\(\int (3+b \sin (e+f x)) (c+d \sin (e+f x))^{5/2} \, dx\) [723]

Optimal result

Integrand size = 25, antiderivative size = 293 \[ \int (3+b \sin (e+f x)) (c+d \sin (e+f x))^{5/2} \, dx=-\frac {2 \left (15 b c^2+168 c d+25 b d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 f}-\frac {2 (5 b c+21 d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac {2 b \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f}+\frac {2 \left (15 b c^3+483 c^2 d+145 b c d^2+189 d^3\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{105 d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 \left (c^2-d^2\right ) \left (15 b c^2+168 c d+25 b d^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{105 d f \sqrt {c+d \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 298, 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 = {2832, 2831, 2742, 2740, 2734, 2732} \[ \int (3+b \sin (e+f x)) (c+d \sin (e+f x))^{5/2} \, dx=-\frac {2 \left (56 a c d+15 b c^2+25 b d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{105 f}-\frac {2 \left (c^2-d^2\right ) \left (56 a c d+15 b c^2+25 b d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{105 d f \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (161 a c^2 d+63 a d^3+15 b c^3+145 b c d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{105 d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 (7 a d+5 b c) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{35 f}-\frac {2 b \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{7 f} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 1.76 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.89 \[ \int (3+b \sin (e+f x)) (c+d \sin (e+f x))^{5/2} \, dx=\frac {-2 d \left (315 c^3+135 b c^2 d+357 c d^2+25 b d^3\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}-2 \left (15 b c^3+483 c^2 d+145 b c d^2+189 d^3\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}-d \cos (e+f x) (c+d \sin (e+f x)) \left (90 b c^2+462 c d+65 b d^2-15 b d^2 \cos (2 (e+f x))+18 d (5 b c+7 d) \sin (e+f x)\right )}{105 d f \sqrt {c+d \sin (e+f x)}} \]

[In]

Integrate[(3 + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(-2*d*(315*c^3 + 135*b*c^2*d + 357*c*d^2 + 25*b*d^3)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c +
d*Sin[e + f*x])/(c + d)] - 2*(15*b*c^3 + 483*c^2*d + 145*b*c*d^2 + 189*d^3)*((c + d)*EllipticE[(-2*e + Pi - 2
*f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*Sqrt[(c + d*Sin[e + f*x])/(c + d)
] - d*Cos[e + f*x]*(c + d*Sin[e + f*x])*(90*b*c^2 + 462*c*d + 65*b*d^2 - 15*b*d^2*Cos[2*(e + f*x)] + 18*d*(5*b
*c + 7*d)*Sin[e + f*x]))/(105*d*f*Sqrt[c + d*Sin[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1834\) vs. \(2(340)=680\).

Time = 9.67 (sec) , antiderivative size = 1835, normalized size of antiderivative = 6.26

method	result	size
parts	\(\text {Expression too large to display}\)	\(1835\)
default	\(\text {Expression too large to display}\)	\(1839\)

[In]

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

[Out]

2/15*a*(15*c^4*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(c-d))^(1/2)*
EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+8*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1
)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))
*c^3*d-6*c^2*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(c-d))^(1/2)*El
lipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^2-8*c*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+
e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1
/2))*d^3-9*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(c-d))^(1/2)*Elli
pticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*d^4-23*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-
1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)
)*c^4+14*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(c-d))^(1/2)*Ellipt
icE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^2*d^2+9*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)
-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2
))*d^4+3*sin(f*x+e)^4*d^4+14*sin(f*x+e)^3*c*d^3+11*sin(f*x+e)^2*c^2*d^2-3*d^4*sin(f*x+e)^2-14*c*d^3*sin(f*x+e)
-11*c^2*d^2)/d/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f+2/21*b*(3*sin(f*x+e)^5*d^5+3*((c+d*sin(f*x+e))/(c-d))^(1/2)
*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-
d)/(c+d))^(1/2))*c^4*d+24*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(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+2*((c+d*sin(f*x+e))/(c-d))^(1
/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),(
(c-d)/(c+d))^(1/2))*c^2*d^3-24*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+
1)/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c*d^4-5*((c+d*sin(f*x+e))/(c-d))
^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(c-d))^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2
),((c-d)/(c+d))^(1/2))*d^5-3*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)
/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c^5-26*((c+d*sin(f*x+e))/(c-d))^(1
/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+1)/(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+29*((c+d*sin(f*x+e))/(c-d))^(1/2)*(-(sin(f*x+e)-1)*d/(c+d))^(1/2)*(-d*(sin(f*x+e)+
1)/(c-d))^(1/2)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))*c*d^4+12*sin(f*x+e)^4*c*d^4+18*s
in(f*x+e)^3*c^2*d^3+2*d^5*sin(f*x+e)^3+9*sin(f*x+e)^2*c^3*d^2-7*c*d^4*sin(f*x+e)^2-18*sin(f*x+e)*c^2*d^3-5*sin
(f*x+e)*d^5-9*c^3*d^2-5*c*d^4)/d^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 = 0.13 (sec) , antiderivative size = 579, normalized size of antiderivative = 1.98 \[ \int (3+b \sin (e+f x)) (c+d \sin (e+f x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (30 \, b c^{4} + 7 \, a c^{3} d - 115 \, b c^{2} d^{2} - 231 \, a c d^{3} - 75 \, b 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 \, b c^{4} + 7 \, a c^{3} d - 115 \, b c^{2} d^{2} - 231 \, a c d^{3} - 75 \, b 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 \, b c^{3} d + 161 i \, a c^{2} d^{2} + 145 i \, b 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 \, b c^{3} d - 161 i \, a c^{2} d^{2} - 145 i \, b 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 \, b d^{4} \cos \left (f x + e\right )^{3} - 3 \, {\left (15 \, b c d^{3} + 7 \, a d^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (45 \, b c^{2} d^{2} + 77 \, a c d^{3} + 40 \, b d^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c}}{315 \, d^{2} f} \]

[In]

integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int (3+b \sin (e+f x)) (c+d \sin (e+f x))^{5/2} \, dx=\int \left (a + b \sin {\left (e + f x \right )}\right ) \left (c + d \sin {\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx \]

[In]

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

[Out]

Integral((a + b*sin(e + f*x))*(c + d*sin(e + f*x))**(5/2), x)

Maxima [F]

\[ \int (3+b \sin (e+f x)) (c+d \sin (e+f x))^{5/2} \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int (3+b \sin (e+f x)) (c+d \sin (e+f x))^{5/2} \, dx=\int { {\left (b \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (3+b \sin (e+f x)) (c+d \sin (e+f x))^{5/2} \, dx=\int \left (a+b\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

[In]

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

[Out]

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