\(\int (a+a \sec (c+d x))^3 (e \tan (c+d x))^m \, dx\) [209]

Optimal result

Integrand size = 23, antiderivative size = 243 \[ \int (a+a \sec (c+d x))^3 (e \tan (c+d x))^m \, dx=\frac {3 a^3 (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {a^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {3 a^3 \cos ^2(c+d x)^{\frac {2+m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {2+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{1+m}}{d e (1+m)}+\frac {a^3 \cos ^2(c+d x)^{\frac {4+m}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {4+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{1+m}}{d e (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 1.61 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.93 \[ \int (a+a \sec (c+d x))^3 (e \tan (c+d x))^m \, dx=\frac {a^3 e (e \tan (c+d x))^{-1+m} \left (-\tan ^2(c+d x)\right )^{-m/2} \left (9 \sqrt {-\tan ^2(c+d x)}+9 (1+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3}{2},\sec ^2(c+d x)\right ) \sec (c+d x) \sqrt {-\tan ^2(c+d x)}+(1+m) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1-m}{2},\frac {5}{2},\sec ^2(c+d x)\right ) \sec ^3(c+d x) \sqrt {-\tan ^2(c+d x)}-9 \left (-\tan ^2(c+d x)\right )^{\frac {2+m}{2}}-3 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(c+d x)\right ) \left (-\tan ^2(c+d x)\right )^{\frac {2+m}{2}}\right )}{3 d (1+m)} \] Input:

Integrate[(a + a*Sec[c + d*x])^3*(e*Tan[c + d*x])^m,x]
 

Output:

(a^3*e*(e*Tan[c + d*x])^(-1 + m)*(9*Sqrt[-Tan[c + d*x]^2] + 9*(1 + m)*Hype 
rgeometric2F1[1/2, (1 - m)/2, 3/2, Sec[c + d*x]^2]*Sec[c + d*x]*Sqrt[-Tan[ 
c + d*x]^2] + (1 + m)*Hypergeometric2F1[3/2, (1 - m)/2, 5/2, Sec[c + d*x]^ 
2]*Sec[c + d*x]^3*Sqrt[-Tan[c + d*x]^2] - 9*(-Tan[c + d*x]^2)^((2 + m)/2) 
- 3*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*(-Tan[c + 
d*x]^2)^((2 + m)/2)))/(3*d*(1 + m)*(-Tan[c + d*x]^2)^(m/2))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 243, 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.

\(\displaystyle \int (a \sec (c+d x)+a)^3 (e \tan (c+d x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^mdx\)

\(\Big \downarrow \) 4374

\(\displaystyle \int \left (a^3 (e \tan (c+d x))^m+a^3 \sec ^3(c+d x) (e \tan (c+d x))^m+3 a^3 \sec ^2(c+d x) (e \tan (c+d x))^m+3 a^3 \sec (c+d x) (e \tan (c+d x))^m\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 (e \tan (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac {a^3 \sec ^3(c+d x) \cos ^2(c+d x)^{\frac {m+4}{2}} (e \tan (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},\frac {m+4}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {3 a^3 \sec (c+d x) \cos ^2(c+d x)^{\frac {m+2}{2}} (e \tan (c+d x))^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},\frac {m+2}{2},\frac {m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {3 a^3 (e \tan (c+d x))^{m+1}}{d e (m+1)}\)

Input:

Int[(a + a*Sec[c + d*x])^3*(e*Tan[c + d*x])^m,x]
 

Output:

(3*a^3*(e*Tan[c + d*x])^(1 + m))/(d*e*(1 + m)) + (a^3*Hypergeometric2F1[1, 
 (1 + m)/2, (3 + m)/2, -Tan[c + d*x]^2]*(e*Tan[c + d*x])^(1 + m))/(d*e*(1 
+ m)) + (3*a^3*(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)) + (a^3*(Cos[c + d*x]^2)^((4 + m)/2)*Hypergeometric2F1[(1 + 
 m)/2, (4 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]^3*(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]

\[\int \left (a +a \sec \left (d x +c \right )\right )^{3} \left (e \tan \left (d x +c \right )\right )^{m}d x\]

Input:

int((a+a*sec(d*x+c))^3*(e*tan(d*x+c))^m,x)
 

Output:

int((a+a*sec(d*x+c))^3*(e*tan(d*x+c))^m,x)
 

Fricas [F]

\[ \int (a+a \sec (c+d x))^3 (e \tan (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \left (e \tan \left (d x + c\right )\right )^{m} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^3*(e*tan(d*x+c))^m,x, algorithm="fricas")
 

Output:

integral((a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x + c)^2 + 3*a^3*sec(d*x + c) + 
 a^3)*(e*tan(d*x + c))^m, x)
 

Sympy [F]

\[ \int (a+a \sec (c+d x))^3 (e \tan (c+d x))^m \, dx=a^{3} \left (\int \left (e \tan {\left (c + d x \right )}\right )^{m}\, dx + \int 3 \left (e \tan {\left (c + d x \right )}\right )^{m} \sec {\left (c + d x \right )}\, dx + \int 3 \left (e \tan {\left (c + d x \right )}\right )^{m} \sec ^{2}{\left (c + d x \right )}\, dx + \int \left (e \tan {\left (c + d x \right )}\right )^{m} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \] Input:

integrate((a+a*sec(d*x+c))**3*(e*tan(d*x+c))**m,x)
 

Output:

a**3*(Integral((e*tan(c + d*x))**m, x) + Integral(3*(e*tan(c + d*x))**m*se 
c(c + d*x), x) + Integral(3*(e*tan(c + d*x))**m*sec(c + d*x)**2, x) + Inte 
gral((e*tan(c + d*x))**m*sec(c + d*x)**3, x))
 

Maxima [F]

\[ \int (a+a \sec (c+d x))^3 (e \tan (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \left (e \tan \left (d x + c\right )\right )^{m} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^3*(e*tan(d*x+c))^m,x, algorithm="maxima")
 

Output:

integrate((a*sec(d*x + c) + a)^3*(e*tan(d*x + c))^m, x)
 

Giac [F]

\[ \int (a+a \sec (c+d x))^3 (e \tan (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{3} \left (e \tan \left (d x + c\right )\right )^{m} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^3*(e*tan(d*x+c))^m,x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)^3*(e*tan(d*x + c))^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^3 (e \tan (c+d x))^m \, dx=\int {\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3 \,d x \] Input:

int((e*tan(c + d*x))^m*(a + a/cos(c + d*x))^3,x)
 

Output:

int((e*tan(c + d*x))^m*(a + a/cos(c + d*x))^3, x)
 

Reduce [F]

\[ \int (a+a \sec (c+d x))^3 (e \tan (c+d x))^m \, dx=e^{m} a^{3} \left (\int \tan \left (d x +c \right )^{m}d x +\int \tan \left (d x +c \right )^{m} \sec \left (d x +c \right )^{3}d x +3 \left (\int \tan \left (d x +c \right )^{m} \sec \left (d x +c \right )^{2}d x \right )+3 \left (\int \tan \left (d x +c \right )^{m} \sec \left (d x +c \right )d x \right )\right ) \] Input:

int((a+a*sec(d*x+c))^3*(e*tan(d*x+c))^m,x)
 

Output:

e**m*a**3*(int(tan(c + d*x)**m,x) + int(tan(c + d*x)**m*sec(c + d*x)**3,x) 
 + 3*int(tan(c + d*x)**m*sec(c + d*x)**2,x) + 3*int(tan(c + d*x)**m*sec(c 
+ d*x),x))