\(\int \sqrt {\cos (c+d x)} \sin ^2(a+b x) \, dx\) [222]

Optimal result

Integrand size = 19, antiderivative size = 258 \[ \int \sqrt {\cos (c+d x)} \sin ^2(a+b x) \, dx=\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {i e^{-\frac {1}{2} i (4 a+c)-\frac {1}{2} i (4 b+d) x+\frac {1}{2} i (c+d x)} \sqrt {\cos (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (-1-\frac {4 b}{d}\right ),\frac {1}{4} \left (3-\frac {4 b}{d}\right ),-e^{2 i (c+d x)}\right )}{2 (4 b+d) \sqrt {1+e^{2 i (c+d x)}}}+\frac {i e^{\frac {1}{2} i (4 a-c)+\frac {1}{2} i (4 b-d) x+\frac {1}{2} i (c+d x)} \sqrt {\cos (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (-1+\frac {4 b}{d}\right ),\frac {1}{4} \left (3+\frac {4 b}{d}\right ),-e^{2 i (c+d x)}\right )}{2 (4 b-d) \sqrt {1+e^{2 i (c+d x)}}} \] Output:

EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d-1/2*I*exp(-1/2*I*(4*a+c)-1/2*I*(4* 
b+d)*x+1/2*I*(d*x+c))*cos(d*x+c)^(1/2)*hypergeom([-1/2, -1/4-b/d],[3/4-b/d 
],-exp(2*I*(d*x+c)))/(4*b+d)/(1+exp(2*I*(d*x+c)))^(1/2)+1/2*I*exp(1/2*I*(4 
*a-c)+1/2*I*(4*b-d)*x+1/2*I*(d*x+c))*cos(d*x+c)^(1/2)*hypergeom([-1/2, -1/ 
4+b/d],[3/4+b/d],-exp(2*I*(d*x+c)))/(4*b-d)/(1+exp(2*I*(d*x+c)))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 5.07 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.39 \[ \int \sqrt {\cos (c+d x)} \sin ^2(a+b x) \, dx=\frac {i e^{-i (2 a+c+(2 b+d) x)} \left (-\left ((4 b-d) e^{2 i (a+b x)} \sqrt {1+e^{2 i (c+d x)}} \left (4 b \left (-1+e^{2 i (c+d x)}\right )-d \left (1+e^{2 i (a+b x)}-e^{2 i (c+d x)}+e^{2 i (a+c+(b+d) x)}\right )\right )\right )-(4 b-d) d \left (1+e^{2 i (c+d x)}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4}-\frac {b}{d},\frac {3}{4}-\frac {b}{d},-e^{2 i (c+d x)}\right )+2 d^2 e^{4 i (a+b x)} \left (1+e^{2 i (c+d x)}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{4}+\frac {b}{d},\frac {3}{4}+\frac {b}{d},-e^{2 i (c+d x)}\right )\right )}{4 \left (16 b^2 d-d^3\right ) \sqrt {1+e^{2 i (c+d x)}} \sqrt {\cos (c+d x)}}-\frac {\cos (c) \csc (d x+\arctan (\tan (c))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}}{2 d \sqrt {\cos (c+d x)}} \] Input:

Integrate[Sqrt[Cos[c + d*x]]*Sin[a + b*x]^2,x]
 

Output:

((I/4)*(-((4*b - d)*E^((2*I)*(a + b*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*(4*b 
*(-1 + E^((2*I)*(c + d*x))) - d*(1 + E^((2*I)*(a + b*x)) - E^((2*I)*(c + d 
*x)) + E^((2*I)*(a + c + (b + d)*x))))) - (4*b - d)*d*(1 + E^((2*I)*(c + d 
*x)))*Hypergeometric2F1[-1/2, -1/4 - b/d, 3/4 - b/d, -E^((2*I)*(c + d*x))] 
 + 2*d^2*E^((4*I)*(a + b*x))*(1 + E^((2*I)*(c + d*x)))*Hypergeometric2F1[1 
/2, -1/4 + b/d, 3/4 + b/d, -E^((2*I)*(c + d*x))]))/((16*b^2*d - d^3)*E^(I* 
(2*a + c + (2*b + d)*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[Cos[c + d*x]]) 
 - (Cos[c]*Csc[d*x + ArcTan[Tan[c]]]*HypergeometricPFQ[{-1/2, -1/4}, {3/4} 
, Cos[d*x + ArcTan[Tan[c]]]^2]*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c] 
]]^2])/(2*d*Sqrt[Cos[c + d*x]])
 

Rubi [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(646\) vs. \(2(258)=516\).

