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

Optimal. Leaf size=173 \[ \frac {(A+i B) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {3}{2};1,-n;\frac {5}{2};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(A-i B) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {3}{2};1,-n;\frac {5}{2};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

[Out]

1/3*(A+I*B)*AppellF1(3/2,1,-n,5/2,-I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^n/d/cot(d*x+c)^(3/2)/((1+b*t
an(d*x+c)/a)^n)+1/3*(A-I*B)*AppellF1(3/2,1,-n,5/2,I*tan(d*x+c),-b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^n/d/cot(d*x+c
)^(3/2)/((1+b*tan(d*x+c)/a)^n)

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Rubi [A]  time = 0.46, antiderivative size = 173, 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} \[ \frac {(A+i B) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {3}{2};1,-n;\frac {5}{2};-i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(A-i B) (a+b \tan (c+d x))^n \left (\frac {b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac {3}{2};1,-n;\frac {5}{2};i \tan (c+d x),-\frac {b \tan (c+d x)}{a}\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + I*B)*AppellF1[3/2, 1, -n, 5/2, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*(a + b*Tan[c + d*x])^n)/(3*d*Co
t[c + d*x]^(3/2)*(1 + (b*Tan[c + d*x])/a)^n) + ((A - I*B)*AppellF1[3/2, 1, -n, 5/2, I*Tan[c + d*x], -((b*Tan[c
+ d*x])/a)]*(a + b*Tan[c + d*x])^n)/(3*d*Cot[c + d*x]^(3/2)*(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 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 \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\\ &=\frac {1}{2} \left ((A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int (1+i \tan (c+d x)) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \, dx+\frac {1}{2} \left ((A+i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int (1-i \tan (c+d x)) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^n \, dx\\ &=\frac {\left ((A-i B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {x} (a+b x)^n}{1-i x} \, 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 {\sqrt {x} (a+b x)^n}{1+i x} \, 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 {x^2 \left (a+b x^2\right )^n}{1-i x^2} \, 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 {x^2 \left (a+b x^2\right )^n}{1+i x^2} \, 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 {x^2 \left (1+\frac {b x^2}{a}\right )^n}{1-i x^2} \, 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 {x^2 \left (1+\frac {b x^2}{a}\right )^n}{1+i x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {(A+i B) F_1\left (\frac {3}{2};1,-n;\frac {5}{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}}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {(A-i B) F_1\left (\frac {3}{2};1,-n;\frac {5}{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}}{3 d \cot ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [F]  time = 13.31, size = 0, normalized size = 0.00 \[ \int \frac {(a+b \tan (c+d x))^n (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\cot \left (d x + c\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^n*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((B*tan(d*x + c) + A)*(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 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt {\cot \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^n*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.53, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^n*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**n*(A+B*tan(d*x+c))/cot(d*x+c)**(1/2),x)

[Out]

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

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