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

Optimal result

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

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)
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.06 \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx=\frac {a^2 e (e \tan (c+d x))^{-1+m} \left (-\tan ^2(c+d x)\right )^{-m/2} \left (\sqrt {-\tan ^2(c+d x)}+2 (1+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3}{2},\sec ^2(c+d x)\right ) \sec (c+d x) \sqrt {-\tan ^2(c+d x)}-\left (-\tan ^2(c+d x)\right )^{\frac {2+m}{2}}-\operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \left (-\tan ^2(c+d x)\right )^{\frac {2+m}{2}}\right )}{d (1+m)} \] Input:

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

Output:

(a^2*e*(e*Tan[c + d*x])^(-1 + m)*(Sqrt[-Tan[c + d*x]^2] + 2*(1 + m)*Hyperg 
eometric2F1[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))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4374, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4374

\(\displaystyle \int \left (a^2 (e \tan (c+d x))^m+a^2 \sec ^2(c+d x) (e \tan (c+d x))^m+2 a^2 \sec (c+d x) (e \tan (c+d x))^m\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(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))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4374

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Int[ExpandIntegrand[(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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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