\(\int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx\) [331]

Optimal result

Integrand size = 23, antiderivative size = 95 \[ \int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx=-\frac {2^{-\frac {1}{2}+\frac {p}{2}} (e \cos (c+d x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {3-p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (-1-p)}}{a d e (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2767, 71} \[ \int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx=-\frac {2^{\frac {p}{2}-\frac {1}{2}} (\sin (c+d x)+1)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {3-p}{2},\frac {p+1}{2},\frac {p+3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{a d e (p+1)} \]

[In]

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

[Out]

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

Rule 71

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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 2767

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left ((e \cos (c+d x))^{1+p} (1-\sin (c+d x))^{\frac {1}{2} (-1-p)} (1+\sin (c+d x))^{\frac {1}{2} (-1-p)}\right ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-1+p)} (1+x)^{-1+\frac {1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{a d e} \\ & = -\frac {2^{-\frac {1}{2}+\frac {p}{2}} (e \cos (c+d x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {3-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 {1}{2} (-1-p)}}{a d e (1+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99 \[ \int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx=-\frac {2^{\frac {1}{2} (-1+p)} \cos (c+d x) (e \cos (c+d x))^p \operatorname {Hypergeometric2F1}\left (\frac {3-p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (-1-p)}}{a d (1+p)} \]

[In]

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

[Out]

-((2^((-1 + 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]*(1 + Sin[c + d*x])^((-1 - p)/2))/(a*d*(1 + p)))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((e*cos(d*x + c))^p/(a*sin(d*x + c) + a), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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