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

Optimal. Leaf size=318 \[ -\frac {2 a \cos (e+f x)}{5 (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {2 a (3 c-5 d) \cos (e+f x)}{15 (c-d) (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 a \left (3 c^2-20 c d+9 d^2\right ) \cos (e+f x)}{15 (c-d)^2 (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {2 a \left (3 c^2-20 c d+9 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 (c-d)^2 d (c+d)^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 a (3 c-5 d) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 (c-d) d (c+d)^2 f \sqrt {c+d \sin (e+f x)}} \]

[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]
time = 0.34, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2833, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\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}} F\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)}} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {a+a \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx &=-\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) F\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] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 6.79, size = 2815, normalized size = 8.85 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*(((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, ...

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1045\) vs. \(2(360)=720\).
time = 23.27, size = 1046, normalized size = 3.29

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

Verification of antiderivative is not currently implemented for this CAS.

[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-sin(f*x+e))*d/(c-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-sin(f*x+e))*d/(c-
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*d*cos(f*x+e)^2/(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-sin(f*
x+e))/(c+d))^(1/2)*((-1-sin(f*x+e))*d/(c-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-sin(f*x+e))*d/(c-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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.23, size = 1489, normalized size = 4.68 \begin {gather*} \frac {{\left (3 \, \sqrt {2} {\left (6 \, a c^{4} d^{2} + 5 \, a c^{3} d^{3} - 18 \, a c^{2} d^{4} + 15 \, a c d^{5}\right )} \cos \left (f x + e\right )^{2} + {\left (\sqrt {2} {\left (6 \, a c^{3} d^{3} + 5 \, a c^{2} d^{4} - 18 \, a c d^{5} + 15 \, a d^{6}\right )} \cos \left (f x + e\right )^{2} - \sqrt {2} {\left (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}\right )}\right )} \sin \left (f x + e\right ) - \sqrt {2} {\left (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}\right )}\right )} \sqrt {i \, d} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) - 3 i \, d \sin \left (f x + e\right ) - 2 i \, c}{3 \, d}\right ) + {\left (3 \, \sqrt {2} {\left (6 \, a c^{4} d^{2} + 5 \, a c^{3} d^{3} - 18 \, a c^{2} d^{4} + 15 \, a c d^{5}\right )} \cos \left (f x + e\right )^{2} + {\left (\sqrt {2} {\left (6 \, a c^{3} d^{3} + 5 \, a c^{2} d^{4} - 18 \, a c d^{5} + 15 \, a d^{6}\right )} \cos \left (f x + e\right )^{2} - \sqrt {2} {\left (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}\right )}\right )} \sin \left (f x + e\right ) - \sqrt {2} {\left (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}\right )}\right )} \sqrt {-i \, d} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) + 3 i \, d \sin \left (f x + e\right ) + 2 i \, c}{3 \, d}\right ) - 3 \, {\left (3 \, \sqrt {2} {\left (-3 i \, a c^{3} d^{3} + 20 i \, a c^{2} d^{4} - 9 i \, a c d^{5}\right )} \cos \left (f x + e\right )^{2} + {\left (\sqrt {2} {\left (-3 i \, a c^{2} d^{4} + 20 i \, a c d^{5} - 9 i \, a d^{6}\right )} \cos \left (f x + e\right )^{2} + \sqrt {2} {\left (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}\right )}\right )} \sin \left (f x + e\right ) + \sqrt {2} {\left (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}\right )}\right )} \sqrt {i \, d} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (8 i \, c^{3} - 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) - 3 i \, d \sin \left (f x + e\right ) - 2 i \, c}{3 \, d}\right )\right ) - 3 \, {\left (3 \, \sqrt {2} {\left (3 i \, a c^{3} d^{3} - 20 i \, a c^{2} d^{4} + 9 i \, a c d^{5}\right )} \cos \left (f x + e\right )^{2} + {\left (\sqrt {2} {\left (3 i \, a c^{2} d^{4} - 20 i \, a c d^{5} + 9 i \, a d^{6}\right )} \cos \left (f x + e\right )^{2} + \sqrt {2} {\left (-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}\right )}\right )} \sin \left (f x + e\right ) + \sqrt {2} {\left (-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}\right )}\right )} \sqrt {-i \, d} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, c^{2} - 3 \, d^{2}\right )}}{3 \, d^{2}}, -\frac {8 \, {\left (-8 i \, c^{3} + 9 i \, c d^{2}\right )}}{27 \, d^{3}}, \frac {3 \, d \cos \left (f x + e\right ) + 3 i \, d \sin \left (f x + e\right ) + 2 i \, c}{3 \, d}\right )\right ) - 6 \, {\left ({\left (3 \, a c^{2} d^{4} - 20 \, a c d^{5} + 9 \, a d^{6}\right )} \cos \left (f x + e\right )^{3} - {\left (9 \, a c^{3} d^{3} - 45 \, a c^{2} d^{4} + 15 \, a c d^{5} + 5 \, a d^{6}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (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}\right )} \cos \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{45 \, {\left (3 \, {\left (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}\right )} f \cos \left (f x + e\right )^{2} - {\left (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}\right )} f + {\left ({\left (c^{5} d^{5} + c^{4} d^{6} - 2 \, c^{3} d^{7} - 2 \, c^{2} d^{8} + c d^{9} + d^{10}\right )} f \cos \left (f x + e\right )^{2} - {\left (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}\right )} f\right )} \sin \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+a\,\sin \left (e+f\,x\right )}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________