\(\int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx\) [210]

Optimal result

Integrand size = 23, antiderivative size = 161 \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx=\frac {a^2 (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {a^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {2 a^2 \cos ^2(c+d x)^{\frac {2+m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {2+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{1+m}}{d e (1+m)} \]

[Out]

a^2*(e*tan(d*x+c))^(1+m)/d/e/(1+m)+a^2*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)+2*a^2*(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] (verified)

Time = 0.20 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3971, 3557, 371, 2697, 2687, 32} \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx=\frac {a^2 (e \tan (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac {2 a^2 \sec (c+d x) \cos ^2(c+d x)^{\frac {m+2}{2}} (e \tan (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},\frac {m+2}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {a^2 (e \tan (c+d x))^{m+1}}{d e (m+1)} \]

[In]

Int[(a + a*Sec[c + d*x])^2*(e*Tan[c + d*x])^m,x]

[Out]

(a^2*(e*Tan[c + d*x])^(1 + m))/(d*e*(1 + m)) + (a^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2
]*(e*Tan[c + d*x])^(1 + m))/(d*e*(1 + m)) + (2*a^2*(Cos[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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2697

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)/(b*f*(n + 1)))*Hypergeometric2F1[(n + 1)/2,
(m + n + 1)/2, (n + 3)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rule 3557

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 3971

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a^2 (e \tan (c+d x))^m+2 a^2 \sec (c+d x) (e \tan (c+d x))^m+a^2 \sec ^2(c+d x) (e \tan (c+d x))^m\right ) \, dx \\ & = a^2 \int (e \tan (c+d x))^m \, dx+a^2 \int \sec ^2(c+d x) (e \tan (c+d x))^m \, dx+\left (2 a^2\right ) \int \sec (c+d x) (e \tan (c+d x))^m \, dx \\ & = \frac {2 a^2 \cos ^2(c+d x)^{\frac {2+m}{2}} \operatorname {Hypergeometric2F1}\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^2 \text {Subst}\left (\int (e x)^m \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (a^2 e\right ) \text {Subst}\left (\int \frac {x^m}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{d} \\ & = \frac {a^2 (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {a^2 \operatorname {Hypergeometric2F1}\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 {2 a^2 \cos ^2(c+d x)^{\frac {2+m}{2}} \operatorname {Hypergeometric2F1}\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] (verified)

Time = 1.23 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.06 \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx=\frac {a^2 e (e \tan (c+d x))^{-1+m} \left (-\tan ^2(c+d x)\right )^{-m/2} \left (\sqrt {-\tan ^2(c+d x)}+2 (1+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3}{2},\sec ^2(c+d x)\right ) \sec (c+d x) \sqrt {-\tan ^2(c+d x)}-\left (-\tan ^2(c+d x)\right )^{\frac {2+m}{2}}-\operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \left (-\tan ^2(c+d x)\right )^{\frac {2+m}{2}}\right )}{d (1+m)} \]

[In]

Integrate[(a + a*Sec[c + d*x])^2*(e*Tan[c + d*x])^m,x]

[Out]

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

Maple [F]

\[\int \left (a +a \sec \left (d x +c \right )\right )^{2} \left (e \tan \left (d x +c \right )\right )^{m}d x\]

[In]

int((a+a*sec(d*x+c))^2*(e*tan(d*x+c))^m,x)

[Out]

int((a+a*sec(d*x+c))^2*(e*tan(d*x+c))^m,x)

Fricas [F]

\[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \tan \left (d x + c\right )\right )^{m} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^2*(e*tan(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2)*(e*tan(d*x + c))^m, x)

Sympy [F]

\[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx=a^{2} \left (\int \left (e \tan {\left (c + d x \right )}\right )^{m}\, dx + \int 2 \left (e \tan {\left (c + d x \right )}\right )^{m} \sec {\left (c + d x \right )}\, dx + \int \left (e \tan {\left (c + d x \right )}\right )^{m} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate((a+a*sec(d*x+c))**2*(e*tan(d*x+c))**m,x)

[Out]

a**2*(Integral((e*tan(c + d*x))**m, x) + Integral(2*(e*tan(c + d*x))**m*sec(c + d*x), x) + Integral((e*tan(c +
d*x))**m*sec(c + d*x)**2, x))

Maxima [F]

\[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \tan \left (d x + c\right )\right )^{m} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^2*(e*tan(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^2*(e*tan(d*x + c))^m, x)

Giac [F]

\[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \tan \left (d x + c\right )\right )^{m} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^2*(e*tan(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^2*(e*tan(d*x + c))^m, x)

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx=\int {\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2 \,d x \]

[In]

int((e*tan(c + d*x))^m*(a + a/cos(c + d*x))^2,x)

[Out]

int((e*tan(c + d*x))^m*(a + a/cos(c + d*x))^2, x)