3.5.82 \(\int (a+a \sin (e+f x)) (c+d \sin (e+f x))^{5/2} \, dx\) [482]

Optimal. Leaf size=290 \[ -\frac {2 a \left (15 c^2+56 c d+25 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 f}-\frac {2 a (5 c+7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac {2 a \left (15 c^3+161 c^2 d+145 c d^2+63 d^3\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{105 d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 a \left (c^2-d^2\right ) \left (15 c^2+56 c d+25 d^2\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{105 d f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-2/35*a*(5*c+7*d)*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/f-2/7*a*cos(f*x+e)*(c+d*sin(f*x+e))^(5/2)/f-2/105*a*(15*c^
2+56*c*d+25*d^2)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f-2/105*a*(15*c^3+161*c^2*d+145*c*d^2+63*d^3)*(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)/d/f/((c+d*sin(f*x+e))/(c+d))^(1/2)+2/105*a*(c^2-d^2)*(15*c^2+56*c*d+25*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)*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)/d/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 290, 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 = {2832, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\frac {2 a \left (15 c^2+56 c d+25 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 f}-\frac {2 a \left (c^2-d^2\right ) \left (15 c^2+56 c d+25 d^2\right ) \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 )}{105 d f \sqrt {c+d \sin (e+f x)}}+\frac {2 a \left (15 c^3+161 c^2 d+145 c d^2+63 d^3\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 )}{105 d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}-\frac {2 a (5 c+7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(-2*a*(15*c^2 + 56*c*d + 25*d^2)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(105*f) - (2*a*(5*c + 7*d)*Cos[e + f*x
]*(c + d*Sin[e + f*x])^(3/2))/(35*f) - (2*a*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/(7*f) + (2*a*(15*c^3 + 16
1*c^2*d + 145*c*d^2 + 63*d^3)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(105*d*f*
Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (2*a*(c^2 - d^2)*(15*c^2 + 56*c*d + 25*d^2)*EllipticF[(e - Pi/2 + f*x)/2
, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(105*d*f*Sqrt[c + d*Sin[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 2832

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

Rubi steps

\begin {align*} \int (a+a \sin (e+f x)) (c+d \sin (e+f x))^{5/2} \, dx &=-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac {2}{7} \int (c+d \sin (e+f x))^{3/2} \left (\frac {1}{2} a (7 c+5 d)+\frac {1}{2} a (5 c+7 d) \sin (e+f x)\right ) \, dx\\ &=-\frac {2 a (5 c+7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac {4}{35} \int \sqrt {c+d \sin (e+f x)} \left (\frac {1}{4} a \left (35 c^2+40 c d+21 d^2\right )+\frac {1}{4} a \left (15 c^2+56 c d+25 d^2\right ) \sin (e+f x)\right ) \, dx\\ &=-\frac {2 a \left (15 c^2+56 c d+25 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 f}-\frac {2 a (5 c+7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac {8}{105} \int \frac {\frac {1}{8} a \left (105 c^3+135 c^2 d+119 c d^2+25 d^3\right )+\frac {1}{8} a \left (15 c^3+161 c^2 d+145 c d^2+63 d^3\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx\\ &=-\frac {2 a \left (15 c^2+56 c d+25 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 f}-\frac {2 a (5 c+7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}-\frac {\left (a \left (c^2-d^2\right ) \left (15 c^2+56 c d+25 d^2\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{105 d}+\frac {\left (a \left (15 c^3+161 c^2 d+145 c d^2+63 d^3\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{105 d}\\ &=-\frac {2 a \left (15 c^2+56 c d+25 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 f}-\frac {2 a (5 c+7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac {\left (a \left (15 c^3+161 c^2 d+145 c d^2+63 d^3\right ) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{105 d \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {\left (a \left (c^2-d^2\right ) \left (15 c^2+56 c d+25 d^2\right ) \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}{105 d \sqrt {c+d \sin (e+f x)}}\\ &=-\frac {2 a \left (15 c^2+56 c d+25 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 f}-\frac {2 a (5 c+7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac {2 a \left (15 c^3+161 c^2 d+145 c d^2+63 d^3\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)}}{105 d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 a \left (c^2-d^2\right ) \left (15 c^2+56 c d+25 d^2\right ) 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}}}{105 d f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 6.54, size = 3531, normalized size = 12.18 \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])^(5/2),x]

[Out]

a*((c^3*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 + Co
t[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 - Ar
cTan[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]]]*Sq
rt[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*Co
s[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[Co
t[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]]))/(7*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) + (23*c^2*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 - 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]]))/(15*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) + (29*c*d^2*Sec[e]*(1 + Sin[e + f*x])*(-((A
ppellF1[-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 + C
ot[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[Co
t[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*Sq
rt[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]]))/(21
*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2) + (3*d^3*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*S
qrt[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*C
sc[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 - Ar
cTan[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]]])/S
qrt[1 + Cot[e]^2])/Sqrt[c + d*Cos[f*x - ArcTan[Cot[e]]]*Sqrt[1 + Cot[e]^2]*Sin[e]]))/(5*f*(Cos[e/2 + (f*x)/2]
+ Sin[e/2 + (f*x)/2])^2) + ((1 + Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]*(-1/210*((180*c^2 + 308*c*d + 115*d^2)
*Cos[e]*Cos[f*x])/f + (d^2*Cos[3*e]*Cos[3*f*x])/(14*f) - (d*(15*c + 7*d)*Cos[2*f*x]*Sin[2*e])/(35*f) + ((180*c
^2 + 308*c*d + 115*d^2)*Sin[e]*Sin[f*x])/(210*f) - (d*(15*c + 7*d)*Cos[2*e]*Sin[2*f*x])/(35*f) - (d^2*Sin[3*e]
*Sin[3*f*x])/(14*f) + (2*(15*c^3 + 161*c^2*d + 145*c*d^2 + 63*d^3)*Tan[e])/(105*d*f)))/(Cos[e/2 + (f*x)/2] + S
in[e/2 + (f*x)/2])^2 + (18*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])))), -((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]])/(7*f*(Cos[e/2 + (f*x)/2] + Sin...

