3.3.16 \(\int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^3 \, dx\) [216]

3.3.16.1 Optimal result
3.3.16.2 Mathematica [C] (verified)
3.3.16.3 Rubi [A] (verified)
3.3.16.4 Maple [A] (verified)
3.3.16.5 Fricas [C] (verification not implemented)
3.3.16.6 Sympy [F(-1)]
3.3.16.7 Maxima [F]
3.3.16.8 Giac [F]
3.3.16.9 Mupad [F(-1)]

3.3.16.1 Optimal result

Integrand size = 25, antiderivative size = 172 \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^3 \, dx=-\frac {26 a^3 (e \cos (c+d x))^{5/2}}{35 d e}+\frac {26 a^3 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {e \cos (c+d x)}}+\frac {26 a^3 e \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac {2 a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}{9 d e}-\frac {26 (e \cos (c+d x))^{5/2} \left (a^3+a^3 \sin (c+d x)\right )}{63 d e} \]

output
-26/35*a^3*(e*cos(d*x+c))^(5/2)/d/e-2/9*a*(e*cos(d*x+c))^(5/2)*(a+a*sin(d* 
x+c))^2/d/e-26/63*(e*cos(d*x+c))^(5/2)*(a^3+a^3*sin(d*x+c))/d/e+26/21*a^3* 
e^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+ 
1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(e*cos(d*x+c))^(1/2)+26/21*a^3*e*sin(d* 
x+c)*(e*cos(d*x+c))^(1/2)/d
 
3.3.16.2 Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.38 \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^3 \, dx=-\frac {32 \sqrt [4]{2} a^3 (e \cos (c+d x))^{5/2} \operatorname {Hypergeometric2F1}\left (-\frac {13}{4},\frac {5}{4},\frac {9}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{5 d e (1+\sin (c+d x))^{5/4}} \]

input
Integrate[(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^3,x]
 
output
(-32*2^(1/4)*a^3*(e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[-13/4, 5/4, 9/4, 
 (1 - Sin[c + d*x])/2])/(5*d*e*(1 + Sin[c + d*x])^(5/4))
 
3.3.16.3 Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3157, 3042, 3157, 3042, 3148, 3042, 3115, 3042, 3121, 3042, 3120}

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.

\(\displaystyle \int (a \sin (c+d x)+a)^3 (e \cos (c+d x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3157

\(\displaystyle \frac {13}{9} a \int (e \cos (c+d x))^{3/2} (\sin (c+d x) a+a)^2dx-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}}{9 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {13}{9} a \int (e \cos (c+d x))^{3/2} (\sin (c+d x) a+a)^2dx-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}}{9 d e}\)

\(\Big \downarrow \) 3157

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

\(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \left (a \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )-\frac {2 a (e \cos (c+d x))^{5/2}}{5 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{5/2}}{7 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}}{9 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \left (a \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )-\frac {2 a (e \cos (c+d x))^{5/2}}{5 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{5/2}}{7 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}}{9 d e}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \left (a \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )-\frac {2 a (e \cos (c+d x))^{5/2}}{5 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{5/2}}{7 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}}{9 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \left (a \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )-\frac {2 a (e \cos (c+d x))^{5/2}}{5 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{5/2}}{7 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}}{9 d e}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {13}{9} a \left (\frac {9}{7} a \left (a \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )-\frac {2 a (e \cos (c+d x))^{5/2}}{5 d e}\right )-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{5/2}}{7 d e}\right )-\frac {2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}}{9 d e}\)

input
Int[(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^3,x]
 
output
(-2*a*(e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])^2)/(9*d*e) + (13*a*((-2* 
(e*Cos[c + d*x])^(5/2)*(a^2 + a^2*Sin[c + d*x]))/(7*d*e) + (9*a*((-2*a*(e* 
Cos[c + d*x])^(5/2))/(5*d*e) + a*((2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + 
 d*x)/2, 2])/(3*d*Sqrt[e*Cos[c + d*x]]) + (2*e*Sqrt[e*Cos[c + d*x]]*Sin[c 
+ d*x])/(3*d))))/7))/9
 

