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

Optimal. Leaf size=159 \[ \frac {(a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {5}{2};1,-n;\frac {7}{2};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {(a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {5}{2};1,-n;\frac {7}{2};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)} \]

[Out]

1/5*AppellF1(5/2,1,-n,7/2,-I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^n/d/cot(d*x+c)^(5/2)/((1+b*tan(d*x+c
)/a)^n)+1/5*AppellF1(5/2,1,-n,7/2,I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^n/d/cot(d*x+c)^(5/2)/((1+b*ta
n(d*x+c)/a)^n)

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4241, 3575, 912, 130, 511, 510} \[ \frac {(a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {5}{2};1,-n;\frac {7}{2};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {(a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {5}{2};1,-n;\frac {7}{2};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(AppellF1[5/2, 1, -n, 7/2, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*(a + b*Tan[c + d*x])^n)/(5*d*Cot[c + d*x]
^(5/2)*(1 + (b*Tan[c + d*x])/a)^n) + (AppellF1[5/2, 1, -n, 7/2, I*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*(a + b*
Tan[c + d*x])^n)/(5*d*Cot[c + d*x]^(5/2)*(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 912

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

Rule 3575

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Wit
h[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((a + b*ff*x)^m*(c + d*ff*x)^n)/(1 + ff^2*x^2), x]
, x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^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 \frac {(a+b \tan (c+d x))^n}{\cot ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^n \, dx\\ &=\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {x^{3/2} (a+b x)^n}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \left (\frac {i x^{3/2} (a+b x)^n}{2 (i-x)}+\frac {i x^{3/2} (a+b x)^n}{2 (i+x)}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\left (i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {x^{3/2} (a+b x)^n}{i-x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {x^{3/2} (a+b x)^n}{i+x} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {\left (i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^n}{i-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}+\frac {\left (i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^n}{i+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {\left (i \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 {x^4 \left (1+\frac {b x^2}{a}\right )^n}{i-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}+\frac {\left (i \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 {x^4 \left (1+\frac {b x^2}{a}\right )^n}{i+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {F_1\left (\frac {5}{2};1,-n;\frac {7}{2};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {F_1\left (\frac {5}{2};1,-n;\frac {7}{2};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^n \left (1+\frac {b \tan (c+d x)}{a}\right )^{-n}}{5 d \cot ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 5.57, size = 0, normalized size = 0.00 \[ \int \frac {(a+b \tan (c+d x))^n}{\cot ^{\frac {3}{2}}(c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{\cot \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 1.05, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \tan \left (d x +c \right )\right )^{n}}{\cot \left (d x +c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**n/cot(d*x+c)**(3/2),x)

[Out]

Integral((a + b*tan(c + d*x))**n/cot(c + d*x)**(3/2), x)

________________________________________________________________________________________