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1314\) vs. \(2(332)=664\).
time = 4.82, size = 1315, normalized size = 4.53

method	result	size
default	\(\text {Expression too large to display}\)	\(1315\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/105*a*(60*c*d^4*sin(f*x+e)^4+90*c^2*d^3*sin(f*x+e)^3+98*c*d^4*sin(f*x+e)^3+45*c^3*d^2*sin(f*x+e)^2+77*c^2*d^
3*sin(f*x+e)^2-35*c*d^4*sin(f*x+e)^2-90*c^2*d^3*sin(f*x+e)-98*c*d^4*sin(f*x+e)+98*((c+d*sin(f*x+e))/(c-d))^(1/
2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((
c-d)/(c+d))^(1/2))*c^2*d^3+145*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e
))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c*d^4+15*d^5*sin(f*x+e)^5+21*d^5
*sin(f*x+e)^4+10*d^5*sin(f*x+e)^3-21*d^5*sin(f*x+e)^2-25*d^5*sin(f*x+e)-77*c^2*d^3-25*c*d^4-45*c^3*d^2+63*((c+
d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin
(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^5-88*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/
2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^5-15*((c+d*
sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f
*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^5-32*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)
*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^2*d^3-176*((c
+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*si
n(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c*d^4-161*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))
^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^4*d+120
*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+
d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^4*d+176*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c
+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^3*d
^2-130*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^(1/2)*Elliptic
E(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^3*d^2)/d^2/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f

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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))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.19, size = 592, normalized size = 2.04 \begin {gather*} -\frac {\sqrt {2} {\left (30 \, a c^{4} + 7 \, a c^{3} d - 115 \, a c^{2} d^{2} - 231 \, a c d^{3} - 75 \, a d^{4}\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 ) + \sqrt {2} {\left (30 \, a c^{4} + 7 \, a c^{3} d - 115 \, a c^{2} d^{2} - 231 \, a c d^{3} - 75 \, a d^{4}\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 \, \sqrt {2} {\left (15 i \, a c^{3} d + 161 i \, a c^{2} d^{2} + 145 i \, a c d^{3} + 63 i \, a d^{4}\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 \, \sqrt {2} {\left (-15 i \, a c^{3} d - 161 i \, a c^{2} d^{2} - 145 i \, a c d^{3} - 63 i \, a d^{4}\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 (15 \, a d^{4} \cos \left (f x + e\right )^{3} - 3 \, {\left (15 \, a c d^{3} + 7 \, a d^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (45 \, a c^{2} d^{2} + 77 \, a c d^{3} + 40 \, a d^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{315 \, d^{2} f} \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))^(5/2),x, algorithm="fricas")

[Out]

-1/315*(sqrt(2)*(30*a*c^4 + 7*a*c^3*d - 115*a*c^2*d^2 - 231*a*c*d^3 - 75*a*d^4)*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) + sqrt(2)*(30*a*c^4 + 7*a*c^3*d - 115*a*c^2*d^2 - 231*a*c*d^3 - 75*a*d^4)*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)*(15*I*a*c^3*d + 161*I*a*c^2*d^2 + 145*I*a*c*d^3 + 63*I*a*d^4)*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*sqrt(2)*(-15*I*a*c^3*d - 161
*I*a*c^2*d^2 - 145*I*a*c*d^3 - 63*I*a*d^4)*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*c
os(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d)) - 6*(15*a*d^4*cos(f*x + e)^3 - 3*(15*a*c*d^3 + 7*a*d^4)*cos(f*x
+ e)*sin(f*x + e) - (45*a*c^2*d^2 + 77*a*c*d^3 + 40*a*d^4)*cos(f*x + e))*sqrt(d*sin(f*x + e) + c))/(d^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int c^{2} \sqrt {c + d \sin {\left (e + f x \right )}}\, dx + \int c^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx + \int d^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx + \int d^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{3}{\left (e + f x \right )}\, dx + \int 2 c d \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx + \int 2 c d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}\, dx\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))**(5/2),x)

[Out]

a*(Integral(c**2*sqrt(c + d*sin(e + f*x)), x) + Integral(c**2*sqrt(c + d*sin(e + f*x))*sin(e + f*x), x) + Inte
gral(d**2*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2, x) + Integral(d**2*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**
3, x) + Integral(2*c*d*sqrt(c + d*sin(e + f*x))*sin(e + f*x), x) + Integral(2*c*d*sqrt(c + d*sin(e + f*x))*sin
(e + f*x)**2, x))

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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))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2} \,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))^(5/2),x)

[Out]

int((a + a*sin(e + f*x))*(c + d*sin(e + f*x))^(5/2), x)

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