\(\int (d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)} \, dx\) [258]

Optimal result

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

1/3*d*(d*cos(b*x+a))^(3/2)*(c*sin(b*x+a))^(3/2)/b/c-1/2*d^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 = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int (d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)} \, dx=\frac {2 d^2 \sqrt {d \cos (a+b x)} \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{4},\frac {7}{4},\sin ^2(a+b x)\right ) \sqrt {c \sin (a+b x)} \tan (a+b x)}{3 b} \] Input:

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

Output:

(2*d^2*Sqrt[d*Cos[a + b*x]]*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[-3/4, 
 3/4, 7/4, Sin[a + b*x]^2]*Sqrt[c*Sin[a + b*x]]*Tan[a + b*x])/(3*b)
 

Rubi [A] (verified)

Time = 0.40 (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, 3049, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3049

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3052

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

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

Output:

(d*(d*Cos[a + b*x])^(3/2)*(c*Sin[a + b*x])^(3/2))/(3*b*c) + (d^2*Sqrt[d*Co 
s[a + b*x]]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[c*Sin[a + b*x]])/(2*b*Sqrt[S 
in[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 3049

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[a*(b*Sin[e + f*x])^(n + 1)*((a*Cos[e + f*x])^(m - 1)/ 
(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n))   Int[(b*Sin[e + f*x])^n*(a 
*Cos[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 = 9.16 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.62

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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