\(\int \cos ^3(a+b x) \sqrt {\cos (c+d x)} \, dx\) [353]

Optimal result

Integrand size = 19, antiderivative size = 485 \[ \int \cos ^3(a+b x) \sqrt {\cos (c+d x)} \, dx=-\frac {3 i e^{\frac {1}{2} i (2 a-c)+\frac {1}{2} i (2 b-d) x+\frac {1}{2} i (c+d x)} \sqrt {\cos (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (-1+\frac {2 b}{d}\right ),\frac {1}{4} \left (3+\frac {2 b}{d}\right ),-e^{2 i (c+d x)}\right )}{4 (2 b-d) \sqrt {1+e^{2 i (c+d x)}}}-\frac {i e^{\frac {1}{2} i (6 a-c)+\frac {1}{2} i (6 b-d) x+\frac {1}{2} i (c+d x)} \sqrt {\cos (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (-1+\frac {6 b}{d}\right ),\frac {3 (2 b+d)}{4 d},-e^{2 i (c+d x)}\right )}{4 (6 b-d) \sqrt {1+e^{2 i (c+d x)}}}+\frac {3 i e^{-\frac {1}{2} i (2 a+c)-\frac {1}{2} i (2 b+d) x+\frac {1}{2} i (c+d x)} \sqrt {\cos (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {2 b+d}{4 d},\frac {1}{4} \left (3-\frac {2 b}{d}\right ),-e^{2 i (c+d x)}\right )}{4 (2 b+d) \sqrt {1+e^{2 i (c+d x)}}}+\frac {i e^{-\frac {1}{2} i (6 a+c)-\frac {1}{2} i (6 b+d) x+\frac {1}{2} i (c+d x)} \sqrt {\cos (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {6 b+d}{4 d},\frac {3}{4} \left (1-\frac {2 b}{d}\right ),-e^{2 i (c+d x)}\right )}{4 (6 b+d) \sqrt {1+e^{2 i (c+d x)}}} \] Output:

-3/4*I*exp(1/2*I*(2*a-c)+1/2*I*(2*b-d)*x+1/2*I*(d*x+c))*cos(d*x+c)^(1/2)*h 
ypergeom([-1/2, -1/4+1/2*b/d],[3/4+1/2*b/d],-exp(2*I*(d*x+c)))/(2*b-d)/(1+ 
exp(2*I*(d*x+c)))^(1/2)-1/4*I*exp(1/2*I*(6*a-c)+1/2*I*(6*b-d)*x+1/2*I*(d*x 
+c))*cos(d*x+c)^(1/2)*hypergeom([-1/2, -1/4+3/2*b/d],[3/4*(2*b+d)/d],-exp( 
2*I*(d*x+c)))/(6*b-d)/(1+exp(2*I*(d*x+c)))^(1/2)+3/4*I*exp(-1/2*I*(2*a+c)- 
1/2*I*(2*b+d)*x+1/2*I*(d*x+c))*cos(d*x+c)^(1/2)*hypergeom([-1/2, -1/4*(2*b 
+d)/d],[3/4-1/2*b/d],-exp(2*I*(d*x+c)))/(2*b+d)/(1+exp(2*I*(d*x+c)))^(1/2) 
+1/4*I*exp(-1/2*I*(6*a+c)-1/2*I*(6*b+d)*x+1/2*I*(d*x+c))*cos(d*x+c)^(1/2)* 
hypergeom([-1/2, -1/4*(6*b+d)/d],[3/4-3/2*b/d],-exp(2*I*(d*x+c)))/(6*b+d)/ 
(1+exp(2*I*(d*x+c)))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(16601\) vs. \(2(485)=970\).

Time = 42.73 (sec) , antiderivative size = 16601, normalized size of antiderivative = 34.23 \[ \int \cos ^3(a+b x) \sqrt {\cos (c+d x)} \, dx=\text {Result too large to show} \] Input:

Integrate[Cos[a + b*x]^3*Sqrt[Cos[c + d*x]],x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 566, normalized size of antiderivative = 1.17, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {5065, 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.

\(\displaystyle \int \cos ^3(a+b x) \sqrt {\cos (c+d x)} \, dx\)

