3.2.34 \(\int (a+a \sec (c+d x))^3 (e \sin (c+d x))^m \, dx\) [134]

3.2.34.1 Optimal result

Integrand size = 23, antiderivative size = 247 \[ \int (a+a \sec (c+d x))^3 (e \sin (c+d x))^m \, dx=\frac {a^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac {a^3 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac {3 a^3 \sqrt {\cos ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{1+m}}{d e (1+m)} \]

output

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

3.2.34.2 Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.65 \[ \int (a+a \sec (c+d x))^3 (e \sin (c+d x))^m \, dx=\frac {a^3 (e \sin (c+d x))^m \left (3 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sin (c+d x)+\operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right ) \sin (c+d x)+\sqrt {\cos ^2(c+d x)} \left (\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right )+3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(c+d x)\right )\right ) \tan (c+d x)\right )}{d (1+m)} \]

input

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

output

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

3.2.34.3 Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4360, 25, 25, 3042, 3352, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

\(\displaystyle \int \sec ^3(c+d x) \left (-(a (-\cos (c+d x))-a)^3\right ) (e \sin (c+d x))^mdx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int -(\cos (c+d x) a+a)^3 \sec ^3(c+d x) (e \sin (c+d x))^mdx\)

\(\Big \downarrow \) 25

\(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^3 (e \sin (c+d x))^mdx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^m}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\)

\(\Big \downarrow \) 3352

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

\(\Big \downarrow \) 2009

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

input

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

output

(a^3*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x 
]^2]*(e*Sin[c + d*x])^(1 + m))/(d*e*(1 + m)*Sqrt[Cos[c + d*x]^2]) + (3*a^3 
*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x 
])^(1 + m))/(d*e*(1 + m)) + (a^3*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2 
, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(1 + m))/(d*e*(1 + m)) + (3*a^3*Sqrt[Co 
s[c + d*x]^2]*Hypergeometric2F1[3/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2] 
*Sec[c + d*x]*(e*Sin[c + d*x])^(1 + m))/(d*e*(1 + m))
 

3.2.34.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3352

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig 
[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F 
reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
 

rule 4360

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 

3.2.34.4 Maple [F]

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

input

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

output

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

3.2.34.5 Fricas [F]

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

input

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

3.2.34.6 Sympy [F]

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

input

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

output

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

3.2.34.7 Maxima [F]

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

input

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

output

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

3.2.34.8 Giac [F]

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

input

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

output

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

3.2.34.9 Mupad [F(-1)]

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

input

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

output

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