\(\int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx\) [222]

Optimal result

Integrand size = 25, antiderivative size = 127 \[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=-\frac {2 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d e^4 \sqrt {e \cos (c+d x)}}+\frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^6 \sqrt {e \cos (c+d x)}}{21 d e^5 \left (a^3-a^3 \sin (c+d x)\right )} \]

[Out]

-2/21*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^4/(e*cos(d*x+c))^(1/2)+4/7*a^5*(e*cos(d*x+c))^(1/2)/d/e^5/(a-a*sin(d*x+c))^2-2/21*a^6*(e*cos(d*x+c))^(1
/2)/d/e^5/(a^3-a^3*sin(d*x+c))

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 127, 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 = {2749, 2759, 2762, 2721, 2720} \[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d e^4 \sqrt {e \cos (c+d x)}}-\frac {2 a^6 \sqrt {e \cos (c+d x)}}{21 d e^5 \left (a^3-a^3 \sin (c+d x)\right )} \]

[In]

Int[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(9/2),x]

[Out]

(-2*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*e^4*Sqrt[e*Cos[c + d*x]]) + (4*a^5*Sqrt[e*Cos[c +
d*x]])/(7*d*e^5*(a - a*Sin[c + d*x])^2) - (2*a^6*Sqrt[e*Cos[c + d*x]])/(21*d*e^5*(a^3 - a^3*Sin[c + d*x]))

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 2749

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a/g)^
(2*m), Int[(g*Cos[e + f*x])^(2*m + p)/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 -
b^2, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]

Rule 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2762

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

Rubi steps \begin{align*} \text {integral}& = \frac {a^6 \int \frac {(e \cos (c+d x))^{3/2}}{(a-a \sin (c+d x))^3} \, dx}{e^6} \\ & = \frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {a^4 \int \frac {1}{\sqrt {e \cos (c+d x)} (a-a \sin (c+d x))} \, dx}{7 e^4} \\ & = \frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^4 \sqrt {e \cos (c+d x)}}{21 d e^5 (a-a \sin (c+d x))}-\frac {a^3 \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{21 e^4} \\ & = \frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^4 \sqrt {e \cos (c+d x)}}{21 d e^5 (a-a \sin (c+d x))}-\frac {\left (a^3 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 e^4 \sqrt {e \cos (c+d x)}} \\ & = -\frac {2 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d e^4 \sqrt {e \cos (c+d x)}}+\frac {4 a^5 \sqrt {e \cos (c+d x)}}{7 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^4 \sqrt {e \cos (c+d x)}}{21 d e^5 (a-a \sin (c+d x))} \\ \end{align*}

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 = 0.52 \[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\frac {4 \sqrt [4]{2} a^3 \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},-\frac {1}{4},-\frac {3}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{7/4}}{7 d e (e \cos (c+d x))^{7/2}} \]

[In]

Integrate[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(9/2),x]

[Out]

(4*2^(1/4)*a^3*Hypergeometric2F1[-7/4, -1/4, -3/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(7/4))/(7*d*e*(e*C
os[c + d*x])^(7/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(139)=278\).

Time = 7.94 (sec) , antiderivative size = 401, normalized size of antiderivative = 3.16

method	result	size
default	\(\frac {2 \left (8 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+28 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-22 \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 )-28 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{21 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \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}\, e^{4} d}\)	\(401\)
parts	\(\text {Expression too large to display}\)	\(860\)

[In]

int((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(9/2),x,method=_RETURNVERBOSE)

[Out]

2/21/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*
d*x+1/2*c)^2*e+e)^(1/2)/e^4*(8*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^6+8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(sin(1/2*d*x+1/2*c)^2)^
(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-8*sin(1/2*d*
x+1/2*c)^4*cos(1/2*d*x+1/2*c)+6*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*
d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+28*sin(1/2*d*x+1/2*c)^5-22*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-
(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))-28*sin(1/2
*d*x+1/2*c)^3-5*sin(1/2*d*x+1/2*c))*a^3/d

Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.47 \[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\frac {{\left (i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} a^{3} \sin \left (d x + c\right ) - 2 i \, \sqrt {2} a^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (-i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} a^{3} \sin \left (d x + c\right ) + 2 i \, \sqrt {2} a^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (a^{3} \sin \left (d x + c\right ) + 5 \, a^{3}\right )} \sqrt {e \cos \left (d x + c\right )}}{21 \, {\left (d e^{5} \cos \left (d x + c\right )^{2} + 2 \, d e^{5} \sin \left (d x + c\right ) - 2 \, d e^{5}\right )}} \]

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(9/2),x, algorithm="fricas")

[Out]

1/21*((I*sqrt(2)*a^3*cos(d*x + c)^2 + 2*I*sqrt(2)*a^3*sin(d*x + c) - 2*I*sqrt(2)*a^3)*sqrt(e)*weierstrassPInve
rse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (-I*sqrt(2)*a^3*cos(d*x + c)^2 - 2*I*sqrt(2)*a^3*sin(d*x + c) + 2*
I*sqrt(2)*a^3)*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(a^3*sin(d*x + c) + 5*a^3
)*sqrt(e*cos(d*x + c)))/(d*e^5*cos(d*x + c)^2 + 2*d*e^5*sin(d*x + c) - 2*d*e^5)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(d*x+c))**3/(e*cos(d*x+c))**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(9/2), x)

Giac [F]

\[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(9/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(9/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{9/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}} \,d x \]

[In]

int((a + a*sin(c + d*x))^3/(e*cos(c + d*x))^(9/2),x)

[Out]

int((a + a*sin(c + d*x))^3/(e*cos(c + d*x))^(9/2), x)