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

Optimal. Leaf size=292 \[ \frac{\left (2 a^2 A-2 a b B n+A b^2 (1-n) n\right ) (a+b \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{b \tan (c+d x)}{a}+1\right )}{2 a^3 d (n+1)}-\frac{(B+i A) (a+b \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}-\frac{(A+i B) (a+b \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}-\frac{\cot (c+d x) (2 a B-A b (1-n)) (a+b \tan (c+d x))^{n+1}}{2 a^2 d}-\frac{A \cot ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{2 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.805643, antiderivative size = 292, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {3609, 3649, 3653, 3539, 3537, 68, 3634, 65} \[ \frac{\left (2 a^2 A-2 a b B n+A b^2 (1-n) n\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b \tan (c+d x)}{a}+1\right )}{2 a^3 d (n+1)}-\frac{\cot (c+d x) (2 a B-A b (1-n)) (a+b \tan (c+d x))^{n+1}}{2 a^2 d}-\frac{(B+i A) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}-\frac{(A+i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}-\frac{A \cot ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3653

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[((c + d*Tan[e + f*x])^n*(1 + Tan[e + f*x]^2))/(a + b*Tan[e + f*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[n, 0] &&  !LeQ[n, -
1]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3537

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3634

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

Rule 65

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

Rubi steps

\begin{align*} \int \cot ^3(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=-\frac{A \cot ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{2 a d}-\frac{\int \cot ^2(c+d x) (a+b \tan (c+d x))^n \left (-2 a B+A (b-b n)+2 a A \tan (c+d x)+A b (1-n) \tan ^2(c+d x)\right ) \, dx}{2 a}\\ &=-\frac{(2 a B-A b (1-n)) \cot (c+d x) (a+b \tan (c+d x))^{1+n}}{2 a^2 d}-\frac{A \cot ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{2 a d}+\frac{\int \cot (c+d x) (a+b \tan (c+d x))^n \left (-2 a^2 A+2 a b B n-A b^2 (1-n) n-2 a^2 B \tan (c+d x)+b n (2 a B-A (b-b n)) \tan ^2(c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac{(2 a B-A b (1-n)) \cot (c+d x) (a+b \tan (c+d x))^{1+n}}{2 a^2 d}-\frac{A \cot ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{2 a d}+\frac{\int \left (-2 a^2 B+2 a^2 A \tan (c+d x)\right ) (a+b \tan (c+d x))^n \, dx}{2 a^2}+\frac{\left (-2 a^2 A+2 a b B n-A b^2 (1-n) n\right ) \int \cot (c+d x) (a+b \tan (c+d x))^n \left (1+\tan ^2(c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac{(2 a B-A b (1-n)) \cot (c+d x) (a+b \tan (c+d x))^{1+n}}{2 a^2 d}-\frac{A \cot ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{2 a d}+\frac{1}{2} (-i A-B) \int (1+i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx+\frac{1}{2} (i A-B) \int (1-i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx-\frac{\left (2 a^2 A-2 a b B n+A b^2 (1-n) n\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^n}{x} \, dx,x,\tan (c+d x)\right )}{2 a^2 d}\\ &=-\frac{(2 a B-A b (1-n)) \cot (c+d x) (a+b \tan (c+d x))^{1+n}}{2 a^2 d}-\frac{A \cot ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{2 a d}+\frac{\left (2 a^2 A-2 a b B n+A b^2 (1-n) n\right ) \, _2F_1\left (1,1+n;2+n;1+\frac{b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{2 a^3 d (1+n)}+\frac{(A-i B) \operatorname{Subst}\left (\int \frac{(a-i b x)^n}{-1+x} \, dx,x,i \tan (c+d x)\right )}{2 d}+\frac{(A+i B) \operatorname{Subst}\left (\int \frac{(a+i b x)^n}{-1+x} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac{(2 a B-A b (1-n)) \cot (c+d x) (a+b \tan (c+d x))^{1+n}}{2 a^2 d}-\frac{A \cot ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{2 a d}-\frac{(A-i B) \, _2F_1\left (1,1+n;2+n;\frac{a+b \tan (c+d x)}{a-i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a-i b) d (1+n)}-\frac{(A+i B) \, _2F_1\left (1,1+n;2+n;\frac{a+b \tan (c+d x)}{a+i b}\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) d (1+n)}+\frac{\left (2 a^2 A-2 a b B n+A b^2 (1-n) n\right ) \, _2F_1\left (1,1+n;2+n;1+\frac{b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{2 a^3 d (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.474324, size = 230, normalized size = 0.79 \[ -\frac{(a+b \tan (c+d x))^{n+1} \left (a^3 (a+i b) (A-i B) \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a-i b}\right )+(a-i b) \left (a^3 (A+i B) \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a+i b}\right )-2 (a+i b) \left (a^2 A \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{b \tan (c+d x)}{a}+1\right )+b \left (a B \text{Hypergeometric2F1}\left (2,n+1,n+2,\frac{b \tan (c+d x)}{a}+1\right )-A b \text{Hypergeometric2F1}\left (3,n+1,n+2,\frac{b \tan (c+d x)}{a}+1\right )\right )\right )\right )\right )}{2 a^3 d (n+1) (a-i b) (a+i b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*(a + I*b)*(A - I*B)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - I*b)] + (a - I*b)*(a^3
*(A + I*B)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + I*b)] - 2*(a + I*b)*(a^2*A*Hypergeomet
ric2F1[1, 1 + n, 2 + n, 1 + (b*Tan[c + d*x])/a] + b*(a*B*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (b*Tan[c + d*x
])/a] - A*b*Hypergeometric2F1[3, 1 + n, 2 + n, 1 + (b*Tan[c + d*x])/a]))))*(a + b*Tan[c + d*x])^(1 + n))/(2*a^
3*(a - I*b)*(a + I*b)*d*(1 + n))

________________________________________________________________________________________

Maple [F]  time = 0.653, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( dx+c \right ) \right ) ^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{n} \left ( A+B\tan \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B \cot \left (d x + c\right )^{3} \tan \left (d x + c\right ) + A \cot \left (d x + c\right )^{3}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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