3.8.13 \(\int \cot ^2(c+d x) (a+b \tan (c+d x))^n \, dx\) [713]

Optimal. Leaf size=245 \[ -\frac {\cot (c+d x) (a+b \tan (c+d x))^{1+n}}{a d}-\frac {b \, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right ) d (1+n)}+\frac {b \, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right ) d (1+n)}-\frac {b n \, _2F_1\left (1,1+n;2+n;1+\frac {b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{a^2 d (1+n)} \]

[Out]

-cot(d*x+c)*(a+b*tan(d*x+c))^(1+n)/a/d-b*n*hypergeom([1, 1+n],[2+n],1+b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^(1+n)/a
^2/d/(1+n)-1/2*b*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a-(-b^2)^(1/2)))*(a+b*tan(d*x+c))^(1+n)/d/(1+n)/(a
-(-b^2)^(1/2))/(-b^2)^(1/2)+1/2*b*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a+(-b^2)^(1/2)))*(a+b*tan(d*x+c))
^(1+n)/d/(1+n)/(-b^2)^(1/2)/(a+(-b^2)^(1/2))

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Rubi [A]
time = 0.32, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3650, 3734, 12, 3566, 726, 70, 3715, 67} \begin {gather*} -\frac {b n (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 )}{a^2 d (n+1)}-\frac {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-\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} d (n+1) \left (a-\sqrt {-b^2}\right )}+\frac {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+\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} d (n+1) \left (a+\sqrt {-b^2}\right )}-\frac {\cot (c+d x) (a+b \tan (c+d x))^{n+1}}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cot[c + d*x]*(a + b*Tan[c + d*x])^(1 + n))/(a*d)) - (b*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*
x])/(a - Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(2*Sqrt[-b^2]*(a - Sqrt[-b^2])*d*(1 + n)) + (b*Hypergeomet
ric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(2*Sqrt[-b^2]*(a
+ Sqrt[-b^2])*d*(1 + n)) - (b*n*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b*Tan[c + d*x])/a]*(a + b*Tan[c + d*x]
)^(1 + n))/(a^2*d*(1 + n))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 726

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 3566

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

Rule 3650

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 3715

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 3734

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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.85, size = 190, normalized size = 0.78 \begin {gather*} -\frac {(b+a \cot (c+d x)) \left (a^2 (-i a+b) \, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a-i b}\right )+(a-i b) \left (i a^2 \, _2F_1\left (1,1+n;2+n;\frac {a+b \tan (c+d x)}{a+i b}\right )+2 (a+i b) \left (a (1+n) \cot (c+d x)+b n \, _2F_1\left (1,1+n;2+n;1+\frac {b \tan (c+d x)}{a}\right )\right )\right )\right ) \tan (c+d x) (a+b \tan (c+d x))^n}{2 a^2 (a-i b) (a+i b) d (1+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.29, size = 0, normalized size = 0.00 \[\int \left (\cot ^{2}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2*(a+b*tan(d*x+c))^n,x)

[Out]

int(cot(d*x+c)^2*(a+b*tan(d*x+c))^n,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^n*cot(d*x + c)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((b*tan(d*x + c) + a)^n*cot(d*x + c)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \cot ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(a+b*tan(d*x+c))**n,x)

[Out]

Integral((a + b*tan(c + d*x))**n*cot(c + d*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)^n*cot(d*x + c)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {cot}\left (c+d\,x\right )}^2\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^2*(a + b*tan(c + d*x))^n,x)

[Out]

int(cot(c + d*x)^2*(a + b*tan(c + d*x))^n, x)

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