\(\int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx\) [488]

Optimal result

Integrand size = 25, antiderivative size = 313 \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx=-\frac {6 \cos (e+f x)}{5 (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {2 (3 c-5 d) \cos (e+f x)}{5 (c-d) (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 \left (3 c^2-20 c d+9 d^2\right ) \cos (e+f x)}{5 (c-d)^2 (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {2 \left (3 c^2-20 c d+9 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{5 (c-d)^2 d (c+d)^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 (3 c-5 d) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{5 (c-d) d (c+d)^2 f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-2/5*a*cos(f*x+e)/(c+d)/f/(c+d*sin(f*x+e))^(5/2)-2/15*a*(3*c-5*d)*cos(f*x+e)/(c-d)/(c+d)^2/f/(c+d*sin(f*x+e))^
(3/2)-2/15*a*(3*c^2-20*c*d+9*d^2)*cos(f*x+e)/(c-d)^2/(c+d)^3/f/(c+d*sin(f*x+e))^(1/2)+2/15*a*(3*c^2-20*c*d+9*d
^2)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*
(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/(c-d)^2/d/(c+d)^3/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-2/15*a*(3*c-5*d)*(s
in(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+
d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/(c-d)/d/(c+d)^2/f/(c+d*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2833, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx=-\frac {2 a \left (3 c^2-20 c d+9 d^2\right ) \cos (e+f x)}{15 f (c-d)^2 (c+d)^3 \sqrt {c+d \sin (e+f x)}}-\frac {2 a \left (3 c^2-20 c d+9 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d f (c-d)^2 (c+d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 a (3 c-5 d) \cos (e+f x)}{15 f (c-d) (c+d)^2 (c+d \sin (e+f x))^{3/2}}-\frac {2 a \cos (e+f x)}{5 f (c+d) (c+d \sin (e+f x))^{5/2}}+\frac {2 a (3 c-5 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{15 d f (c-d) (c+d)^2 \sqrt {c+d \sin (e+f x)}} \]

[In]

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

[Out]

(-2*a*Cos[e + f*x])/(5*(c + d)*f*(c + d*Sin[e + f*x])^(5/2)) - (2*a*(3*c - 5*d)*Cos[e + f*x])/(15*(c - d)*(c +
d)^2*f*(c + d*Sin[e + f*x])^(3/2)) - (2*a*(3*c^2 - 20*c*d + 9*d^2)*Cos[e + f*x])/(15*(c - d)^2*(c + d)^3*f*Sq
rt[c + d*Sin[e + f*x]]) - (2*a*(3*c^2 - 20*c*d + 9*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c +
d*Sin[e + f*x]])/(15*(c - d)^2*d*(c + d)^3*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (2*a*(3*c - 5*d)*EllipticF[
(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*(c - d)*d*(c + d)^2*f*Sqrt[c + d*Si
n[e + f*x]])