Time = 1.14 (sec) , antiderivative size = 646, normalized size of antiderivative = 2.50, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {5066, 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.

\(\displaystyle \int \sin ^2(a+b x) \sqrt {\cos (c+d x)} \, dx\)

\(\Big \downarrow \) 5066

\(\displaystyle \frac {\int \left (-e^{-2 i a-2 i b x} \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}}-e^{2 i a+2 i b x} \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}}+2 \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}}\right )dx}{4 \sqrt {2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {2 i \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (-\frac {4 b}{d}-1\right ),\frac {1}{4} \left (3-\frac {4 b}{d}\right ),-e^{2 i (c+d x)}\right ) \exp \left (-\frac {1}{2} i (4 a+c)-\frac {1}{2} i x (4 b+d)+\frac {1}{2} i (c+d x)\right )}{(4 b+d) \sqrt {1+e^{2 i c+2 i d x}}}+\frac {2 i \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (\frac {4 b}{d}-1\right ),\frac {1}{4} \left (\frac {4 b}{d}+3\right ),-e^{2 i (c+d x)}\right ) \exp \left (\frac {1}{2} i (4 a-c)+\frac {1}{2} i x (4 b-d)+\frac {1}{2} i (c+d x)\right )}{(4 b-d) \sqrt {1+e^{2 i c+2 i d x}}}-\frac {4 i \sqrt {e^{i (c+d x)}} \left (1+e^{i (c+d x)}\right ) \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}} \sqrt {\frac {1+e^{2 i (c+d x)}}{\left (1+e^{i (c+d x)}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {e^{i (c+d x)}}\right ),\frac {1}{2}\right )}{d \left (1+e^{2 i (c+d x)}\right )}+\frac {8 i \sqrt {e^{i (c+d x)}} \left (1+e^{i (c+d x)}\right ) \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}} \sqrt {\frac {1+e^{2 i (c+d x)}}{\left (1+e^{i (c+d x)}\right )^2}} E\left (2 \arctan \left (\sqrt {e^{i (c+d x)}}\right )|\frac {1}{2}\right )}{d \left (1+e^{2 i (c+d x)}\right )}+\frac {4 i \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}}}{d}-\frac {8 i e^{i (c+d x)} \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}}}{d \left (1+e^{i (c+d x)}\right )}}{4 \sqrt {2}}\)

Input:

Int[Sqrt[Cos[c + d*x]]*Sin[a + b*x]^2,x]
 

Output:

