\(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [225]

Optimal result

Integrand size = 34, antiderivative size = 185 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=-\frac {(2 B+i A n) \cot (c+d x) (a+i a \tan (c+d x))^n}{2 d}-\frac {A \cot ^2(c+d x) (a+i a \tan (c+d x))^n}{2 d}-\frac {(A-i B) \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^n}{2 d n}-\frac {\left (2 i B n-A \left (2-n+n^2\right )\right ) \operatorname {Hypergeometric2F1}(1,n,1+n,1+i \tan (c+d x)) (a+i a \tan (c+d x))^n}{2 d n} \]

[Out]

-1/2*(2*B+I*A*n)*cot(d*x+c)*(a+I*a*tan(d*x+c))^n/d-1/2*A*cot(d*x+c)^2*(a+I*a*tan(d*x+c))^n/d-1/2*(A-I*B)*hyper
geom([1, n],[1+n],1/2+1/2*I*tan(d*x+c))*(a+I*a*tan(d*x+c))^n/d/n-1/2*(2*I*B*n-A*(n^2-n+2))*hypergeom([1, n],[1
+n],1+I*tan(d*x+c))*(a+I*a*tan(d*x+c))^n/d/n

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3679, 3681, 3562, 70, 3680, 67} \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=-\frac {\left (-A \left (n^2-n+2\right )+2 i B n\right ) (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}(1,n,n+1,i \tan (c+d x)+1)}{2 d n}-\frac {(A-i B) (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}\left (1,n,n+1,\frac {1}{2} (i \tan (c+d x)+1)\right )}{2 d n}-\frac {(2 B+i A n) \cot (c+d x) (a+i a \tan (c+d x))^n}{2 d}-\frac {A \cot ^2(c+d x) (a+i a \tan (c+d x))^n}{2 d} \]

[In]

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

[Out]

-1/2*((2*B + I*A*n)*Cot[c + d*x]*(a + I*a*Tan[c + d*x])^n)/d - (A*Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^n)/(2*
d) - ((A - I*B)*Hypergeometric2F1[1, n, 1 + n, (1 + I*Tan[c + d*x])/2]*(a + I*a*Tan[c + d*x])^n)/(2*d*n) - (((
2*I)*B*n - A*(2 - n + n^2))*Hypergeometric2F1[1, n, 1 + n, 1 + I*Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(2*d*
n)

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3562

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[-b/d, Subst[Int[(a + x)^(n - 1)/(a - x), x]
, x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

Rule 3679

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

Rule 3680

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b*(B/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 3681

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

Rubi steps \begin{align*} \text {integral}& = -\frac {A \cot ^2(c+d x) (a+i a \tan (c+d x))^n}{2 d}+\frac {\int \cot ^2(c+d x) (a+i a \tan (c+d x))^n (a (2 B+i A n)-a A (2-n) \tan (c+d x)) \, dx}{2 a} \\ & = -\frac {(2 B+i A n) \cot (c+d x) (a+i a \tan (c+d x))^n}{2 d}-\frac {A \cot ^2(c+d x) (a+i a \tan (c+d x))^n}{2 d}+\frac {\int \cot (c+d x) (a+i a \tan (c+d x))^n \left (a^2 \left (2 i B n-A \left (2-n+n^2\right )\right )-a^2 (1-n) (2 B+i A n) \tan (c+d x)\right ) \, dx}{2 a^2} \\ & = -\frac {(2 B+i A n) \cot (c+d x) (a+i a \tan (c+d x))^n}{2 d}-\frac {A \cot ^2(c+d x) (a+i a \tan (c+d x))^n}{2 d}+(-i A-B) \int (a+i a \tan (c+d x))^n \, dx+\frac {\left (2 i B n-A \left (2-n+n^2\right )\right ) \int \cot (c+d x) (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^n \, dx}{2 a} \\ & = -\frac {(2 B+i A n) \cot (c+d x) (a+i a \tan (c+d x))^n}{2 d}-\frac {A \cot ^2(c+d x) (a+i a \tan (c+d x))^n}{2 d}-\frac {(a (A-i B)) \text {Subst}\left (\int \frac {(a+x)^{-1+n}}{a-x} \, dx,x,i a \tan (c+d x)\right )}{d}+\frac {\left (a \left (2 i B n-A \left (2-n+n^2\right )\right )\right ) \text {Subst}\left (\int \frac {(a+i a x)^{-1+n}}{x} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = -\frac {(2 B+i A n) \cot (c+d x) (a+i a \tan (c+d x))^n}{2 d}-\frac {A \cot ^2(c+d x) (a+i a \tan (c+d x))^n}{2 d}-\frac {(A-i B) \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^n}{2 d n}-\frac {\left (2 i B n-A \left (2-n+n^2\right )\right ) \operatorname {Hypergeometric2F1}(1,n,1+n,1+i \tan (c+d x)) (a+i a \tan (c+d x))^n}{2 d n} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.17 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.19 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\frac {(a+i a \tan (c+d x))^n \left (-i (A-i B) n \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {1}{2} (1+i \tan (c+d x))\right ) (-i+\tan (c+d x))+2 (A+i B+A n+i B n+2 A n \operatorname {Hypergeometric2F1}(3,1+n,2+n,1+i \tan (c+d x))+2 A n \operatorname {Hypergeometric2F1}(1,1+n,2+n,1+i \tan (c+d x)) (1+i \tan (c+d x))+2 i A n \operatorname {Hypergeometric2F1}(3,1+n,2+n,1+i \tan (c+d x)) \tan (c+d x)-2 B n \operatorname {Hypergeometric2F1}(2,1+n,2+n,1+i \tan (c+d x)) (-i+\tan (c+d x)))\right )}{4 d n (1+n)} \]

[In]

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

[Out]

((a + I*a*Tan[c + d*x])^n*((-I)*(A - I*B)*n*Hypergeometric2F1[1, 1 + n, 2 + n, (1 + I*Tan[c + d*x])/2]*(-I + T
an[c + d*x]) + 2*(A + I*B + A*n + I*B*n + 2*A*n*Hypergeometric2F1[3, 1 + n, 2 + n, 1 + I*Tan[c + d*x]] + 2*A*n
*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + I*Tan[c + d*x]]*(1 + I*Tan[c + d*x]) + (2*I)*A*n*Hypergeometric2F1[3,
1 + n, 2 + n, 1 + I*Tan[c + d*x]]*Tan[c + d*x] - 2*B*n*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + I*Tan[c + d*x]]*
(-I + Tan[c + d*x]))))/(4*d*n*(1 + n))

Maple [F]

\[\int \left (\cot ^{3}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

integral(((-I*A - B)*e^(6*I*d*x + 6*I*c) + (-3*I*A - B)*e^(4*I*d*x + 4*I*c) + (-3*I*A + B)*e^(2*I*d*x + 2*I*c)
- I*A + B)*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n/(e^(6*I*d*x + 6*I*c) - 3*e^(4*I*d*x + 4*I*c)
+ 3*e^(2*I*d*x + 2*I*c) - 1), x)

Sympy [F]

\[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \left (A + B \tan {\left (c + d x \right )}\right ) \cot ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**n*(A + B*tan(c + d*x))*cot(c + d*x)**3, x)

Maxima [F]

\[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*cot(d*x + c)^3, x)

Giac [F]

\[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n*cot(d*x + c)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^n (A+B \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^3\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]

[In]

int(cot(c + d*x)^3*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n,x)

[Out]

int(cot(c + d*x)^3*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n, x)