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

Optimal. Leaf size=173 \[ \frac{\tan ^{m+1}(c+d x) \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac{1}{2},1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt{a+b \tan (c+d x)}}+\frac{\tan ^{m+1}(c+d x) \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac{1}{2},1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt{a+b \tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.182699, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3575, 912, 135, 133} \[ \frac{\tan ^{m+1}(c+d x) \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac{1}{2},1;m+2;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right )}{2 d (m+1) \sqrt{a+b \tan (c+d x)}}+\frac{\tan ^{m+1}(c+d x) \sqrt{\frac{b \tan (c+d x)}{a}+1} F_1\left (m+1;\frac{1}{2},1;m+2;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right )}{2 d (m+1) \sqrt{a+b \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^m/Sqrt[a + b*Tan[c + d*x]],x]

[Out]

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

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 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^m(c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^m}{\sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{i x^m}{2 (i-x) \sqrt{a+b x}}+\frac{i x^m}{2 (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{x^m}{(i-x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{i \operatorname{Subst}\left (\int \frac{x^m}{(i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{\left (i \sqrt{1+\frac{b \tan (c+d x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{x^m}{(i-x) \sqrt{1+\frac{b x}{a}}} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt{a+b \tan (c+d x)}}+\frac{\left (i \sqrt{1+\frac{b \tan (c+d x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{x^m}{(i+x) \sqrt{1+\frac{b x}{a}}} \, dx,x,\tan (c+d x)\right )}{2 d \sqrt{a+b \tan (c+d x)}}\\ &=\frac{F_1\left (1+m;\frac{1}{2},1;2+m;-\frac{b \tan (c+d x)}{a},-i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt{1+\frac{b \tan (c+d x)}{a}}}{2 d (1+m) \sqrt{a+b \tan (c+d x)}}+\frac{F_1\left (1+m;\frac{1}{2},1;2+m;-\frac{b \tan (c+d x)}{a},i \tan (c+d x)\right ) \tan ^{1+m}(c+d x) \sqrt{1+\frac{b \tan (c+d x)}{a}}}{2 d (1+m) \sqrt{a+b \tan (c+d x)}}\\ \end{align*}

Mathematica [F]  time = 4.32905, size = 0, normalized size = 0. \[ \int \frac{\tan ^m(c+d x)}{\sqrt{a+b \tan (c+d x)}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Tan[c + d*x]^m/Sqrt[a + b*Tan[c + d*x]],x]

[Out]

Integrate[Tan[c + d*x]^m/Sqrt[a + b*Tan[c + d*x]], x]

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Maple [F]  time = 0.316, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( dx+c \right ) \right ) ^{m}{\frac{1}{\sqrt{a+b\tan \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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