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2831

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2833

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b
^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a \cos (e+f x)}{5 (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \int \frac {-\frac {5}{2} a (c-d)-\frac {3}{2} a (c-d) \sin (e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 \left (c^2-d^2\right )} \\ & = -\frac {2 a \cos (e+f x)}{5 (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {2 a (3 c-5 d) \cos (e+f x)}{15 (c-d) (c+d)^2 f (c+d \sin (e+f x))^{3/2}}+\frac {4 \int \frac {\frac {3}{4} a (5 c-3 d) (c-d)+\frac {1}{4} a (3 c-5 d) (c-d) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 \left (c^2-d^2\right )^2} \\ & = -\frac {2 a \cos (e+f x)}{5 (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {2 a (3 c-5 d) \cos (e+f x)}{15 (c-d) (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 a \left (3 c^2-20 c d+9 d^2\right ) \cos (e+f x)}{15 (c-d)^2 (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {8 \int \frac {-\frac {1}{8} a (c-d) \left (15 c^2-12 c d+5 d^2\right )+\frac {1}{8} a (c-d) \left (3 c^2-20 c d+9 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 \left (c^2-d^2\right )^3} \\ & = -\frac {2 a \cos (e+f x)}{5 (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {2 a (3 c-5 d) \cos (e+f x)}{15 (c-d) (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 a \left (3 c^2-20 c d+9 d^2\right ) \cos (e+f x)}{15 (c-d)^2 (c+d)^3 f \sqrt {c+d \sin (e+f x)}}+\frac {(a (3 c-5 d)) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 (c-d) d (c+d)^2}-\frac {\left (a \left (3 c^2-20 c d+9 d^2\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{15 (c-d)^2 d (c+d)^3} \\ & = -\frac {2 a \cos (e+f x)}{5 (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {2 a (3 c-5 d) \cos (e+f x)}{15 (c-d) (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 a \left (3 c^2-20 c d+9 d^2\right ) \cos (e+f x)}{15 (c-d)^2 (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (a \left (3 c^2-20 c d+9 d^2\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{15 (c-d)^2 d (c+d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (a (3 c-5 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right ) \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}} \, dx}{15 (c-d) d (c+d)^2 \sqrt {c+d \sin (e+f x)}} \\ & = -\frac {2 a \cos (e+f x)}{5 (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {2 a (3 c-5 d) \cos (e+f x)}{15 (c-d) (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 a \left (3 c^2-20 c d+9 d^2\right ) \cos (e+f x)}{15 (c-d)^2 (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {2 a \left (3 c^2-20 c d+9 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 (c-d)^2 d (c+d)^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 a (3 c-5 d) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 (c-d) d (c+d)^2 f \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 7.42 (sec) , antiderivative size = 2815, normalized size of antiderivative = 8.99 \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

