3.657 \(\int \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=169 \[ -\frac{(A+i B) \sqrt{\cot (c+d x)} (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (-\frac{1}{2};1,-n;\frac{1}{2};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{d}-\frac{(A-i B) \sqrt{\cot (c+d x)} (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (-\frac{1}{2};1,-n;\frac{1}{2};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{d} \]

[Out]

-(((A + I*B)*AppellF1[-1/2, 1, -n, 1/2, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Sqrt[Cot[c + d*x]]*(a + b*Ta
n[c + d*x])^n)/(d*(1 + (b*Tan[c + d*x])/a)^n)) - ((A - I*B)*AppellF1[-1/2, 1, -n, 1/2, I*Tan[c + d*x], -((b*Ta
n[c + d*x])/a)]*Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^n)/(d*(1 + (b*Tan[c + d*x])/a)^n)

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Rubi [A]  time = 0.487656, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4241, 3603, 3602, 130, 511, 510} \[ -\frac{(A+i B) \sqrt{\cot (c+d x)} (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (-\frac{1}{2};1,-n;\frac{1}{2};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{d}-\frac{(A-i B) \sqrt{\cot (c+d x)} (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (-\frac{1}{2};1,-n;\frac{1}{2};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((A + I*B)*AppellF1[-1/2, 1, -n, 1/2, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Sqrt[Cot[c + d*x]]*(a + b*Ta
n[c + d*x])^n)/(d*(1 + (b*Tan[c + d*x])/a)^n)) - ((A - I*B)*AppellF1[-1/2, 1, -n, 1/2, I*Tan[c + d*x], -((b*Ta
n[c + d*x])/a)]*Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^n)/(d*(1 + (b*Tan[c + d*x])/a)^n)

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rule 3603

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

Rule 3602

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

Rule 130

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + (b*x^k)/e)^m*(c + (d*x^k)/e)^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{1}{2} \left ((A-i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(1+i \tan (c+d x)) (a+b \tan (c+d x))^n}{\tan ^{\frac{3}{2}}(c+d x)} \, dx+\frac{1}{2} \left ((A+i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(1-i \tan (c+d x)) (a+b \tan (c+d x))^n}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{\left ((A-i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^n}{(1-i x) x^{3/2}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left ((A+i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^n}{(1+i x) x^{3/2}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{\left ((A-i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^n}{x^2 \left (1-i x^2\right )} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}+\frac{\left ((A+i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^n}{x^2 \left (1+i x^2\right )} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{\left ((A-i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b x^2}{a}\right )^n}{x^2 \left (1-i x^2\right )} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}+\frac{\left ((A+i B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b x^2}{a}\right )^n}{x^2 \left (1+i x^2\right )} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{(A+i B) F_1\left (-\frac{1}{2};1,-n;\frac{1}{2};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right ) \sqrt{\cot (c+d x)} (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}}{d}-\frac{(A-i B) F_1\left (-\frac{1}{2};1,-n;\frac{1}{2};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right ) \sqrt{\cot (c+d x)} (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}}{d}\\ \end{align*}

Mathematica [F]  time = 8.40279, size = 0, normalized size = 0. \[ \int \cot ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.4, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \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)^(3/2)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

int(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

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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 )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(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)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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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)**(3/2)*(a+b*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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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 )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(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)^(3/2), x)