Integrand size = 23, antiderivative size = 93 \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^3} \, dx=-\frac {2^{\frac {1}{2} (-5+p)} (e \cos (c+d x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {7-p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (-1-p)}}{a^3 d e (1+p)} \] Output:
-2^(-5/2+1/2*p)*(e*cos(d*x+c))^(p+1)*hypergeom([1/2*p+1/2, 7/2-1/2*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^3/d/e/(p+1)
Time = 0.17 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.01 \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^3} \, dx=-\frac {2^{\frac {1}{2} (-5+p)} \cos (c+d x) (e \cos (c+d x))^p \operatorname {Hypergeometric2F1}\left (\frac {7-p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (-1-p)}}{a^3 d (1+p)} \] Input:
Integrate[(e*Cos[c + d*x])^p/(a + a*Sin[c + d*x])^3,x]
Output:
-((2^((-5 + p)/2)*Cos[c + d*x]*(e*Cos[c + d*x])^p*Hypergeometric2F1[(7 - p )/2, (1 + p)/2, (3 + p)/2, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^((-1 - p)/2))/(a^3*d*(1 + p)))
Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.30, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3167, 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.
\(\displaystyle \int \frac {(e \cos (c+d x))^p}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(e \cos (c+d x))^p}{(a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3167 |
\(\displaystyle \frac {(1-\sin (c+d x))^{\frac {1}{2} (-p-1)} (\sin (c+d x)+1)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \int (1-\sin (c+d x))^{\frac {p-1}{2}} (\sin (c+d x)+1)^{\frac {p-7}{2}}d\sin (c+d x)}{a^3 d e}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {2^{\frac {p-5}{2}} (1-\sin (c+d x))^{\frac {1}{2} (-p-1)+\frac {p+1}{2}} (\sin (c+d x)+1)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {7-p}{2},\frac {p+1}{2},\frac {p+3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{a^3 d e (p+1)}\) |
Input:
Int[(e*Cos[c + d*x])^p/(a + a*Sin[c + d*x])^3,x]
Output:
-((2^((-5 + p)/2)*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[(7 - p)/2, (1 + p)/2, (3 + p)/2, (1 - Sin[c + d*x])/2]*(1 - Sin[c + d*x])^((-1 - p)/2 + (1 + p)/2)*(1 + Sin[c + d*x])^((-1 - p)/2))/(a^3*d*e*(1 + p)))
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]))
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.), x_Symbol] :> Simp[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]
\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +a \sin \left (d x +c \right )\right )^{3}}d x\]
Input:
int((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^3,x)
Output:
int((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^3,x)
\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:
integrate((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^3,x, algorithm="fricas")
Output:
integral(-(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)
\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\left (e \cos {\left (c + d x \right )}\right )^{p}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate((e*cos(d*x+c))**p/(a+a*sin(d*x+c))**3,x)
Output:
Integral((e*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
\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:
integrate((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^3,x, algorithm="maxima")
Output:
integrate((e*cos(d*x + c))^p/(a*sin(d*x + c) + a)^3, x)
\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:
integrate((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
integrate((e*cos(d*x + c))^p/(a*sin(d*x + c) + a)^3, x)
Timed out. \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^3} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \] Input:
int((e*cos(c + d*x))^p/(a + a*sin(c + d*x))^3,x)
Output:
int((e*cos(c + d*x))^p/(a + a*sin(c + d*x))^3, x)
\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^3} \, dx=\frac {e^{p} \left (\int \frac {\cos \left (d x +c \right )^{p}}{\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:
int((e*cos(d*x+c))^p/(a+a*sin(d*x+c))^3,x)
Output:
(e**p*int(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