3*(((1 + Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]*((-2*(3*c^2 - 20*c*d + 9*d^2)*Csc[e]*Sec[e])/(15*(c - d)^2*d*(
c + d)^3*f) + (2*Csc[e]*(c*Cos[e] + d*Sin[f*x]))/(5*d*(c + d)*f*(c + d*Sin[e + f*x])^3) - (2*Csc[e]*(5*c*Cos[e
] - 3*d*Cos[e] - 3*c*Sin[f*x] + 5*d*Sin[f*x]))/(15*(c - d)*(c + d)^2*f*(c + d*Sin[e + f*x])^2) - (2*Csc[e]*(15
*c^2*Cos[e] - 12*c*d*Cos[e] + 5*d^2*Cos[e] - 3*c^2*Sin[f*x] + 20*c*d*Sin[f*x] - 9*d^2*Sin[f*x]))/(15*(c - d)^2
*(c + d)^3*f*(c + d*Sin[e + f*x]))))/(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2 - (c^2*Sec[e]*(1 + Sin[e + f*
x])*(-((AppellF1[-1/2, -1/2, -1/2, 1/2, -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))
/(d*Sqrt[1 + Cot[e]^2]*(1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2])))), -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*S
qrt[1 + Cot[e]^2]*Sin[e]))/(d*Sqrt[1 + Cot[e]^2]*(-1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2]))))]*Cot[e]*Sin[f*x -
ArcTan[Cot[e]]])/(Sqrt[1 + Cot[e]^2]*Sqrt[(d*Sqrt[1 + Cot[e]^2] + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^
2])/(d*Sqrt[1 + Cot[e]^2] - c*Csc[e])]*Sqrt[(d*Sqrt[1 + Cot[e]^2] - d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e
]^2])/(d*Sqrt[1 + Cot[e]^2] + c*Csc[e])]*Sqrt[c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]])) - (
(2*d*Sin[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d^2*Cos[e]^2 + d^2*Sin[e]^2) - (Cot[
e]*Sin[f*x - ArcTan[Cot[e]]])/Sqrt[1 + Cot[e]^2])/Sqrt[c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[
e]]))/(5*(c - d)^2*(c + d)^3*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) + (4*c*d*Sec[e]*(1 + Sin[e + f*x])
*(-((AppellF1[-1/2, -1/2, -1/2, 1/2, -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d
*Sqrt[1 + Cot[e]^2]*(1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2])))), -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt
[1 + Cot[e]^2]*Sin[e]))/(d*Sqrt[1 + Cot[e]^2]*(-1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2]))))]*Cot[e]*Sin[f*x - Arc
Tan[Cot[e]]])/(Sqrt[1 + Cot[e]^2]*Sqrt[(d*Sqrt[1 + Cot[e]^2] + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2])
/(d*Sqrt[1 + Cot[e]^2] - c*Csc[e])]*Sqrt[(d*Sqrt[1 + Cot[e]^2] - d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2
])/(d*Sqrt[1 + Cot[e]^2] + c*Csc[e])]*Sqrt[c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]])) - ((2*
d*Sin[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d^2*Cos[e]^2 + d^2*Sin[e]^2) - (Cot[e]*
Sin[f*x - ArcTan[Cot[e]]])/Sqrt[1 + Cot[e]^2])/Sqrt[c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]]
))/(3*(c - d)^2*(c + d)^3*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) - (3*d^2*Sec[e]*(1 + Sin[e + f*x])*(-
((AppellF1[-1/2, -1/2, -1/2, 1/2, -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d*Sq
rt[1 + Cot[e]^2]*(1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2])))), -((Csc[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1
+ Cot[e]^2]*Sin[e]))/(d*Sqrt[1 + Cot[e]^2]*(-1 - (c*Csc[e])/(d*Sqrt[1 + Cot[e]^2]))))]*Cot[e]*Sin[f*x - ArcTan
[Cot[e]]])/(Sqrt[1 + Cot[e]^2]*Sqrt[(d*Sqrt[1 + Cot[e]^2] + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2])/(d
*Sqrt[1 + Cot[e]^2] - c*Csc[e])]*Sqrt[(d*Sqrt[1 + Cot[e]^2] - d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2])/
(d*Sqrt[1 + Cot[e]^2] + c*Csc[e])]*Sqrt[c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]])) - ((2*d*S
in[e]*(c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]))/(d^2*Cos[e]^2 + d^2*Sin[e]^2) - (Cot[e]*Sin
[f*x - ArcTan[Cot[e]]])/Sqrt[1 + Cot[e]^2])/Sqrt[c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]]))/
(5*(c - d)^2*(c + d)^3*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) - (8*c*AppellF1[1/2, 1/2, 1/2, 3/2, -((S
ec[e]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]))/(d*Sqrt[1 + Tan[e]^2]*(1 - (c*Sec[e])/(d*Sq
rt[1 + Tan[e]^2])))), -((Sec[e]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]))/(d*Sqrt[1 + Tan[e
]^2]*(-1 - (c*Sec[e])/(d*Sqrt[1 + Tan[e]^2]))))]*Sec[e]*Sec[f*x + ArcTan[Tan[e]]]*(1 + Sin[e + f*x])*Sqrt[(d*S
qrt[1 + Tan[e]^2] - d*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2])/(c*Sec[e] + d*Sqrt[1 + Tan[e]^2])]*Sqrt[(d
*Sqrt[1 + Tan[e]^2] + d*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2])/(-(c*Sec[e]) + d*Sqrt[1 + Tan[e]^2])]*Sq
rt[c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]])/(5*(c - d)^2*(c + d)^3*f*(Cos[e/2 + (f*x)/2] +
Sin[e/2 + (f*x)/2])^2*Sqrt[1 + Tan[e]^2]) + (2*c^2*AppellF1[1/2, 1/2, 1/2, 3/2, -((Sec[e]*(c + d*Cos[e]*Sin[f*
x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]))/(d*Sqrt[1 + Tan[e]^2]*(1 - (c*Sec[e])/(d*Sqrt[1 + Tan[e]^2])))), -((S
ec[e]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]))/(d*Sqrt[1 + Tan[e]^2]*(-1 - (c*Sec[e])/(d*S
qrt[1 + Tan[e]^2]))))]*Sec[e]*Sec[f*x + ArcTan[Tan[e]]]*(1 + Sin[e + f*x])*Sqrt[(d*Sqrt[1 + Tan[e]^2] - d*Sin[
f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2])/(c*Sec[e] + d*Sqrt[1 + Tan[e]^2])]*Sqrt[(d*Sqrt[1 + Tan[e]^2] + d*Si
n[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2])/(-(c*Sec[e]) + d*Sqrt[1 + Tan[e]^2])]*Sqrt[c + d*Cos[e]*Sin[f*x +
ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]])/((c - d)^2*d*(c + d)^3*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2*Sqrt
[1 + Tan[e]^2]) + (2*d*AppellF1[1/2, 1/2, 1/2, 3/2, -((Sec[e]*(c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 +
Tan[e]^2]))/(d*Sqrt[1 + Tan[e]^2]*(1 - (c*Sec[e])/(d*Sqrt[1 + Tan[e]^2])))), -((Sec[e]*(c + d*Cos[e]*Sin[f*x
+ ArcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]))/(d*Sqrt[1 + Tan[e]^2]*(-1 - (c*Sec[e])/(d*Sqrt[1 + Tan[e]^2]))))]*Sec[e
]*Sec[f*x + ArcTan[Tan[e]]]*(1 + Sin[e + f*x])*Sqrt[(d*Sqrt[1 + Tan[e]^2] - d*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1
+ Tan[e]^2])/(c*Sec[e] + d*Sqrt[1 + Tan[e]^2])]*Sqrt[(d*Sqrt[1 + Tan[e]^2] + d*Sin[f*x + ArcTan[Tan[e]]]*Sqrt
[1 + Tan[e]^2])/(-(c*Sec[e]) + d*Sqrt[1 + Tan[e]^2])]*Sqrt[c + d*Cos[e]*Sin[f*x + ArcTan[Tan[e]]]*Sqrt[1 + Tan
[e]^2]])/(3*(c - d)^2*(c + d)^3*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2*Sqrt[1 + Tan[e]^2]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1045\) vs. \(2(360)=720\).

