3.3.11 \(\int (d \cos (a+b x))^{9/2} \sin ^4(a+b x) \, dx\) [211]

3.3.11.1 Optimal result
3.3.11.2 Mathematica [C] (verified)
3.3.11.3 Rubi [A] (verified)
3.3.11.4 Maple [A] (verified)
3.3.11.5 Fricas [C] (verification not implemented)
3.3.11.6 Sympy [F(-1)]
3.3.11.7 Maxima [F]
3.3.11.8 Giac [F]
3.3.11.9 Mupad [F(-1)]

3.3.11.1 Optimal result

Integrand size = 21, antiderivative size = 156 \[ \int (d \cos (a+b x))^{9/2} \sin ^4(a+b x) \, dx=\frac {56 d^4 \sqrt {d \cos (a+b x)} E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{1105 b \sqrt {\cos (a+b x)}}+\frac {56 d^3 (d \cos (a+b x))^{3/2} \sin (a+b x)}{3315 b}+\frac {8 d (d \cos (a+b x))^{7/2} \sin (a+b x)}{663 b}-\frac {12 (d \cos (a+b x))^{11/2} \sin (a+b x)}{221 b d}-\frac {2 (d \cos (a+b x))^{11/2} \sin ^3(a+b x)}{17 b d} \]

output
56/3315*d^3*(d*cos(b*x+a))^(3/2)*sin(b*x+a)/b+8/663*d*(d*cos(b*x+a))^(7/2) 
*sin(b*x+a)/b-12/221*(d*cos(b*x+a))^(11/2)*sin(b*x+a)/b/d-2/17*(d*cos(b*x+ 
a))^(11/2)*sin(b*x+a)^3/b/d+56/1105*d^4*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1 
/2*a+1/2*b*x)*EllipticE(sin(1/2*a+1/2*b*x),2^(1/2))*(d*cos(b*x+a))^(1/2)/b 
/cos(b*x+a)^(1/2)
 
3.3.11.2 Mathematica [C] (verified)

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

Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.37 \[ \int (d \cos (a+b x))^{9/2} \sin ^4(a+b x) \, dx=\frac {(d \cos (a+b x))^{9/2} \sqrt [4]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {5}{2},\frac {7}{2},\sin ^2(a+b x)\right ) \tan ^5(a+b x)}{5 b} \]

input
Integrate[(d*Cos[a + b*x])^(9/2)*Sin[a + b*x]^4,x]
 
output
((d*Cos[a + b*x])^(9/2)*(Cos[a + b*x]^2)^(1/4)*Hypergeometric2F1[-7/4, 5/2 
, 7/2, Sin[a + b*x]^2]*Tan[a + b*x]^5)/(5*b)
 
3.3.11.3 Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3048, 3042, 3048, 3042, 3115, 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.

\(\displaystyle \int \sin ^4(a+b x) (d \cos (a+b x))^{9/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (a+b x)^4 (d \cos (a+b x))^{9/2}dx\)

\(\Big \downarrow \) 3048

\(\displaystyle \frac {6}{17} \int (d \cos (a+b x))^{9/2} \sin ^2(a+b x)dx-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{17} \int (d \cos (a+b x))^{9/2} \sin (a+b x)^2dx-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}\)

\(\Big \downarrow \) 3048

\(\displaystyle \frac {6}{17} \left (\frac {2}{13} \int (d \cos (a+b x))^{9/2}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{11/2}}{13 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{17} \left (\frac {2}{13} \int \left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{9/2}dx-\frac {2 \sin (a+b x) (d \cos (a+b x))^{11/2}}{13 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {6}{17} \left (\frac {2}{13} \left (\frac {7}{9} d^2 \int (d \cos (a+b x))^{5/2}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{11/2}}{13 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{17} \left (\frac {2}{13} \left (\frac {7}{9} d^2 \int \left (d \sin \left (a+b x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{11/2}}{13 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {6}{17} \left (\frac {2}{13} \left (\frac {7}{9} d^2 \left (\frac {3}{5} d^2 \int \sqrt {d \cos (a+b x)}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{11/2}}{13 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{17} \left (\frac {2}{13} \left (\frac {7}{9} d^2 \left (\frac {3}{5} d^2 \int \sqrt {d \sin \left (a+b x+\frac {\pi }{2}\right )}dx+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{11/2}}{13 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {6}{17} \left (\frac {2}{13} \left (\frac {7}{9} d^2 \left (\frac {3 d^2 \sqrt {d \cos (a+b x)} \int \sqrt {\cos (a+b x)}dx}{5 \sqrt {\cos (a+b x)}}+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{11/2}}{13 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{17} \left (\frac {2}{13} \left (\frac {7}{9} d^2 \left (\frac {3 d^2 \sqrt {d \cos (a+b x)} \int \sqrt {\sin \left (a+b x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (a+b x)}}+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{11/2}}{13 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {6}{17} \left (\frac {2}{13} \left (\frac {7}{9} d^2 \left (\frac {6 d^2 E\left (\left .\frac {1}{2} (a+b x)\right |2\right ) \sqrt {d \cos (a+b x)}}{5 b \sqrt {\cos (a+b x)}}+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{3/2}}{5 b}\right )+\frac {2 d \sin (a+b x) (d \cos (a+b x))^{7/2}}{9 b}\right )-\frac {2 \sin (a+b x) (d \cos (a+b x))^{11/2}}{13 b d}\right )-\frac {2 \sin ^3(a+b x) (d \cos (a+b x))^{11/2}}{17 b d}\)

input
Int[(d*Cos[a + b*x])^(9/2)*Sin[a + b*x]^4,x]
 
output
(-2*(d*Cos[a + b*x])^(11/2)*Sin[a + b*x]^3)/(17*b*d) + (6*((-2*(d*Cos[a + 
b*x])^(11/2)*Sin[a + b*x])/(13*b*d) + (2*((2*d*(d*Cos[a + b*x])^(7/2)*Sin[ 
a + b*x])/(9*b) + (7*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))/13))/17
 

