3.212 \(\int \frac{(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=130 \[ \frac{e (e \tan (c+d x))^{m-1} \text{Hypergeometric2F1}\left (1,\frac{m-1}{2},\frac{m+1}{2},-\tan ^2(c+d x)\right )}{a d (1-m)}-\frac{e \sec (c+d x) \cos ^2(c+d x)^{m/2} (e \tan (c+d x))^{m-1} \text{Hypergeometric2F1}\left (\frac{m-1}{2},\frac{m}{2},\frac{m+1}{2},\sin ^2(c+d x)\right )}{a d (1-m)} \]

[Out]

(e*Hypergeometric2F1[1, (-1 + m)/2, (1 + m)/2, -Tan[c + d*x]^2]*(e*Tan[c + d*x])^(-1 + m))/(a*d*(1 - m)) - (e*
(Cos[c + d*x]^2)^(m/2)*Hypergeometric2F1[(-1 + m)/2, m/2, (1 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*(e*Tan[c + d
*x])^(-1 + m))/(a*d*(1 - m))

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Rubi [A]  time = 0.159216, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3888, 3884, 3476, 364, 2617} \[ \frac{e (e \tan (c+d x))^{m-1} \, _2F_1\left (1,\frac{m-1}{2};\frac{m+1}{2};-\tan ^2(c+d x)\right )}{a d (1-m)}-\frac{e \sec (c+d x) \cos ^2(c+d x)^{m/2} (e \tan (c+d x))^{m-1} \, _2F_1\left (\frac{m-1}{2},\frac{m}{2};\frac{m+1}{2};\sin ^2(c+d x)\right )}{a d (1-m)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x]),x]

[Out]

(e*Hypergeometric2F1[1, (-1 + m)/2, (1 + m)/2, -Tan[c + d*x]^2]*(e*Tan[c + d*x])^(-1 + m))/(a*d*(1 - m)) - (e*
(Cos[c + d*x]^2)^(m/2)*Hypergeometric2F1[(-1 + m)/2, m/2, (1 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*(e*Tan[c + d
*x])^(-1 + m))/(a*d*(1 - m))

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rule 3884

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(e*
Cot[c + d*x])^m, x], x] + Dist[b, Int[(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

\begin{align*} \int \frac{(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx &=\frac{e^2 \int (-a+a \sec (c+d x)) (e \tan (c+d x))^{-2+m} \, dx}{a^2}\\ &=-\frac{e^2 \int (e \tan (c+d x))^{-2+m} \, dx}{a}+\frac{e^2 \int \sec (c+d x) (e \tan (c+d x))^{-2+m} \, dx}{a}\\ &=-\frac{e \cos ^2(c+d x)^{m/2} \, _2F_1\left (\frac{1}{2} (-1+m),\frac{m}{2};\frac{1+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-1+m}}{a d (1-m)}-\frac{e^3 \operatorname{Subst}\left (\int \frac{x^{-2+m}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a d}\\ &=\frac{e \, _2F_1\left (1,\frac{1}{2} (-1+m);\frac{1+m}{2};-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-1+m}}{a d (1-m)}-\frac{e \cos ^2(c+d x)^{m/2} \, _2F_1\left (\frac{1}{2} (-1+m),\frac{m}{2};\frac{1+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-1+m}}{a d (1-m)}\\ \end{align*}

Mathematica [F]  time = 0.472587, size = 0, normalized size = 0. \[ \int \frac{(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x]),x]

[Out]

Integrate[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x]), x]

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Maple [F]  time = 0.684, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( e\tan \left ( dx+c \right ) \right ) ^{m}}{a+a\sec \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c)),x)

[Out]

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((e*tan(d*x + c))^m/(a*sec(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((e*tan(d*x + c))^m/(a*sec(d*x + c) + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\left (e \tan{\left (c + d x \right )}\right )^{m}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))**m/(a+a*sec(d*x+c)),x)

[Out]

Integral((e*tan(c + d*x))**m/(sec(c + d*x) + 1), x)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*tan(d*x + c))^m/(a*sec(d*x + c) + a), x)