Time = 9.15 (sec) , antiderivative size = 1046, normalized size of antiderivative = 3.34

method	result	size
default	\(\text {Expression too large to display}\)	\(1046\)
parts	\(\text {Expression too large to display}\)	\(1628\)

[In]

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

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*a*(1/d*(2/3/(c^2-d^2)/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f
*x+e)+c/d)^2+8/3*d*cos(f*x+e)^2/(c^2-d^2)^2*c/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*(3*c^2+d^2)/(3*c^4-6*c
^2*d^2+3*d^4)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(1/(c-d)*(-sin(f*x+e)-1)*d
)^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+
8/3*c*d/(c^2-d^2)^2*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(1/(c-d)*(-sin(f*x+e
)-1)*d)^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d
)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))+(-c+d)/d*(2/5/(c^2-d^2)/d^2*(-
(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^3+16/15*c/(c^2-d^2)^2/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^
2)^(1/2)/(sin(f*x+e)+c/d)^2+2/15*cos(f*x+e)^2*d/(c^2-d^2)^3*(23*c^2+9*d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(
1/2)+2*(15*c^3+17*c*d^2)/(15*c^6-45*c^4*d^2+45*c^2*d^4-15*d^6)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-si
n(f*x+e))/(c+d))^(1/2)*(1/(c-d)*(-sin(f*x+e)-1)*d)^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c
+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+2/15*d*(23*c^2+9*d^2)/(c^2-d^2)^3*(c/d-1)*((c+d*sin(f*x+e))/(
c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(1/(c-d)*(-sin(f*x+e)-1)*d)^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)
^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-
d))^(1/2),((c-d)/(c+d))^(1/2)))))/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f

Fricas [C] (verification not implemented)

