Integrand size = 25, antiderivative size = 123 \[ \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^m \, dx=-\frac {4\ 2^{3/4} e \operatorname {AppellF1}\left (\frac {7}{4},-\frac {3}{4},-m,\frac {11}{4},\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) (e \cos (c+d x))^{3/2} (1-\sin (c+d x)) (a+b \sin (c+d x))^m \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-m}}{7 d (1+\sin (c+d x))^{3/4}} \] Output:
-4/7*2^(3/4)*e*AppellF1(7/4,-m,-3/4,11/4,b*(1-sin(d*x+c))/(a+b),1/2-1/2*si n(d*x+c))*(e*cos(d*x+c))^(3/2)*(1-sin(d*x+c))*(a+b*sin(d*x+c))^m/d/(1+sin( d*x+c))^(3/4)/(((a+b*sin(d*x+c))/(a+b))^m)
\[ \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^m \, dx=\int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^m \, dx \] Input:
Integrate[(e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x])^m,x]
Output:
Integrate[(e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x])^m, x]
Time = 0.32 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3042, 3183, 155}
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.
\(\displaystyle \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^mdx\) |
\(\Big \downarrow \) 3183 |
\(\displaystyle \frac {e (e \cos (c+d x))^{3/2} \int (a+b \sin (c+d x))^m \left (-\frac {\sin (c+d x) b}{a-b}-\frac {b}{a-b}\right )^{3/4} \left (\frac {b}{a+b}-\frac {b \sin (c+d x)}{a+b}\right )^{3/4}d\sin (c+d x)}{d \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{3/4} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{3/4}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {e (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {3}{4},-\frac {3}{4},m+2,\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{b d (m+1) \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{3/4} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{3/4}}\) |
Input:
Int[(e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x])^m,x]
Output:
(e*AppellF1[1 + m, -3/4, -3/4, 2 + m, (a + b*Sin[c + d*x])/(a - b), (a + b *Sin[c + d*x])/(a + b)]*(e*Cos[c + d*x])^(3/2)*(a + b*Sin[c + d*x])^(1 + m ))/(b*d*(1 + m)*(1 - (a + b*Sin[c + d*x])/(a - b))^(3/4)*(1 - (a + b*Sin[c + d*x])/(a + b))^(3/4))
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ (b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(f*(1 - (a + b*Sin [e + f*x])/(a - b))^((p - 1)/2)*(1 - (a + b*Sin[e + f*x])/(a + b))^((p - 1) /2))) Subst[Int[(-b/(a - b) - b*(x/(a - b)))^((p - 1)/2)*(b/(a + b) - b*( x/(a + b)))^((p - 1)/2)*(a + b*x)^m, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] && !IGtQ[m, 0]
\[\int \left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}} \left (a +b \sin \left (d x +c \right )\right )^{m}d x\]
Input:
int((e*cos(d*x+c))^(5/2)*(a+b*sin(d*x+c))^m,x)
Output:
int((e*cos(d*x+c))^(5/2)*(a+b*sin(d*x+c))^m,x)
\[ \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \] Input:
integrate((e*cos(d*x+c))^(5/2)*(a+b*sin(d*x+c))^m,x, algorithm="fricas")
Output:
integral(sqrt(e*cos(d*x + c))*(b*sin(d*x + c) + a)^m*e^2*cos(d*x + c)^2, x )
Timed out. \[ \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^m \, dx=\text {Timed out} \] Input:
integrate((e*cos(d*x+c))**(5/2)*(a+b*sin(d*x+c))**m,x)
Output:
Timed out
\[ \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \] Input:
integrate((e*cos(d*x+c))^(5/2)*(a+b*sin(d*x+c))^m,x, algorithm="maxima")
Output:
integrate((e*cos(d*x + c))^(5/2)*(b*sin(d*x + c) + a)^m, x)
\[ \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \] Input:
integrate((e*cos(d*x+c))^(5/2)*(a+b*sin(d*x+c))^m,x, algorithm="giac")
Output:
integrate((e*cos(d*x + c))^(5/2)*(b*sin(d*x + c) + a)^m, x)
Timed out. \[ \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^m \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m \,d x \] Input:
int((e*cos(c + d*x))^(5/2)*(a + b*sin(c + d*x))^m,x)
Output:
int((e*cos(c + d*x))^(5/2)*(a + b*sin(c + d*x))^m, x)
\[ \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^m \, dx=\sqrt {e}\, \left (\int \left (\sin \left (d x +c \right ) b +a \right )^{m} \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) e^{2} \] Input:
int((e*cos(d*x+c))^(5/2)*(a+b*sin(d*x+c))^m,x)
Output:
sqrt(e)*int((sin(c + d*x)*b + a)**m*sqrt(cos(c + d*x))*cos(c + d*x)**2,x)* e**2