3.717 \(\int \frac{(a+b \tan (c+d x))^n}{\sqrt{\tan (c+d x)}} \, dx\)

Optimal. Leaf size=153 \[ \frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac{1}{2};1,-n;\frac{3}{2};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac{1}{2};1,-n;\frac{3}{2};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{d} \]

[Out]

(AppellF1[1/2, 1, -n, 3/2, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^n
)/(d*(1 + (b*Tan[c + d*x])/a)^n) + (AppellF1[1/2, 1, -n, 3/2, I*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Sqrt[Tan[
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 = 0.190311, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3575, 912, 130, 430, 429} \[ \frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac{1}{2};1,-n;\frac{3}{2};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac{1}{2};1,-n;\frac{3}{2};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(AppellF1[1/2, 1, -n, 3/2, (-I)*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^n
)/(d*(1 + (b*Tan[c + d*x])/a)^n) + (AppellF1[1/2, 1, -n, 3/2, I*Tan[c + d*x], -((b*Tan[c + d*x])/a)]*Sqrt[Tan[
c + d*x]]*(a + b*Tan[c + d*x])^n)/(d*(1 + (b*Tan[c + d*x])/a)^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 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 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 430

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

Rule 429

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

Rubi steps

\begin{align*} \int \frac{(a+b \tan (c+d x))^n}{\sqrt{\tan (c+d x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^n}{\sqrt{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{i (a+b x)^n}{2 (i-x) \sqrt{x}}+\frac{i (a+b x)^n}{2 \sqrt{x} (i+x)}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{(a+b x)^n}{(i-x) \sqrt{x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{i \operatorname{Subst}\left (\int \frac{(a+b x)^n}{\sqrt{x} (i+x)} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^n}{i-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}+\frac{i \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^n}{i+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{\left (i (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}{i-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}+\frac{\left (i (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}{i+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{F_1\left (\frac{1}{2};1,-n;\frac{3}{2};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}}{d}+\frac{F_1\left (\frac{1}{2};1,-n;\frac{3}{2};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right ) \sqrt{\tan (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 = 0.979583, size = 0, normalized size = 0. \[ \int \frac{(a+b \tan (c+d x))^n}{\sqrt{\tan (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.243, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\tan \left ( dx+c \right ) \right ) ^{n}{\frac{1}{\sqrt{\tan \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt{\tan \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*tan(d*x + c) + a)^n/sqrt(tan(d*x + c)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \tan{\left (c + d x \right )}\right )^{n}}{\sqrt{\tan{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \tan \left (d x + c\right ) + a\right )}^{n}}{\sqrt{\tan \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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