3.211 \(\int (a+a \sec (c+d x)) (e \tan (c+d x))^m \, dx\)

Optimal. Leaf size=129 \[ \frac{a (e \tan (c+d x))^{m+1} \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac{a \sec (c+d x) \cos ^2(c+d x)^{\frac{m+2}{2}} (e \tan (c+d x))^{m+1} \text{Hypergeometric2F1}\left (\frac{m+1}{2},\frac{m+2}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)} \]

[Out]

(a*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*(e*Tan[c + d*x])^(1 + m))/(d*e*(1 + m)) + (a*(C
os[c + d*x]^2)^((2 + m)/2)*Hypergeometric2F1[(1 + m)/2, (2 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*(e*
Tan[c + d*x])^(1 + m))/(d*e*(1 + m))

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Rubi [A]  time = 0.0824419, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3884, 3476, 364, 2617} \[ \frac{a (e \tan (c+d x))^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac{a \sec (c+d x) \cos ^2(c+d x)^{\frac{m+2}{2}} (e \tan (c+d x))^{m+1} \, _2F_1\left (\frac{m+1}{2},\frac{m+2}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*(e*Tan[c + d*x])^(1 + m))/(d*e*(1 + m)) + (a*(C
os[c + d*x]^2)^((2 + m)/2)*Hypergeometric2F1[(1 + m)/2, (2 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*(e*
Tan[c + d*x])^(1 + m))/(d*e*(1 + m))

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

Mathematica [A]  time = 0.702312, size = 105, normalized size = 0.81 \[ \frac{a (e \tan (c+d x))^m \left (\frac{\tan (c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{m+1}+\csc (c+d x) \left (-\tan ^2(c+d x)\right )^{\frac{1-m}{2}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-m}{2},\frac{3}{2},\sec ^2(c+d x)\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(e*Tan[c + d*x])^m*((Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x])/(1 + m) + Cs
c[c + d*x]*Hypergeometric2F1[1/2, (1 - m)/2, 3/2, Sec[c + d*x]^2]*(-Tan[c + d*x]^2)^((1 - m)/2)))/d

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Maple [F]  time = 0.68, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) \left ( e\tan \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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