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

Time = 0.17 (sec) , antiderivative size = 1461, normalized size of antiderivative = 4.67 \[ \int \frac {3+3 \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/45*((3*sqrt(2)*(6*a*c^4*d^2 + 5*a*c^3*d^3 - 18*a*c^2*d^4 + 15*a*c*d^5)*cos(f*x + e)^2 + (sqrt(2)*(6*a*c^3*d^
3 + 5*a*c^2*d^4 - 18*a*c*d^5 + 15*a*d^6)*cos(f*x + e)^2 - sqrt(2)*(18*a*c^5*d + 15*a*c^4*d^2 - 48*a*c^3*d^3 +
50*a*c^2*d^4 - 18*a*c*d^5 + 15*a*d^6))*sin(f*x + e) - sqrt(2)*(6*a*c^6 + 5*a*c^5*d + 30*a*c^3*d^3 - 54*a*c^2*d
^4 + 45*a*c*d^5))*sqrt(I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3
*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d) + (3*sqrt(2)*(6*a*c^4*d^2 + 5*a*c^3*d^3 - 18*a*c^2*d^4 + 1
5*a*c*d^5)*cos(f*x + e)^2 + (sqrt(2)*(6*a*c^3*d^3 + 5*a*c^2*d^4 - 18*a*c*d^5 + 15*a*d^6)*cos(f*x + e)^2 - sqrt
(2)*(18*a*c^5*d + 15*a*c^4*d^2 - 48*a*c^3*d^3 + 50*a*c^2*d^4 - 18*a*c*d^5 + 15*a*d^6))*sin(f*x + e) - sqrt(2)*
(6*a*c^6 + 5*a*c^5*d + 30*a*c^3*d^3 - 54*a*c^2*d^4 + 45*a*c*d^5))*sqrt(-I*d)*weierstrassPInverse(-4/3*(4*c^2 -
3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d) - 3*(3*s
qrt(2)*(-3*I*a*c^3*d^3 + 20*I*a*c^2*d^4 - 9*I*a*c*d^5)*cos(f*x + e)^2 + (sqrt(2)*(-3*I*a*c^2*d^4 + 20*I*a*c*d^
5 - 9*I*a*d^6)*cos(f*x + e)^2 + sqrt(2)*(9*I*a*c^4*d^2 - 60*I*a*c^3*d^3 + 30*I*a*c^2*d^4 - 20*I*a*c*d^5 + 9*I*
a*d^6))*sin(f*x + e) + sqrt(2)*(3*I*a*c^5*d - 20*I*a*c^4*d^2 + 18*I*a*c^3*d^3 - 60*I*a*c^2*d^4 + 27*I*a*c*d^5)
)*sqrt(I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, weierstrassPInverse(-4/
3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d)
) - 3*(3*sqrt(2)*(3*I*a*c^3*d^3 - 20*I*a*c^2*d^4 + 9*I*a*c*d^5)*cos(f*x + e)^2 + (sqrt(2)*(3*I*a*c^2*d^4 - 20*
I*a*c*d^5 + 9*I*a*d^6)*cos(f*x + e)^2 + sqrt(2)*(-9*I*a*c^4*d^2 + 60*I*a*c^3*d^3 - 30*I*a*c^2*d^4 + 20*I*a*c*d
^5 - 9*I*a*d^6))*sin(f*x + e) + sqrt(2)*(-3*I*a*c^5*d + 20*I*a*c^4*d^2 - 18*I*a*c^3*d^3 + 60*I*a*c^2*d^4 - 27*
I*a*c*d^5))*sqrt(-I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, weierstrass
PInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e
) + 2*I*c)/d)) - 6*((3*a*c^2*d^4 - 20*a*c*d^5 + 9*a*d^6)*cos(f*x + e)^3 - (9*a*c^3*d^3 - 45*a*c^2*d^4 + 15*a*c
*d^5 + 5*a*d^6)*cos(f*x + e)*sin(f*x + e) - (9*a*c^4*d^2 - 25*a*c^3*d^3 + 3*a*c^2*d^4 - 15*a*c*d^5 + 12*a*d^6)
*cos(f*x + e))*sqrt(d*sin(f*x + e) + c))/(3*(c^6*d^4 + c^5*d^5 - 2*c^4*d^6 - 2*c^3*d^7 + c^2*d^8 + c*d^9)*f*co
s(f*x + e)^2 - (c^8*d^2 + c^7*d^3 + c^6*d^4 + c^5*d^5 - 5*c^4*d^6 - 5*c^3*d^7 + 3*c^2*d^8 + 3*c*d^9)*f + ((c^5
*d^5 + c^4*d^6 - 2*c^3*d^7 - 2*c^2*d^8 + c*d^9 + d^10)*f*cos(f*x + e)^2 - (3*c^7*d^3 + 3*c^6*d^4 - 5*c^5*d^5 -
5*c^4*d^6 + c^3*d^7 + c^2*d^8 + c*d^9 + d^10)*f)*sin(f*x + e))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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