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

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

[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 = 0.44, 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 = {2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 a \left (15 c^2+56 c d+25 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 f}-\frac {2 a \left (c^2-d^2\right ) \left (15 c^2+56 c d+25 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} F\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{105 d f \sqrt {c+d \sin (e+f x)}}+\frac {2 a \left (161 c^2 d+15 c^3+145 c d^2+63 d^3\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{105 d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 a \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}-\frac {2 a (5 c+7 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f} \]

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 2653

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

Rule 2655

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rule 2663

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

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 2753

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)*Simp[
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]  time = 6.76, size = 3531, normalized size = 12.18 \[ \text {Result too large to show} \]

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[e/2 + (f*x)/2])^2*Sqrt[1 + Tan[e]^2]) + (2*c^3*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 + T
an[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]])/(d*f*(Cos[e/2 + (f*x)/2] + Si
n[e/2 + (f*x)/2])^2*Sqrt[1 + Tan[e]^2]) + (34*c*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])))), -((Se
c[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]*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 + A
rcTan[Tan[e]]]*Sqrt[1 + Tan[e]^2]])/(15*f*(Cos[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2*Sqrt[1 + Tan[e]^2]) + (1
0*d^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*S
qrt[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 + ArcTa
n[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]])/(21*f*(C
os[e/2 + (f*x)/2] + Sin[e/2 + (f*x)/2])^2*Sqrt[1 + Tan[e]^2]))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a c^{2} + 2 \, a c d + a d^{2} - {\left (2 \, a c d + a d^{2}\right )} \cos \left (f x + e\right )^{2} - {\left (a d^{2} \cos \left (f x + e\right )^{2} - a c^{2} - 2 \, a c d - a d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}, x\right ) \]

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]

integral((a*c^2 + 2*a*c*d + a*d^2 - (2*a*c*d + a*d^2)*cos(f*x + e)^2 - (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d
- a*d^2)*sin(f*x + e))*sqrt(d*sin(f*x + e) + c), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}\,{d x} \]

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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maple [B]  time = 1.72, size = 1315, normalized size = 4.53 \[ \text {result too large to display} \]

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)

[Out]

2/105*a*(-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)*E
llipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^4*d-25*c*d^4-77*c^2*d^3-45*c^3*d^2-15*((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+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*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)+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*s
in(f*x+e)-98*c*d^4*sin(f*x+e)+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)^3-130*((c+d*si
n(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^3*d^2+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*s
in(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^3*d^2-32*((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*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c*d^4+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)*Ellip
ticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c*d^4)/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 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}\,{d x} \]

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

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

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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