3.3.18 \(\int \frac {(a+a \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx\) [218]

3.3.18.1 Optimal result
3.3.18.2 Mathematica [C] (verified)
3.3.18.3 Rubi [A] (verified)
3.3.18.4 Maple [A] (verified)
3.3.18.5 Fricas [C] (verification not implemented)
3.3.18.6 Sympy [F(-1)]
3.3.18.7 Maxima [F]
3.3.18.8 Giac [F]
3.3.18.9 Mupad [F(-1)]

3.3.18.1 Optimal result

Integrand size = 25, antiderivative size = 136 \[ \int \frac {(a+a \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {6 a^3 \sqrt {e \cos (c+d x)}}{d e}+\frac {6 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d \sqrt {e \cos (c+d x)}}-\frac {2 a \sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^2}{5 d e}-\frac {6 \sqrt {e \cos (c+d x)} \left (a^3+a^3 \sin (c+d x)\right )}{5 d e} \]

output
6*a^3*(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)-6*a^3*(e*cos(d*x 
+c))^(1/2)/d/e-2/5*a*(a+a*sin(d*x+c))^2*(e*cos(d*x+c))^(1/2)/d/e-6/5*(a^3+ 
a^3*sin(d*x+c))*(e*cos(d*x+c))^(1/2)/d/e
 
3.3.18.2 Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.47 \[ \int \frac {(a+a \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {16 \sqrt [4]{2} a^3 \sqrt {e \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},\frac {1}{4},\frac {5}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{d e \sqrt [4]{1+\sin (c+d x)}} \]

input
Integrate[(a + a*Sin[c + d*x])^3/Sqrt[e*Cos[c + d*x]],x]
 
output
(-16*2^(1/4)*a^3*Sqrt[e*Cos[c + d*x]]*Hypergeometric2F1[-9/4, 1/4, 5/4, (1 
 - Sin[c + d*x])/2])/(d*e*(1 + Sin[c + d*x])^(1/4))
 
3.3.18.3 Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3157, 3042, 3157, 3042, 3148, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3157

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3157

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

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

3.3.18.3.1 Defintions of rubi rules used

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

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]
 
3.3.18.4 Maple [A] (verified)

Time = 3.64 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.31

method result size
default \(\frac {2 a^{3} \left (-8 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+34 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-19 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) \(178\)
parts \(\frac {2 a^{3} \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \operatorname {am}^{-1}\left (\frac {d x}{2}+\frac {c}{2}| \sqrt {2}\right )}{d \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}}+\frac {2 a^{3} \left (\frac {\left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-e^{2} \sqrt {e \cos \left (d x +c \right )}\right )}{d \,e^{3}}-\frac {6 a^{3} \sqrt {e \cos \left (d x +c \right )}}{d e}+\frac {4 a^{3} \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-e \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {e \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d}\) \(308\)

input
int((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 
output
2/5/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^3*(-8*sin(1/2 
*d*x+1/2*c)^7+20*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*sin(1/2*d*x+1/ 
2*c)^5-10*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-15*(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))+34*sin(1/2*d*x+1/2*c)^3-19*sin(1/2*d*x+1/2*c))/d
 
3.3.18.5 Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.79 \[ \int \frac {(a+a \sin (c+d x))^3}{\sqrt {e \cos (c+d x)}} \, dx=\frac {-15 i \, \sqrt {2} a^{3} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 i \, \sqrt {2} a^{3} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - 5 \, a^{3} \sin \left (d x + c\right ) - 20 \, a^{3}\right )} \sqrt {e \cos \left (d x + c\right )}}{5 \, d e} \]

input
integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")
 
output
1/5*(-15*I*sqrt(2)*a^3*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) + I 
*sin(d*x + c)) + 15*I*sqrt(2)*a^3*sqrt(e)*weierstrassPInverse(-4, 0, cos(d 
*x + c) - I*sin(d*x + c)) + 2*(a^3*cos(d*x + c)^2 - 5*a^3*sin(d*x + c) - 2 
0*a^3)*sqrt(e*cos(d*x + c)))/(d*e)
 
3.3.18.6 Sympy [F(-1)]

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

input
integrate((a+a*sin(d*x+c))**3/(e*cos(d*x+c))**(1/2),x)
 
output
Timed out
 
3.3.18.7 Maxima [F]

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

input
integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")
 
output
integrate((a*sin(d*x + c) + a)^3/sqrt(e*cos(d*x + c)), x)
 
3.3.18.8 Giac [F]

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

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

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

input
int((a + a*sin(c + d*x))^3/(e*cos(c + d*x))^(1/2),x)
 
output
int((a + a*sin(c + d*x))^3/(e*cos(c + d*x))^(1/2), x)