\(\int (a+a \sec (c+d x)) (e \tan (c+d x))^m \, dx\) [211]

Optimal result

Integrand size = 21, antiderivative size = 129 \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^m \, dx=\frac {a \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {a \cos ^2(c+d x)^{\frac {2+m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {2+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{1+m}}{d e (1+m)} \] Output:

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)
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81 \[ \int (a+a \sec (c+d x)) (e \tan (c+d x))^m \, dx=\frac {a (e \tan (c+d x))^m \left (\frac {\operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)}{1+m}+\csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3}{2},\sec ^2(c+d x)\right ) \left (-\tan ^2(c+d x)\right )^{\frac {1-m}{2}}\right )}{d} \] Input:

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

Output:

(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) + Csc[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
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4372, 3042, 3097, 3957, 278}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^mdx\)

\(\Big \downarrow \) 4372

\(\displaystyle a \int (e \tan (c+d x))^mdx+a \int \sec (c+d x) (e \tan (c+d x))^mdx\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int (e \tan (c+d x))^mdx+a \int \sec (c+d x) (e \tan (c+d x))^mdx\)

\(\Big \downarrow \) 3097

\(\displaystyle a \int (e \tan (c+d x))^mdx+\frac {a \sec (c+d x) \cos ^2(c+d x)^{\frac {m+2}{2}} (e \tan (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},\frac {m+2}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}\)

\(\Big \downarrow \) 3957

\(\displaystyle \frac {a e \int \frac {(e \tan (c+d x))^m}{\tan ^2(c+d x) e^2+e^2}d(e \tan (c+d x))}{d}+\frac {a \sec (c+d x) \cos ^2(c+d x)^{\frac {m+2}{2}} (e \tan (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},\frac {m+2}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {a (e \tan (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac {a \sec (c+d x) \cos ^2(c+d x)^{\frac {m+2}{2}} (e \tan (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},\frac {m+2}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}\)

Input:

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

Output:

(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*(Cos[c + d*x]^2)^((2 + m)/2)*Hypergeomet 
ric2F1[(1 + m)/2, (2 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*(e*Ta 
n[c + d*x])^(1 + m))/(d*e*(1 + m))
 

Defintions of rubi rules used

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3097

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] &&  !IntegerQ[m/2]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[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 4372

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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