Integrand size = 25, antiderivative size = 121 \[ \int \frac {(a+b \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {2 \sqrt [4]{2} e \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},-m,\frac {5}{4},\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) (1-\sin (c+d x)) (1+\sin (c+d x))^{3/4} (a+b \sin (c+d x))^m \left (\frac {a+b \sin (c+d x)}{a+b}\right )^{-m}}{d (e \cos (c+d x))^{3/2}} \] Output:
-2*2^(1/4)*e*AppellF1(1/4,-m,3/4,5/4,b*(1-sin(d*x+c))/(a+b),1/2-1/2*sin(d* x+c))*(1-sin(d*x+c))*(1+sin(d*x+c))^(3/4)*(a+b*sin(d*x+c))^m/d/(e*cos(d*x+ c))^(3/2)/(((a+b*sin(d*x+c))/(a+b))^m)
\[ \int \frac {(a+b \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}} \, dx=\int \frac {(a+b \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}} \, dx \] Input:
Integrate[(a + b*Sin[c + d*x])^m/Sqrt[e*Cos[c + d*x]],x]
Output:
Integrate[(a + b*Sin[c + d*x])^m/Sqrt[e*Cos[c + d*x]], x]
Time = 0.31 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.11, 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.
\(\displaystyle \int \frac {(a+b \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}}dx\) |
\(\Big \downarrow \) 3183 |
\(\displaystyle \frac {e \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{3/4} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{3/4} \int \frac {(a+b \sin (c+d x))^m}{\left (-\frac {\sin (c+d x) b}{a-b}-\frac {b}{a-b}\right )^{3/4} \left (\frac {b}{a+b}-\frac {b \sin (c+d x)}{a+b}\right )^{3/4}}d\sin (c+d x)}{d (e \cos (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {e \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{3/4} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{3/4} (a+b \sin (c+d x))^{m+1} \operatorname {AppellF1}\left (m+1,\frac {3}{4},\frac {3}{4},m+2,\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{b d (m+1) (e \cos (c+d x))^{3/2}}\) |
Input:
Int[(a + b*Sin[c + d*x])^m/Sqrt[e*Cos[c + d*x]],x]
Output:
(e*AppellF1[1 + m, 3/4, 3/4, 2 + m, (a + b*Sin[c + d*x])/(a - b), (a + b*S in[c + d*x])/(a + b)]*(a + b*Sin[c + d*x])^(1 + m)*(1 - (a + b*Sin[c + d*x ])/(a - b))^(3/4)*(1 - (a + b*Sin[c + d*x])/(a + b))^(3/4))/(b*d*(1 + m)*( e*Cos[c + d*x])^(3/2))
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]
\[\int \frac {\left (a +b \sin \left (d x +c \right )\right )^{m}}{\sqrt {e \cos \left (d x +c \right )}}d x\]
Input:
int((a+b*sin(d*x+c))^m/(e*cos(d*x+c))^(1/2),x)
Output:
int((a+b*sin(d*x+c))^m/(e*cos(d*x+c))^(1/2),x)
\[ \int \frac {(a+b \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \] Input:
integrate((a+b*sin(d*x+c))^m/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")
Output:
integral(sqrt(e*cos(d*x + c))*(b*sin(d*x + c) + a)^m/(e*cos(d*x + c)), x)
\[ \int \frac {(a+b \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}} \, dx=\int \frac {\left (a + b \sin {\left (c + d x \right )}\right )^{m}}{\sqrt {e \cos {\left (c + d x \right )}}}\, dx \] Input:
integrate((a+b*sin(d*x+c))**m/(e*cos(d*x+c))**(1/2),x)
Output:
Integral((a + b*sin(c + d*x))**m/sqrt(e*cos(c + d*x)), x)
\[ \int \frac {(a+b \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \] Input:
integrate((a+b*sin(d*x+c))^m/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")
Output:
integrate((b*sin(d*x + c) + a)^m/sqrt(e*cos(d*x + c)), x)
\[ \int \frac {(a+b \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{m}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \] Input:
integrate((a+b*sin(d*x+c))^m/(e*cos(d*x+c))^(1/2),x, algorithm="giac")
Output:
integrate((b*sin(d*x + c) + a)^m/sqrt(e*cos(d*x + c)), x)
Timed out. \[ \int \frac {(a+b \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}} \, dx=\int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m}{\sqrt {e\,\cos \left (c+d\,x\right )}} \,d x \] Input:
int((a + b*sin(c + d*x))^m/(e*cos(c + d*x))^(1/2),x)
Output:
int((a + b*sin(c + d*x))^m/(e*cos(c + d*x))^(1/2), x)
\[ \int \frac {(a+b \sin (c+d x))^m}{\sqrt {e \cos (c+d x)}} \, dx=\frac {\sqrt {e}\, \left (\int \frac {\left (\sin \left (d x +c \right ) b +a \right )^{m} \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right )}{e} \] Input:
int((a+b*sin(d*x+c))^m/(e*cos(d*x+c))^(1/2),x)
Output:
(sqrt(e)*int(((sin(c + d*x)*b + a)**m*sqrt(cos(c + d*x)))/cos(c + d*x),x)) /e