3.3.16.3.1 Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

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 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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]
 

rule 3148
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + 
 Simp[a   Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && 
 (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
 

rule 3157
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + 
f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p))   Int[(g* 
Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && NeQ[m + p, 0] && Integers 
Q[2*m, 2*p]
 
3.3.16.4 Maple [A] (verified)

Time = 6.29 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.46

method result size
default \(\frac {2 a^{3} e^{2} \left (-1120 \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2160 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2800 \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3240 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-784 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+840 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-1624 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+120 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-195 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+1162 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-217 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{315 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) \(251\)
parts \(-\frac {2 a^{3} \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, e^{2} \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}+\frac {2 a^{3} \left (\frac {\left (e \cos \left (d x +c \right )\right )^{\frac {9}{2}}}{9}-\frac {\left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}} e^{2}}{5}\right )}{d \,e^{3}}-\frac {6 a^{3} \left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d e}+\frac {4 a^{3} \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, e^{2} \left (24 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-60 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+50 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+\cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7 \sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}\) \(464\)

input
int((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 
output
2/315/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^3*e^2*(-112 
0*sin(1/2*d*x+1/2*c)^11+2160*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+2800* 
sin(1/2*d*x+1/2*c)^9-3240*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-784*sin( 
1/2*d*x+1/2*c)^7+840*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-1624*sin(1/2* 
d*x+1/2*c)^5+120*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-195*(sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2 
*c),2^(1/2))+1162*sin(1/2*d*x+1/2*c)^3-217*sin(1/2*d*x+1/2*c))/d
 
3.3.16.5 Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.78 \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^3 \, dx=\frac {-195 i \, \sqrt {2} a^{3} e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 195 i \, \sqrt {2} a^{3} e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (35 \, a^{3} e \cos \left (d x + c\right )^{4} - 252 \, a^{3} e \cos \left (d x + c\right )^{2} - 15 \, {\left (9 \, a^{3} e \cos \left (d x + c\right )^{2} - 13 \, a^{3} e\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{315 \, d} \]

input
integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")
 
output
1/315*(-195*I*sqrt(2)*a^3*e^(3/2)*weierstrassPInverse(-4, 0, cos(d*x + c) 
+ I*sin(d*x + c)) + 195*I*sqrt(2)*a^3*e^(3/2)*weierstrassPInverse(-4, 0, c 
os(d*x + c) - I*sin(d*x + c)) + 2*(35*a^3*e*cos(d*x + c)^4 - 252*a^3*e*cos 
(d*x + c)^2 - 15*(9*a^3*e*cos(d*x + c)^2 - 13*a^3*e)*sin(d*x + c))*sqrt(e* 
cos(d*x + c)))/d
 
3.3.16.6 Sympy [F(-1)]

Timed out. \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \]

input
integrate((e*cos(d*x+c))**(3/2)*(a+a*sin(d*x+c))**3,x)
 
output
Timed out
 
3.3.16.7 Maxima [F]

\[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^3 \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{3} \,d x } \]

input
integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^3,x, algorithm="maxima")
 
output
integrate((e*cos(d*x + c))^(3/2)*(a*sin(d*x + c) + a)^3, x)
 
3.3.16.8 Giac [F]

\[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^3 \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}^{3} \,d x } \]

input
integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^3,x, algorithm="giac")
 
output
integrate((e*cos(d*x + c))^(3/2)*(a*sin(d*x + c) + a)^3, x)
 
3.3.16.9 Mupad [F(-1)]

Timed out. \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^3 \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3 \,d x \]

input
int((e*cos(c + d*x))^(3/2)*(a + a*sin(c + d*x))^3,x)
 
output
int((e*cos(c + d*x))^(3/2)*(a + a*sin(c + d*x))^3, x)