Integrand size = 17, antiderivative size = 235 \[ \int \sin (a+b x) \sqrt {\sin (c+d x)} \, dx=-\frac {e^{\frac {1}{2} i (2 a-c)+\frac {1}{2} i (2 b-d) x+\frac {1}{2} i (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (-1+\frac {2 b}{d}\right ),\frac {1}{4} \left (3+\frac {2 b}{d}\right ),e^{2 i (c+d x)}\right ) \sqrt {\sin (c+d x)}}{(2 b-d) \sqrt {1-e^{2 i (c+d x)}}}-\frac {e^{-\frac {1}{2} i (2 a+c)-\frac {1}{2} i (2 b+d) x+\frac {1}{2} i (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {2 b+d}{4 d},\frac {1}{4} \left (3-\frac {2 b}{d}\right ),e^{2 i (c+d x)}\right ) \sqrt {\sin (c+d x)}}{(2 b+d) \sqrt {1-e^{2 i (c+d x)}}} \] Output:
-exp(1/2*I*(2*a-c)+1/2*I*(2*b-d)*x+1/2*I*(d*x+c))*hypergeom([-1/2, -1/4+1/ 2*b/d],[3/4+1/2*b/d],exp(2*I*(d*x+c)))*sin(d*x+c)^(1/2)/(2*b-d)/(1-exp(2*I *(d*x+c)))^(1/2)-exp(-1/2*I*(2*a+c)-1/2*I*(2*b+d)*x+1/2*I*(d*x+c))*hyperge om([-1/2, -1/4*(2*b+d)/d],[3/4-1/2*b/d],exp(2*I*(d*x+c)))*sin(d*x+c)^(1/2) /(2*b+d)/(1-exp(2*I*(d*x+c)))^(1/2)
Time = 10.36 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.83 \[ \int \sin (a+b x) \sqrt {\sin (c+d x)} \, dx=\frac {i e^{-\frac {1}{2} i (2 a-2 c+2 b x-d x)} \left (-i e^{-i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )\right )^{3/2} \left ((2 b-d) e^{\frac {i d x}{2}} \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4}-\frac {b}{2 d},\frac {3}{4}-\frac {b}{2 d},e^{2 i (c+d x)}\right )+(2 b+d) e^{\frac {1}{2} i (4 a+(4 b+d) x)} \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4}+\frac {b}{2 d},\frac {3}{4}+\frac {b}{2 d},e^{2 i (c+d x)}\right )\right )}{\sqrt {2} \left (4 b^2-d^2\right )} \] Input:
Integrate[Sin[a + b*x]*Sqrt[Sin[c + d*x]],x]
Output:
(I*(((-I)*(-1 + E^((2*I)*(c + d*x))))/E^(I*(c + d*x)))^(3/2)*((2*b - d)*E^ ((I/2)*d*x)*Hypergeometric2F1[1, 5/4 - b/(2*d), 3/4 - b/(2*d), E^((2*I)*(c + d*x))] + (2*b + d)*E^((I/2)*(4*a + (4*b + d)*x))*Hypergeometric2F1[1, 5 /4 + b/(2*d), 3/4 + b/(2*d), E^((2*I)*(c + d*x))]))/(Sqrt[2]*(4*b^2 - d^2) *E^((I/2)*(2*a - 2*c + 2*b*x - d*x)))
Time = 0.94 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.28, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {5064, 2009}
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.
