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

Optimal. Leaf size=129 \[ \frac {a (e \tan (c+d x))^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac {a \sec (c+d x) \cos ^2(c+d x)^{\frac {m+2}{2}} (e \tan (c+d x))^{m+1} \, _2F_1\left (\frac {m+1}{2},\frac {m+2}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)} \]

[Out]

a*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-tan(d*x+c)^2)*(e*tan(d*x+c))^(1+m)/d/e/(1+m)+a*(cos(d*x+c)^2)^(1+1/2*m
)*hypergeom([1+1/2*m, 1/2+1/2*m],[3/2+1/2*m],sin(d*x+c)^2)*sec(d*x+c)*(e*tan(d*x+c))^(1+m)/d/e/(1+m)

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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 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]

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 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]

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 = 0.73, size = 105, normalized size = 0.81 \[ \frac {a (e \tan (c+d x))^m \left (\frac {\tan (c+d x) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{m+1}+\csc (c+d x) \left (-\tan ^2(c+d x)\right )^{\frac {1-m}{2}} \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3}{2};\sec ^2(c+d x)\right )\right )}{d} \]

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

________________________________________________________________________________________

fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{m}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{m}\,{d x} \]

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)

________________________________________________________________________________________

maple [F]  time = 2.45, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right ) \left (e \tan \left (d x +c \right )\right )^{m}\, dx \]

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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )} \left (e \tan \left (d x + c\right )\right )^{m}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \left (e \tan {\left (c + d x \right )}\right )^{m}\, dx + \int \left (e \tan {\left (c + d x \right )}\right )^{m} \sec {\left (c + d x \right )}\, dx\right ) \]

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))

________________________________________________________________________________________