\(\int (d \cos (a+b x))^{5/2} \sin ^2(a+b x) \, dx\) [196]

Optimal result

Integrand size = 21, antiderivative size = 98 \[ \int (d \cos (a+b x))^{5/2} \sin ^2(a+b x) \, dx=\frac {4 d^2 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{15 b \sqrt {\cos (a+b x)}}+\frac {4 d (d \cos (a+b x))^{3/2} \sin (a+b x)}{45 b}-\frac {2 (d \cos (a+b x))^{7/2} \sin (a+b x)}{9 b d} \] Output:

4/15*d^2*(d*cos(b*x+a))^(1/2)*EllipticE(sin(1/2*a+1/2*b*x),2^(1/2))/b/cos( 
b*x+a)^(1/2)+4/45*d*(d*cos(b*x+a))^(3/2)*sin(b*x+a)/b-2/9*(d*cos(b*x+a))^( 
7/2)*sin(b*x+a)/b/d
 

Mathematica [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.58 \[ \int (d \cos (a+b x))^{5/2} \sin ^2(a+b x) \, dx=\frac {(d \cos (a+b x))^{5/2} \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {3}{2},\frac {5}{2},\sin ^2(a+b x)\right ) \tan ^3(a+b x)}{3 b} \] Input:

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

Output:

((d*Cos[a + b*x])^(5/2)*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[-3/4, 3/2 
, 5/2, Sin[a + b*x]^2]*Tan[a + b*x]^3)/(3*b)
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3048, 3042, 3115, 3042, 3121, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3048

\(\displaystyle \frac {2}{9} \int (d \cos (a+b x))^{5/2}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{9} \int \left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{5/2}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {2}{9} \left (\frac {3}{5} d^2 \int \sqrt {d \cos (a+b x)}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{9} \left (\frac {3}{5} d^2 \int \sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {2}{9} \left (\frac {3 d^2 \sqrt {d \cos (a+b x)} \int \sqrt {\cos (a+b x)}dx}{5 \sqrt {\cos (a+b x)}}+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2}{9} \left (\frac {3 d^2 \sqrt {d \cos (a+b x)} \int \sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (a+b x)}}+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2}{9} \left (\frac {6 d^2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{5 b \sqrt {\cos (a+b x)}}+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b d}\)

Input:

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

Output:

(-2*(d*Cos[a + b*x])^(7/2)*Sin[a + b*x])/(9*b*d) + (2*((6*d^2*Sqrt[d*Cos[a 
 + b*x]]*EllipticE[(a + b*x)/2, 2])/(5*b*Sqrt[Cos[a + b*x]]) + (2*d*(d*Cos 
[a + b*x])^(3/2)*Sin[a + b*x])/(5*b)))/9
 

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 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

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]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(222\) vs. \(2(86)=172\).

Time = 7.10 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.28

method	result	size
default	\(\frac {4 \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right ) \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}\, d^{3} \left (80 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{11}-240 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{9}+272 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{7}-144 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{5}+35 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}+3 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {-2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )-3 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{45 \sqrt {-d \left (2 \sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\sin \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )}\, b}\)	\(223\)

Input:

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

Output:

4/45*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^3*(80*cos 
(1/2*b*x+1/2*a)^11-240*cos(1/2*b*x+1/2*a)^9+272*cos(1/2*b*x+1/2*a)^7-144*c 
os(1/2*b*x+1/2*a)^5+35*cos(1/2*b*x+1/2*a)^3+3*(sin(1/2*b*x+1/2*a)^2)^(1/2) 
*(-2*cos(1/2*b*x+1/2*a)^2+1)^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))-3 
*cos(1/2*b*x+1/2*a))/(-d*(2*sin(1/2*b*x+1/2*a)^4-sin(1/2*b*x+1/2*a)^2))^(1 
/2)/sin(1/2*b*x+1/2*a)/(d*(2*cos(1/2*b*x+1/2*a)^2-1))^(1/2)/b
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.08 \[ \int (d \cos (a+b x))^{5/2} \sin ^2(a+b x) \, dx=-\frac {2 \, {\left (-6 i \, \sqrt {\frac {1}{2}} d^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 6 i \, \sqrt {\frac {1}{2}} d^{\frac {5}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + {\left (5 \, d^{2} \cos \left (b x + a\right )^{3} - 2 \, d^{2} \cos \left (b x + a\right )\right )} \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{45 \, b} \] Input:

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

Output:

-2/45*(-6*I*sqrt(1/2)*d^(5/2)*weierstrassZeta(-4, 0, weierstrassPInverse(- 
4, 0, cos(b*x + a) + I*sin(b*x + a))) + 6*I*sqrt(1/2)*d^(5/2)*weierstrassZ 
eta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a))) + (5 
*d^2*cos(b*x + a)^3 - 2*d^2*cos(b*x + a))*sqrt(d*cos(b*x + a))*sin(b*x + a 
))/b
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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