3.3.11.3.1 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]
 
3.3.11.4 Maple [A] (verified)

Time = 13.67 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.76

method result size
default \(-\frac {8 \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, d^{5} \left (24960 \left (\cos ^{19}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-124800 \left (\cos ^{17}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+265440 \left (\cos ^{15}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-312960 \left (\cos ^{13}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+222520 \left (\cos ^{11}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-96360 \left (\cos ^{9}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+23866 \left (\cos ^{7}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-2652 \left (\cos ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-35 \left (\cos ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-21 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {1-2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, E\left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )+21 \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3315 \sqrt {-d \left (2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {d \left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right )}\, b}\) \(275\)

input
int((d*cos(b*x+a))^(9/2)*sin(b*x+a)^4,x,method=_RETURNVERBOSE)
 
output
-8/3315*(d*(2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*d^5*(249 
60*cos(1/2*b*x+1/2*a)^19-124800*cos(1/2*b*x+1/2*a)^17+265440*cos(1/2*b*x+1 
/2*a)^15-312960*cos(1/2*b*x+1/2*a)^13+222520*cos(1/2*b*x+1/2*a)^11-96360*c 
os(1/2*b*x+1/2*a)^9+23866*cos(1/2*b*x+1/2*a)^7-2652*cos(1/2*b*x+1/2*a)^5-3 
5*cos(1/2*b*x+1/2*a)^3-21*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(1-2*cos(1/2*b*x+1/ 
2*a)^2)^(1/2)*EllipticE(cos(1/2*b*x+1/2*a),2^(1/2))+21*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
 
3.3.11.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.13 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.85 \[ \int (d \cos (a+b x))^{9/2} \sin ^4(a+b x) \, dx=-\frac {2 \, {\left (-42 i \, \sqrt {2} d^{\frac {9}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 42 i \, \sqrt {2} d^{\frac {9}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) - {\left (195 \, d^{4} \cos \left (b x + a\right )^{7} - 285 \, d^{4} \cos \left (b x + a\right )^{5} + 20 \, d^{4} \cos \left (b x + a\right )^{3} + 28 \, d^{4} \cos \left (b x + a\right )\right )} \sqrt {d \cos \left (b x + a\right )} \sin \left (b x + a\right )\right )}}{3315 \, b} \]

input
integrate((d*cos(b*x+a))^(9/2)*sin(b*x+a)^4,x, algorithm="fricas")
 
output
-2/3315*(-42*I*sqrt(2)*d^(9/2)*weierstrassZeta(-4, 0, weierstrassPInverse( 
-4, 0, cos(b*x + a) + I*sin(b*x + a))) + 42*I*sqrt(2)*d^(9/2)*weierstrassZ 
eta(-4, 0, weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a))) - (1 
95*d^4*cos(b*x + a)^7 - 285*d^4*cos(b*x + a)^5 + 20*d^4*cos(b*x + a)^3 + 2 
8*d^4*cos(b*x + a))*sqrt(d*cos(b*x + a))*sin(b*x + a))/b
 
3.3.11.6 Sympy [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{9/2} \sin ^4(a+b x) \, dx=\text {Timed out} \]

input
integrate((d*cos(b*x+a))**(9/2)*sin(b*x+a)**4,x)
 
output
Timed out
 
3.3.11.7 Maxima [F]

\[ \int (d \cos (a+b x))^{9/2} \sin ^4(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} \sin \left (b x + a\right )^{4} \,d x } \]

input
integrate((d*cos(b*x+a))^(9/2)*sin(b*x+a)^4,x, algorithm="maxima")
 
output
integrate((d*cos(b*x + a))^(9/2)*sin(b*x + a)^4, x)
 
3.3.11.8 Giac [F]

\[ \int (d \cos (a+b x))^{9/2} \sin ^4(a+b x) \, dx=\int { \left (d \cos \left (b x + a\right )\right )^{\frac {9}{2}} \sin \left (b x + a\right )^{4} \,d x } \]

input
integrate((d*cos(b*x+a))^(9/2)*sin(b*x+a)^4,x, algorithm="giac")
 
output
integrate((d*cos(b*x + a))^(9/2)*sin(b*x + a)^4, x)
 
3.3.11.9 Mupad [F(-1)]

Timed out. \[ \int (d \cos (a+b x))^{9/2} \sin ^4(a+b x) \, dx=\int {\sin \left (a+b\,x\right )}^4\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{9/2} \,d x \]

input
int(sin(a + b*x)^4*(d*cos(a + b*x))^(9/2),x)
 
output
int(sin(a + b*x)^4*(d*cos(a + b*x))^(9/2), x)