\(\int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx\) [221]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.26 (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, 2719} \[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\frac {4 a^5 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}+\frac {6 a^3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d e^4 \sqrt {\cos (c+d x)}}-\frac {6 a^6 (e \cos (c+d x))^{3/2}}{5 d e^5 \left (a^3-a^3 \sin (c+d x)\right )} \]

[In]

Int[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(7/2),x]

[Out]

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

Rule 2719

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 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))^{5/2}}{(a-a \sin (c+d x))^3} \, dx}{e^6} \\ & = \frac {4 a^5 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}-\frac {\left (3 a^4\right ) \int \frac {\sqrt {e \cos (c+d x)}}{a-a \sin (c+d x)} \, dx}{5 e^4} \\ & = \frac {4 a^5 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}-\frac {6 a^4 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))}+\frac {\left (3 a^3\right ) \int \sqrt {e \cos (c+d x)} \, dx}{5 e^4} \\ & = \frac {4 a^5 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}-\frac {6 a^4 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))}+\frac {\left (3 a^3 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 e^4 \sqrt {\cos (c+d x)}} \\ & = \frac {6 a^3 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d e^4 \sqrt {\cos (c+d x)}}+\frac {4 a^5 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}-\frac {6 a^4 (e \cos (c+d x))^{3/2}}{5 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 \[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\frac {4\ 2^{3/4} a^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},-\frac {3}{4},-\frac {1}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{5/4}}{5 d e (e \cos (c+d x))^{5/2}} \]

[In]

Integrate[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(7/2),x]

[Out]

(4*2^(3/4)*a^3*Hypergeometric2F1[-5/4, -3/4, -1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(5/4))/(5*d*e*(e*C
os[c + d*x])^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(329\) vs. \(2(139)=278\).

Time = 7.44 (sec) , antiderivative size = 330, normalized size of antiderivative = 2.60

method	result	size
default	\(-\frac {2 \left (24 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-20 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{5 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{3} d}\)	\(330\)
parts	\(\text {Expression too large to display}\)	\(797\)

[In]

int((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/5/(4*sin(1/2*d*x+1/2*c)^4-4*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^3*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*
c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(2
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+
1/2*c)^2+20*sin(1/2*d*x+1/2*c)^5-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-20*sin(1/2*d*x+1/2*c)^3+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 = 0.11 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.50 \[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=-\frac {3 \, {\left (-i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} + i \, \sqrt {2} a^{3} \cos \left (d x + c\right ) + 2 i \, \sqrt {2} a^{3} + {\left (-i \, \sqrt {2} a^{3} \cos \left (d x + c\right ) - 2 i \, \sqrt {2} a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (i \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} - i \, \sqrt {2} a^{3} \cos \left (d x + c\right ) - 2 i \, \sqrt {2} a^{3} + {\left (i \, \sqrt {2} a^{3} \cos \left (d x + c\right ) + 2 i \, \sqrt {2} a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, a^{3} \cos \left (d x + c\right )^{2} + a^{3} \cos \left (d x + c\right ) - 2 \, a^{3} - {\left (3 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{5 \, {\left (d e^{4} \cos \left (d x + c\right )^{2} - d e^{4} \cos \left (d x + c\right ) - 2 \, d e^{4} + {\left (d e^{4} \cos \left (d x + c\right ) + 2 \, d e^{4}\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(d*x+c))**3/(e*cos(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(7/2), x)

Giac [F]

\[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\int { \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \]

[In]

int((a + a*sin(c + d*x))^3/(e*cos(c + d*x))^(7/2),x)

[Out]

int((a + a*sin(c + d*x))^3/(e*cos(c + d*x))^(7/2), x)