\(\int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{5/2} \, dx\) [278]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{5/2} \, dx=-\frac {c (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2}}{3 b d}+\frac {c^2 \sqrt {d \cos (a+b x)} E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {c \sin (a+b x)}}{2 b \sqrt {\sin (2 a+2 b x)}} \] Output:

-1/3*c*(d*cos(b*x+a))^(3/2)*(c*sin(b*x+a))^(3/2)/b/d-1/2*c^2*(d*cos(b*x+a) 
)^(1/2)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))*(c*sin(b*x+a))^(1/2)/b/sin(2* 
b*x+2*a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71 \[ \int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{5/2} \, dx=\frac {2 \sqrt {d \cos (a+b x)} \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {7}{4},\frac {11}{4},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{5/2} \tan (a+b x)}{7 b} \] Input:

Integrate[Sqrt[d*Cos[a + b*x]]*(c*Sin[a + b*x])^(5/2),x]
 

Output:

(2*Sqrt[d*Cos[a + b*x]]*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[1/4, 7/4, 
 11/4, Sin[a + b*x]^2]*(c*Sin[a + b*x])^(5/2)*Tan[a + b*x])/(7*b)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3048, 3042, 3052, 3042, 3119}

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.

\(\displaystyle \int (c \sin (a+b x))^{5/2} \sqrt {d \cos (a+b x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c \sin (a+b x))^{5/2} \sqrt {d \cos (a+b x)}dx\)

\(\Big \downarrow \) 3048

\(\displaystyle \frac {1}{2} c^2 \int \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}dx-\frac {c (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{3/2}}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} c^2 \int \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}dx-\frac {c (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{3/2}}{3 b d}\)

\(\Big \downarrow \) 3052

\(\displaystyle \frac {c^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)} \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)}}-\frac {c (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{3/2}}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {c^2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)} \int \sqrt {\sin (2 a+2 b x)}dx}{2 \sqrt {\sin (2 a+2 b x)}}-\frac {c (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{3/2}}{3 b d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {c^2 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{2 b \sqrt {\sin (2 a+2 b x)}}-\frac {c (c \sin (a+b x))^{3/2} (d \cos (a+b x))^{3/2}}{3 b d}\)

Input:

Int[Sqrt[d*Cos[a + b*x]]*(c*Sin[a + b*x])^(5/2),x]
 

Output:

-1/3*(c*(d*Cos[a + b*x])^(3/2)*(c*Sin[a + b*x])^(3/2))/(b*d) + (c^2*Sqrt[d 
*Cos[a + b*x]]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[c*Sin[a + b*x]])/(2*b*Sqr 
t[Sin[2*a + 2*b*x]])
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3048

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 
1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n))   Int[(b*Cos[e + f*x])^n 
*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] 
 && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
 

rule 3052

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] 
, x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e 
 + 2*f*x]])   Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, x]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(248\) vs. \(2(82)=164\).

Time = 5.52 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.62

method	result	size
default	\(\frac {\left (\left (3 \cos \left (b x +a \right )+3\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+\left (-6 \cos \left (b x +a \right )-6\right ) \sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticE}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+\cos \left (b x +a \right ) \left (4 \cos \left (b x +a \right )^{3}-10 \cos \left (b x +a \right )+6\right )\right ) \sqrt {d \cos \left (b x +a \right )}\, \sqrt {c \sin \left (b x +a \right )}\, c^{2} \sec \left (b x +a \right ) \csc \left (b x +a \right )}{12 b}\)	\(249\)

Input:

int((d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/12/b*((3*cos(b*x+a)+3)*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(2*cot(b*x+a)-2* 
csc(b*x+a)+2)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticF((-cot(b*x+a)+c 
sc(b*x+a)+1)^(1/2),1/2*2^(1/2))+(-6*cos(b*x+a)-6)*(-cot(b*x+a)+csc(b*x+a)+ 
1)^(1/2)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2) 
*EllipticE((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))+cos(b*x+a)*(4*cos 
(b*x+a)^3-10*cos(b*x+a)+6))*(d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(1/2)*c^2* 
sec(b*x+a)*csc(b*x+a)
 

Fricas [F]

\[ \int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{5/2} \, dx=\int { \sqrt {d \cos \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \] Input:

integrate((d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate((d*cos(b*x+a))**(1/2)*(c*sin(b*x+a))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{5/2} \, dx=\int { \sqrt {d \cos \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \] Input:

integrate((d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(5/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(d*cos(b*x + a))*(c*sin(b*x + a))^(5/2), x)
 

Giac [F]

\[ \int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{5/2} \, dx=\int { \sqrt {d \cos \left (b x + a\right )} \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \] Input:

integrate((d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(5/2),x, algorithm="giac")
 

Output:

integrate(sqrt(d*cos(b*x + a))*(c*sin(b*x + a))^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{5/2} \, dx=\int \sqrt {d\,\cos \left (a+b\,x\right )}\,{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2} \,d x \] Input:

int((d*cos(a + b*x))^(1/2)*(c*sin(a + b*x))^(5/2),x)
 

Output:

int((d*cos(a + b*x))^(1/2)*(c*sin(a + b*x))^(5/2), x)
 

Reduce [F]

\[ \int \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{5/2} \, dx=\sqrt {d}\, \sqrt {c}\, \left (\int \sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}\, \sin \left (b x +a \right )^{2}d x \right ) c^{2} \] Input:

int((d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(5/2),x)
 

Output:

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