\(\int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx\) [338]

Optimal result

Integrand size = 25, antiderivative size = 104 \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2^{-2+\frac {p}{2}} (e \cos (c+d x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {6-p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {4-p}{2}}}{d e (1+p) (a+a \sin (c+d x))^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98 \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2^{-2+\frac {p}{2}} \cos (c+d x) (e \cos (c+d x))^p \operatorname {Hypergeometric2F1}\left (3-\frac {p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-p/2}}{a^2 d (1+p) \sqrt {a (1+\sin (c+d x))}} \] Input:

Integrate[(e*Cos[c + d*x])^p/(a + a*Sin[c + d*x])^(5/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.38, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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.

\(\displaystyle \int \frac {(e \cos (c+d x))^p}{(a \sin (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(e \cos (c+d x))^p}{(a \sin (c+d x)+a)^{5/2}}dx\)

\(\Big \downarrow \) 3168

\(\displaystyle \frac {a^2 (a-a \sin (c+d x))^{\frac {1}{2} (-p-1)} (a \sin (c+d x)+a)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \int (a-a \sin (c+d x))^{\frac {p-1}{2}} (\sin (c+d x) a+a)^{\frac {p-6}{2}}d\sin (c+d x)}{d e}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {2^{\frac {p}{2}-3} (\sin (c+d x)+1)^{-p/2} (a-a \sin (c+d x))^{\frac {1}{2} (-p-1)} (a \sin (c+d x)+a)^{\frac {1}{2} (-p-1)+\frac {p}{2}} (e \cos (c+d x))^{p+1} \int \left (\frac {1}{2} \sin (c+d x)+\frac {1}{2}\right )^{\frac {p-6}{2}} (a-a \sin (c+d x))^{\frac {p-1}{2}}d\sin (c+d x)}{a d e}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {2^{\frac {p}{2}-2} (\sin (c+d x)+1)^{-p/2} (a-a \sin (c+d x))^{\frac {1}{2} (-p-1)+\frac {p+1}{2}} (a \sin (c+d x)+a)^{\frac {1}{2} (-p-1)+\frac {p}{2}} (e \cos (c+d x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {6-p}{2},\frac {p+1}{2},\frac {p+3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{a^2 d e (p+1)}\)

Input:

Int[(e*Cos[c + d*x])^p/(a + a*Sin[c + d*x])^(5/2),x]
 

Output:

-((2^(-2 + p/2)*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(6 - p)/2, (1 + 
 p)/2, (3 + p)/2, (1 - Sin[c + d*x])/2]*(a - a*Sin[c + d*x])^((-1 - p)/2 + 
 (1 + p)/2)*(a + a*Sin[c + d*x])^((-1 - p)/2 + p/2))/(a^2*d*e*(1 + p)*(1 + 
 Sin[c + d*x])^(p/2)))
 

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]

\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}d x\]

Input:

int((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^(5/2),x)
 

Output:

int((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^(5/2),x)
 

Fricas [F]

\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

integral(-sqrt(a*sin(d*x + c) + a)*(e*cos(d*x + c))^p/(3*a^3*cos(d*x + c)^ 
2 - 4*a^3 + (a^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x + c)), x)
 

Sympy [F]

\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {\left (e \cos {\left (c + d x \right )}\right )^{p}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((e*cos(d*x+c))**p/(a+a*sin(d*x+c))**(5/2),x)
 

Output:

Integral((e*cos(c + d*x))**p/(a*(sin(c + d*x) + 1))**(5/2), x)
 

Maxima [F]

\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((e*cos(d*x + c))^p/(a*sin(d*x + c) + a)^(5/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:

int((e*cos(c + d*x))^p/(a + a*sin(c + d*x))^(5/2),x)
 

Output:

int((e*cos(c + d*x))^p/(a + a*sin(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {e^{p} \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{p}}{\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:

int((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^(5/2),x)
 

Output:

(e**p*sqrt(a)*int((sqrt(sin(c + d*x) + 1)*cos(c + d*x)**p)/(sin(c + d*x)** 
3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1),x))/a**3