\(\int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx\) [350]

Optimal result

Integrand size = 21, antiderivative size = 78 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {2^{-\frac {3}{2}+m} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^3(c+d x) (1+\sin (c+d x))^{\frac {3}{2}-m} (a+a \sin (c+d x))^m}{3 d} \] Output:

1/3*2^(-3/2+m)*hypergeom([-3/2, 5/2-m],[-1/2],1/2-1/2*sin(d*x+c))*sec(d*x+ 
c)^3*(1+sin(d*x+c))^(3/2-m)*(a+a*sin(d*x+c))^m/d
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00 \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {2^{-\frac {3}{2}+m} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^3(c+d x) (1+\sin (c+d x))^{\frac {3}{2}-m} (a (1+\sin (c+d x)))^m}{3 d} \] Input:

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^m,x]
 

Output:

(2^(-3/2 + m)*Hypergeometric2F1[-3/2, 5/2 - m, -1/2, (1 - Sin[c + d*x])/2] 
*Sec[c + d*x]^3*(1 + Sin[c + d*x])^(3/2 - m)*(a*(1 + Sin[c + d*x]))^m)/(3* 
d)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3168, 80, 79}

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.

\(\displaystyle \int \sec ^4(c+d x) (a \sin (c+d x)+a)^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^m}{\cos (c+d x)^4}dx\)

\(\Big \downarrow \) 3168

\(\displaystyle \frac {a^2 \sec ^3(c+d x) (a-a \sin (c+d x))^{3/2} (a \sin (c+d x)+a)^{3/2} \int \frac {(\sin (c+d x) a+a)^{m-\frac {5}{2}}}{(a-a \sin (c+d x))^{5/2}}d\sin (c+d x)}{d}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {2^{m-\frac {5}{2}} \sec ^3(c+d x) (a-a \sin (c+d x))^{3/2} (\sin (c+d x)+1)^{\frac {1}{2}-m} (a \sin (c+d x)+a)^{m+1} \int \frac {\left (\frac {1}{2} \sin (c+d x)+\frac {1}{2}\right )^{m-\frac {5}{2}}}{(a-a \sin (c+d x))^{5/2}}d\sin (c+d x)}{d}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {2^{m-\frac {3}{2}} \sec ^3(c+d x) (\sin (c+d x)+1)^{\frac {1}{2}-m} (a \sin (c+d x)+a)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2}-m,-\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{3 a d}\)

Input:

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^m,x]
 

Output:

(2^(-3/2 + m)*Hypergeometric2F1[-3/2, 5/2 - m, -1/2, (1 - Sin[c + d*x])/2] 
*Sec[c + d*x]^3*(1 + Sin[c + d*x])^(1/2 - m)*(a + a*Sin[c + d*x])^(1 + m)) 
/(3*a*d)
 

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 3042

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

rule 3168

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.), x_Symbol] :> Simp[a^2*((g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin 
[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)))   Subst[Int[(a + 
b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; Fre 
eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]
 

Maple [F]

\[\int \sec \left (d x +c \right )^{4} \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]

Input:

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x)
 

Output:

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x)
 

Fricas [F]

\[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="fricas")
 

Output:

integral((a*sin(d*x + c) + a)^m*sec(d*x + c)^4, x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**4*(a+a*sin(d*x+c))**m,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="maxima")
 

Output:

integrate((a*sin(d*x + c) + a)^m*sec(d*x + c)^4, x)
 

Giac [F]

\[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x, algorithm="giac")
 

Output:

integrate((a*sin(d*x + c) + a)^m*sec(d*x + c)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m}{{\cos \left (c+d\,x\right )}^4} \,d x \] Input:

int((a + a*sin(c + d*x))^m/cos(c + d*x)^4,x)
 

Output:

int((a + a*sin(c + d*x))^m/cos(c + d*x)^4, x)
 

Reduce [F]

\[ \int \sec ^4(c+d x) (a+a \sin (c+d x))^m \, dx=\int \left (\sin \left (d x +c \right ) a +a \right )^{m} \sec \left (d x +c \right )^{4}d x \] Input:

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^m,x)
 

Output:

int((sin(c + d*x)*a + a)**m*sec(c + d*x)**4,x)