(((4*I)*Sqrt[E^((-I)*(c + d*x)) + E^(I*(c + d*x))])/d - ((8*I)*E^(I*(c + d 
*x))*Sqrt[E^((-I)*(c + d*x)) + E^(I*(c + d*x))])/(d*(1 + E^(I*(c + d*x)))) 
 + ((8*I)*Sqrt[E^(I*(c + d*x))]*(1 + E^(I*(c + d*x)))*Sqrt[E^((-I)*(c + d* 
x)) + E^(I*(c + d*x))]*Sqrt[(1 + E^((2*I)*(c + d*x)))/(1 + E^(I*(c + d*x)) 
)^2]*EllipticE[2*ArcTan[Sqrt[E^(I*(c + d*x))]], 1/2])/(d*(1 + E^((2*I)*(c 
+ d*x)))) - ((4*I)*Sqrt[E^(I*(c + d*x))]*(1 + E^(I*(c + d*x)))*Sqrt[E^((-I 
)*(c + d*x)) + E^(I*(c + d*x))]*Sqrt[(1 + E^((2*I)*(c + d*x)))/(1 + E^(I*( 
c + d*x)))^2]*EllipticF[2*ArcTan[Sqrt[E^(I*(c + d*x))]], 1/2])/(d*(1 + E^( 
(2*I)*(c + d*x)))) - ((2*I)*E^((-1/2*I)*(4*a + c) - (I/2)*(4*b + d)*x + (I 
/2)*(c + d*x))*Sqrt[E^((-I)*(c + d*x)) + E^(I*(c + d*x))]*Hypergeometric2F 
1[-1/2, (-1 - (4*b)/d)/4, (3 - (4*b)/d)/4, -E^((2*I)*(c + d*x))])/((4*b + 
d)*Sqrt[1 + E^((2*I)*c + (2*I)*d*x)]) + ((2*I)*E^((I/2)*(4*a - c) + (I/2)* 
(4*b - d)*x + (I/2)*(c + d*x))*Sqrt[E^((-I)*(c + d*x)) + E^(I*(c + d*x))]* 
Hypergeometric2F1[-1/2, (-1 + (4*b)/d)/4, (3 + (4*b)/d)/4, -E^((2*I)*(c + 
d*x))])/((4*b - d)*Sqrt[1 + E^((2*I)*c + (2*I)*d*x)]))/(4*Sqrt[2])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5066

Int[Cos[(c_.) + (d_.)*(x_)]^(q_.)*Sin[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] 
:> Simp[1/2^(p + q)   Int[ExpandIntegrand[(E^((-I)*(c + d*x)) + 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]
 

Maple [F]

\[\int \sqrt {\cos \left (d x +c \right )}\, \sin \left (b x +a \right )^{2}d x\]

Input:

int(cos(d*x+c)^(1/2)*sin(b*x+a)^2,x)
 

Output:

int(cos(d*x+c)^(1/2)*sin(b*x+a)^2,x)
 

Fricas [F(-2)]

Exception generated. \[ \int \sqrt {\cos (c+d x)} \sin ^2(a+b x) \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(d*x+c)^(1/2)*sin(b*x+a)^2,x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [F]

\[ \int \sqrt {\cos (c+d x)} \sin ^2(a+b x) \, dx=\int \sin ^{2}{\left (a + b x \right )} \sqrt {\cos {\left (c + d x \right )}}\, dx \] Input:

integrate(cos(d*x+c)**(1/2)*sin(b*x+a)**2,x)
 

Output:

Integral(sin(a + b*x)**2*sqrt(cos(c + d*x)), x)
 

Maxima [F]

\[ \int \sqrt {\cos (c+d x)} \sin ^2(a+b x) \, dx=\int { \sqrt {\cos \left (d x + c\right )} \sin \left (b x + a\right )^{2} \,d x } \] Input:

integrate(cos(d*x+c)^(1/2)*sin(b*x+a)^2,x, algorithm="maxima")
 

Output:

integrate(sqrt(cos(d*x + c))*sin(b*x + a)^2, x)
 

Giac [F]

\[ \int \sqrt {\cos (c+d x)} \sin ^2(a+b x) \, dx=\int { \sqrt {\cos \left (d x + c\right )} \sin \left (b x + a\right )^{2} \,d x } \] Input:

integrate(cos(d*x+c)^(1/2)*sin(b*x+a)^2,x, algorithm="giac")
 

Output:

integrate(sqrt(cos(d*x + c))*sin(b*x + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {\cos (c+d x)} \sin ^2(a+b x) \, dx=\int \sqrt {\cos \left (c+d\,x\right )}\,{\sin \left (a+b\,x\right )}^2 \,d x \] Input:

int(cos(c + d*x)^(1/2)*sin(a + b*x)^2,x)
 

Output:

int(cos(c + d*x)^(1/2)*sin(a + b*x)^2, x)
 

Reduce [F]

\[ \int \sqrt {\cos (c+d x)} \sin ^2(a+b x) \, dx=\int \sqrt {\cos \left (d x +c \right )}\, \sin \left (b x +a \right )^{2}d x \] Input:

int(cos(d*x+c)^(1/2)*sin(b*x+a)^2,x)
 

Output:

int(sqrt(cos(c + d*x))*sin(a + b*x)**2,x)