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(d*x+c),-b*tan(d*x+c)/a)*cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^n/d/((1+b*tan
(d*x+c)/a)^n)-(A-I*B)*AppellF1(-1/2,1,-n,1/2,I*tan(d*x+c),-b*tan(d*x+c)/a)*cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^n
/d/((1+b*tan(d*x+c)/a)^n)

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Rubi [A]  time = 0.49, antiderivative size = 169, normalized size of antiderivative = 1.00, 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 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 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])

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 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 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 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]

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*}

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Mathematica [F]  time = 8.15, size = 0, normalized size = 0.00 \[ \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \]

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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fricas [F]  time = 1.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \cot \left (d x + c\right ) \tan \left (d x + c\right ) + A \cot \left (d x + c\right )\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sqrt {\cot \left (d x + c\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

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)

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maple [F]  time = 1.56, size = 0, normalized size = 0.00 \[ \int \left (\cot ^{\frac {3}{2}}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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