[ \frac{20 a \left (a^2 A-3 a b B-3 A b^2\right ) \cot ^3(c+d x)+30 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \cot ^2(c+d x)-60 \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \cot (c+d x)+60 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \log (\tan (c+d x))-15 a^2 (a B+3 A b) \cot ^4(c+d x)-12 a^3 A \cot ^5(c+d x)+30 i (a+i b)^3 (A+i B) \log (-\tan (c+d x)+i)+30 (b+i a)^3 (A-i B) \log (\tan (c+d x)+i)}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(-60*(a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*Cot[c + d*x] + 30*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Cot[c +
d*x]^2 + 20*a*(a^2*A - 3*A*b^2 - 3*a*b*B)*Cot[c + d*x]^3 - 15*a^2*(3*A*b + a*B)*Cot[c + d*x]^4 - 12*a^3*A*Cot
[c + d*x]^5 + (30*I)*(a + I*b)^3*(A + I*B)*Log[I - Tan[c + d*x]] + 60*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*
Log[Tan[c + d*x]] + 30*(I*a + b)^3*(A - I*B)*Log[I + Tan[c + d*x]])/(60*d)

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Maple [A]  time = 0.087, size = 376, normalized size = 1.6 \begin{align*} -{\frac{A\cot \left ( dx+c \right ){a}^{3}}{d}}-A{a}^{3}x-{\frac{B{b}^{3}c}{d}}-B{b}^{3}x-{\frac{A{a}^{3}c}{d}}+{\frac{B{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{A{b}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{Aa{b}^{2}c}{d}}+3\,{\frac{B{a}^{2}bc}{d}}+3\,{\frac{A{a}^{2}b\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-3\,{\frac{Ba{b}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,A{a}^{2}b \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+3\,{\frac{B\cot \left ( dx+c \right ){a}^{2}b}{d}}+3\,{\frac{A\cot \left ( dx+c \right ) a{b}^{2}}{d}}-{\frac{Aa{b}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{3\,Ba{b}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{3\,A{a}^{2}b \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{B{a}^{2}b \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{d}}-{\frac{A{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{5}}{5\,d}}-{\frac{B{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{A{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{B{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{A{b}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{B\cot \left ( dx+c \right ){b}^{3}}{d}}+3\,Aa{b}^{2}x+3\,B{a}^{2}bx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

-1/d*A*cot(d*x+c)*a^3-A*a^3*x-1/d*B*b^3*c-B*b^3*x-1/d*A*a^3*c+1/d*B*a^3*ln(sin(d*x+c))-1/d*A*b^3*ln(sin(d*x+c)
)+3/d*A*a*b^2*c+3/d*B*a^2*b*c+3/d*A*a^2*b*ln(sin(d*x+c))-3/d*B*a*b^2*ln(sin(d*x+c))+3/2/d*A*a^2*b*cot(d*x+c)^2
+3/d*B*cot(d*x+c)*a^2*b+3/d*A*cot(d*x+c)*a*b^2-1/d*A*a*b^2*cot(d*x+c)^3-3/2/d*B*a*b^2*cot(d*x+c)^2-3/4/d*A*a^2
*b*cot(d*x+c)^4-1/d*B*a^2*b*cot(d*x+c)^3-1/5/d*A*a^3*cot(d*x+c)^5-1/4/d*B*a^3*cot(d*x+c)^4+1/3/d*A*a^3*cot(d*x
+c)^3+1/2/d*B*a^3*cot(d*x+c)^2-1/2/d*A*b^3*cot(d*x+c)^2-1/d*B*cot(d*x+c)*b^3+3*A*a*b^2*x+3*B*a^2*b*x

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Maxima [A]  time = 1.48341, size = 338, normalized size = 1.45 \begin{align*} -\frac{60 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )}{\left (d x + c\right )} + 30 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac{60 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \tan \left (d x + c\right )^{4} + 12 \, A a^{3} - 30 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{3} - 20 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} + 15 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(60*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*(d*x + c) + 30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*log(t
an(d*x + c)^2 + 1) - 60*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*log(tan(d*x + c)) + (60*(A*a^3 - 3*B*a^2*b - 3
*A*a*b^2 + B*b^3)*tan(d*x + c)^4 + 12*A*a^3 - 30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*tan(d*x + c)^3 - 20*(
A*a^3 - 3*B*a^2*b - 3*A*a*b^2)*tan(d*x + c)^2 + 15*(B*a^3 + 3*A*a^2*b)*tan(d*x + c))/tan(d*x + c)^5)/d

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Fricas [A]  time = 2.00985, size = 620, normalized size = 2.66 \begin{align*} \frac{30 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\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 )^{5} + 15 \,{\left (3 \, B a^{3} + 9 \, A a^{2} b - 6 \, B a b^{2} - 2 \, A b^{3} - 4 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{5} - 60 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \tan \left (d x + c\right )^{4} - 12 \, A a^{3} + 30 \,{\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \tan \left (d x + c\right )^{3} + 20 \,{\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \tan \left (d x + c\right )^{2} - 15 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{60 \, d \tan \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/60*(30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^5 + 15*
(3*B*a^3 + 9*A*a^2*b - 6*B*a*b^2 - 2*A*b^3 - 4*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*d*x)*tan(d*x + c)^5 - 6
0*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*tan(d*x + c)^4 - 12*A*a^3 + 30*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^
3)*tan(d*x + c)^3 + 20*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2)*tan(d*x + c)^2 - 15*(B*a^3 + 3*A*a^2*b)*tan(d*x + c))/(
d*tan(d*x + c)^5)

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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)**6*(a+b*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 2.10664, size = 905, normalized size = 3.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/960*(6*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 15*B*a^3*tan(1/2*d*x + 1/2*c)^4 - 45*A*a^2*b*tan(1/2*d*x + 1/2*c)^4 -
70*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 120*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 120*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 18
0*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 540*A*a^2*b*tan(1/2*d*x + 1/2*c)^2 - 360*B*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 120
*A*b^3*tan(1/2*d*x + 1/2*c)^2 + 660*A*a^3*tan(1/2*d*x + 1/2*c) - 1800*B*a^2*b*tan(1/2*d*x + 1/2*c) - 1800*A*a*
b^2*tan(1/2*d*x + 1/2*c) + 480*B*b^3*tan(1/2*d*x + 1/2*c) - 960*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2 + B*b^3)*(d*x +
c) - 960*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*log(tan(1/2*d*x + 1/2*c)^2 + 1) + 960*(B*a^3 + 3*A*a^2*b - 3
*B*a*b^2 - A*b^3)*log(abs(tan(1/2*d*x + 1/2*c))) - (2192*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 6576*A*a^2*b*tan(1/2*d
*x + 1/2*c)^5 - 6576*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 2192*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 660*A*a^3*tan(1/2*d*
x + 1/2*c)^4 - 1800*B*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 1800*A*a*b^2*tan(1/2*d*x + 1/2*c)^4 + 480*B*b^3*tan(1/2*d
*x + 1/2*c)^4 - 180*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 540*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 360*B*a*b^2*tan(1/2*d*
x + 1/2*c)^3 + 120*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 70*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 120*B*a^2*b*tan(1/2*d*x +
1/2*c)^2 + 120*A*a*b^2*tan(1/2*d*x + 1/2*c)^2 + 15*B*a^3*tan(1/2*d*x + 1/2*c) + 45*A*a^2*b*tan(1/2*d*x + 1/2*c
) + 6*A*a^3)/tan(1/2*d*x + 1/2*c)^5)/d