\(\Big \downarrow \) 5065

\(\displaystyle \frac {\int \left (3 e^{-i a-i b x} \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}}+3 e^{i a+i b x} \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}}+e^{-3 i a-3 i b x} \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}}+e^{3 i a+3 i b x} \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}}\right )dx}{8 \sqrt {2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {6 i \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (\frac {2 b}{d}-1\right ),\frac {1}{4} \left (\frac {2 b}{d}+3\right ),-e^{2 i (c+d x)}\right ) \exp \left (\frac {1}{2} i (2 a-c)+\frac {1}{2} i x (2 b-d)+\frac {1}{2} i (c+d x)\right )}{(2 b-d) \sqrt {1+e^{2 i c+2 i d x}}}-\frac {2 i \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (\frac {6 b}{d}-1\right ),\frac {3 (2 b+d)}{4 d},-e^{2 i (c+d x)}\right ) \exp \left (\frac {1}{2} i (6 a-c)+\frac {1}{2} i x (6 b-d)+\frac {1}{2} i (c+d x)\right )}{(6 b-d) \sqrt {1+e^{2 i c+2 i d x}}}+\frac {6 i \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {2 b+d}{4 d},\frac {1}{4} \left (3-\frac {2 b}{d}\right ),-e^{2 i (c+d x)}\right ) \exp \left (-\frac {1}{2} i (2 a+c)-\frac {1}{2} i x (2 b+d)+\frac {1}{2} i (c+d x)\right )}{(2 b+d) \sqrt {1+e^{2 i c+2 i d x}}}+\frac {2 i \sqrt {e^{-i (c+d x)}+e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {6 b+d}{4 d},\frac {3}{4} \left (1-\frac {2 b}{d}\right ),-e^{2 i (c+d x)}\right ) \exp \left (-\frac {1}{2} i (6 a+c)-\frac {1}{2} i x (6 b+d)+\frac {1}{2} i (c+d x)\right )}{(6 b+d) \sqrt {1+e^{2 i c+2 i d x}}}}{8 \sqrt {2}}\)

Input:

Int[Cos[a + b*x]^3*Sqrt[Cos[c + d*x]],x]
 

Output:

(((-6*I)*E^((I/2)*(2*a - c) + (I/2)*(2*b - d)*x + (I/2)*(c + d*x))*Sqrt[E^ 
((-I)*(c + d*x)) + 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*I)*E^((I/2)*(6*a - c) + (I/2)*(6*b - d)*x + (I/2)*(c 
 + d*x))*Sqrt[E^((-I)*(c + d*x)) + E^(I*(c + d*x))]*Hypergeometric2F1[-1/2 
, (-1 + (6*b)/d)/4, (3*(2*b + d))/(4*d), -E^((2*I)*(c + d*x))])/((6*b - d) 
*Sqrt[1 + E^((2*I)*c + (2*I)*d*x)]) + ((6*I)*E^((-1/2*I)*(2*a + c) - (I/2) 
*(2*b + d)*x + (I/2)*(c + d*x))*Sqrt[E^((-I)*(c + d*x)) + 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*I)*E^((-1/2*I 
)*(6*a + c) - (I/2)*(6*b + d)*x + (I/2)*(c + d*x))*Sqrt[E^((-I)*(c + d*x)) 
 + E^(I*(c + d*x))]*Hypergeometric2F1[-1/2, -1/4*(6*b + d)/d, (3*(1 - (2*b 
)/d))/4, -E^((2*I)*(c + d*x))])/((6*b + d)*Sqrt[1 + E^((2*I)*c + (2*I)*d*x 
)]))/(8*Sqrt[2])
 

Defintions of rubi rules used

rule 2009

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

rule 5065

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*Cos[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] 
:> Simp[1/2^(p + q)   Int[ExpandIntegrand[(E^((-I)*(c + d*x)) + E^(I*(c + d 
*x)))^q, (E^((-I)*(a + b*x)) + E^(I*(a + b*x)))^p, x], x], x] /; FreeQ[{a, 
b, c, d, q}, x] && IGtQ[p, 0] &&  !IntegerQ[q]
 

Maple [F]

\[\int \cos \left (b x +a \right )^{3} \sqrt {\cos \left (d x +c \right )}d x\]

Input:

int(cos(b*x+a)^3*cos(d*x+c)^(1/2),x)
 

Output:

int(cos(b*x+a)^3*cos(d*x+c)^(1/2),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \cos ^3(a+b x) \sqrt {\cos (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(b*x+a)^3*cos(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(a+b x) \sqrt {\cos (c+d x)} \, dx=\text {Timed out} \] Input:

integrate(cos(b*x+a)**3*cos(d*x+c)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^3(a+b x) \sqrt {\cos (c+d x)} \, dx=\int { \cos \left (b x + a\right )^{3} \sqrt {\cos \left (d x + c\right )} \,d x } \] Input:

integrate(cos(b*x+a)^3*cos(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate(cos(b*x + a)^3*sqrt(cos(d*x + c)), x)
 

Giac [F]

\[ \int \cos ^3(a+b x) \sqrt {\cos (c+d x)} \, dx=\int { \cos \left (b x + a\right )^{3} \sqrt {\cos \left (d x + c\right )} \,d x } \] Input:

integrate(cos(b*x+a)^3*cos(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

integrate(cos(b*x + a)^3*sqrt(cos(d*x + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^3(a+b x) \sqrt {\cos (c+d x)} \, dx=\int {\cos \left (a+b\,x\right )}^3\,\sqrt {\cos \left (c+d\,x\right )} \,d x \] Input:

int(cos(a + b*x)^3*cos(c + d*x)^(1/2),x)
 

Output:

int(cos(a + b*x)^3*cos(c + d*x)^(1/2), x)
 

Reduce [F]

\[ \int \cos ^3(a+b x) \sqrt {\cos (c+d x)} \, dx=\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (b x +a \right )^{3}d x \] Input:

int(cos(b*x+a)^3*cos(d*x+c)^(1/2),x)
 

Output:

int(sqrt(cos(c + d*x))*cos(a + b*x)**3,x)