\(\displaystyle \int \sin (a+b x) \sqrt {\sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 5064 |
\(\displaystyle \frac {\int \left (i e^{-i a-i b x} \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}}-i e^{i a+i b x} \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}}\right )dx}{2 \sqrt {2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2 \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (\frac {2 b}{d}-1\right ),\frac {1}{4} \left (\frac {2 b}{d}+3\right ),e^{2 i (c+d x)}\right ) \exp \left (\frac {1}{2} i (2 a-c)+\frac {1}{2} i x (2 b-d)+\frac {1}{2} i (c+d x)\right )}{(2 b-d) \sqrt {1-e^{2 i c+2 i d x}}}-\frac {2 \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {2 b+d}{4 d},\frac {1}{4} \left (3-\frac {2 b}{d}\right ),e^{2 i (c+d x)}\right ) \exp \left (-\frac {1}{2} i (2 a+c)-\frac {1}{2} i x (2 b+d)+\frac {1}{2} i (c+d x)\right )}{(2 b+d) \sqrt {1-e^{2 i c+2 i d x}}}}{2 \sqrt {2}}\) |
Input:
Int[Sin[a + b*x]*Sqrt[Sin[c + d*x]],x]
Output:
((-2*E^((I/2)*(2*a - c) + (I/2)*(2*b - d)*x + (I/2)*(c + d*x))*Sqrt[I/E^(I *(c + d*x)) - I*E^(I*(c + d*x))]*Hypergeometric2F1[-1/2, (-1 + (2*b)/d)/4, (3 + (2*b)/d)/4, E^((2*I)*(c + d*x))])/((2*b - d)*Sqrt[1 - E^((2*I)*c + ( 2*I)*d*x)]) - (2*E^((-1/2*I)*(2*a + c) - (I/2)*(2*b + d)*x + (I/2)*(c + d* x))*Sqrt[I/E^(I*(c + d*x)) - I*E^(I*(c + d*x))]*Hypergeometric2F1[-1/2, -1 /4*(2*b + d)/d, (3 - (2*b)/d)/4, E^((2*I)*(c + d*x))])/((2*b + d)*Sqrt[1 - E^((2*I)*c + (2*I)*d*x)]))/(2*Sqrt[2])
Int[Sin[(a_.) + (b_.)*(x_)]^(p_.)*Sin[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] :> Simp[1/2^(p + q) Int[ExpandIntegrand[(I/E^(I*(c + d*x)) - I*E^(I*(c + d*x)))^q, (I/E^(I*(a + b*x)) - I*E^(I*(a + b*x)))^p, x], x], x] /; FreeQ[{a , b, c, d, q}, x] && IGtQ[p, 0] && !IntegerQ[q]
\[\int \sin \left (b x +a \right ) \sqrt {\sin \left (d x +c \right )}d x\]
Input:
int(sin(b*x+a)*sin(d*x+c)^(1/2),x)
Output:
int(sin(b*x+a)*sin(d*x+c)^(1/2),x)
Exception generated. \[ \int \sin (a+b x) \sqrt {\sin (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sin(b*x+a)*sin(d*x+c)^(1/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
\[ \int \sin (a+b x) \sqrt {\sin (c+d x)} \, dx=\int \sin {\left (a + b x \right )} \sqrt {\sin {\left (c + d x \right )}}\, dx \] Input:
integrate(sin(b*x+a)*sin(d*x+c)**(1/2),x)
Output:
Integral(sin(a + b*x)*sqrt(sin(c + d*x)), x)
\[ \int \sin (a+b x) \sqrt {\sin (c+d x)} \, dx=\int { \sin \left (b x + a\right ) \sqrt {\sin \left (d x + c\right )} \,d x } \] Input:
integrate(sin(b*x+a)*sin(d*x+c)^(1/2),x, algorithm="maxima")
Output:
integrate(sin(b*x + a)*sqrt(sin(d*x + c)), x)
\[ \int \sin (a+b x) \sqrt {\sin (c+d x)} \, dx=\int { \sin \left (b x + a\right ) \sqrt {\sin \left (d x + c\right )} \,d x } \] Input:
integrate(sin(b*x+a)*sin(d*x+c)^(1/2),x, algorithm="giac")
Output:
integrate(sin(b*x + a)*sqrt(sin(d*x + c)), x)
Timed out. \[ \int \sin (a+b x) \sqrt {\sin (c+d x)} \, dx=\int \sin \left (a+b\,x\right )\,\sqrt {\sin \left (c+d\,x\right )} \,d x \] Input:
int(sin(a + b*x)*sin(c + d*x)^(1/2),x)
Output:
int(sin(a + b*x)*sin(c + d*x)^(1/2), x)
\[ \int \sin (a+b x) \sqrt {\sin (c+d x)} \, dx=\int \sqrt {\sin \left (d x +c \right )}\, \sin \left (b x +a \right )d x \] Input:
int(sin(b*x+a)*sin(d*x+c)^(1/2),x)
Output:
int(sqrt(sin(c + d*x))*sin(a + b*x),x)