\(\int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2 \, dx\) [206]

Optimal result

Integrand size = 25, antiderivative size = 137 \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2 \, dx=-\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}+\frac {6 a^2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 d \sqrt {e \cos (c+d x)}}+\frac {6 a^2 e \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 d}-\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e} \]

[Out]

-18/35*a^2*(e*cos(d*x+c))^(5/2)/d/e-2/7*(e*cos(d*x+c))^(5/2)*(a^2+a^2*sin(d*x+c))/d/e+6/7*a^2*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)+6/7*a^2*e*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2757, 2748, 2715, 2721, 2720} \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2 \, dx=\frac {6 a^2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 d \sqrt {e \cos (c+d x)}}-\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}-\frac {2 \left (a^2 \sin (c+d x)+a^2\right ) (e \cos (c+d x))^{5/2}}{7 d e}+\frac {6 a^2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{7 d} \]

[In]

Int[(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^2,x]

[Out]

(-18*a^2*(e*Cos[c + d*x])^(5/2))/(35*d*e) + (6*a^2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(7*d*Sqrt
[e*Cos[c + d*x]]) + (6*a^2*e*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(7*d) - (2*(e*Cos[c + d*x])^(5/2)*(a^2 + a^2*S
in[c + d*x]))/(7*d*e)

Rule 2715

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] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2720

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 2721

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

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[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 2757

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] + Dist[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] && IntegersQ[2*m, 2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e}+\frac {1}{7} (9 a) \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x)) \, dx \\ & = -\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}-\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e}+\frac {1}{7} \left (9 a^2\right ) \int (e \cos (c+d x))^{3/2} \, dx \\ & = -\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}+\frac {6 a^2 e \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 d}-\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e}+\frac {1}{7} \left (3 a^2 e^2\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx \\ & = -\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}+\frac {6 a^2 e \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 d}-\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e}+\frac {\left (3 a^2 e^2 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{7 \sqrt {e \cos (c+d x)}} \\ & = -\frac {18 a^2 (e \cos (c+d x))^{5/2}}{35 d e}+\frac {6 a^2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 d \sqrt {e \cos (c+d x)}}+\frac {6 a^2 e \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 d}-\frac {2 (e \cos (c+d x))^{5/2} \left (a^2+a^2 \sin (c+d x)\right )}{7 d e} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.48 \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2 \, dx=-\frac {16 \sqrt [4]{2} a^2 (e \cos (c+d x))^{5/2} \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},\frac {5}{4},\frac {9}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{5 d e (1+\sin (c+d x))^{5/4}} \]

[In]

Integrate[(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^2,x]

[Out]

(-16*2^(1/4)*a^2*(e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[-9/4, 5/4, 9/4, (1 - Sin[c + d*x])/2])/(5*d*e*(1 + S
in[c + d*x])^(5/4))

Maple [A] (verified)

Time = 5.94 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.48

method	result	size
default	\(\frac {2 a^{2} e^{2} \left (80 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+112 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-168 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+84 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-14 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(203\)
parts	\(-\frac {2 a^{2} \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 {4 a^{2} \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 )}{21 \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 {4 a^{2} \left (e \cos \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d e}\)	\(425\)

[In]

int((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

2/35/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^2*e^2*(80*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8-120*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+112*sin(1/2*d*x+1/2*c)^7-168*sin(1/2*d*x+1/2*c)^5+20*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))+84*sin(1/2*d*x+1/2*c)^3-14*sin(1/2*d*x+1/2*c))/d

Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.87 \[ \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2 \, dx=\frac {-15 i \, \sqrt {2} a^{2} e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 i \, \sqrt {2} a^{2} e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (14 \, a^{2} e \cos \left (d x + c\right )^{2} + 5 \, {\left (a^{2} e \cos \left (d x + c\right )^{2} - 3 \, a^{2} e\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{35 \, d} \]

[In]

integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/35*(-15*I*sqrt(2)*a^2*e^(3/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*I*sqrt(2)*a^2*e
^(3/2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(14*a^2*e*cos(d*x + c)^2 + 5*(a^2*e*cos(d
*x + c)^2 - 3*a^2*e)*sin(d*x + c))*sqrt(e*cos(d*x + c)))/d

Sympy [F(-1)]

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

[In]

integrate((e*cos(d*x+c))**(3/2)*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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