3.7.57 \(\int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\) [657]

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

[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 = 0.31, 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 = {4326, 3684, 3683, 129, 525, 524} \begin {gather*} -\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} \end {gather*}

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 129

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 3683

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 3684

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 4326

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 = 5.14, size = 0, normalized size = 0.00 \begin {gather*} \int \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx \end {gather*}

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 = 0.53, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, 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)^(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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, 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)^(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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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