\(\int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx\) [729]

Optimal result

Integrand size = 25, antiderivative size = 362 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx=-\frac {2 (b c-3 d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \left (3 b c^2-24 c d+5 b d^2\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 \left (3 b c^3-69 c^2 d+29 b c d^2-27 d^3\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}-\frac {2 \left (3 b c^3-69 c^2 d+29 b c d^2-27 d^3\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{15 d \left (c^2-d^2\right )^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 \left (3 b c^2-24 c d+5 b d^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{15 d \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-2/5*(-a*d+b*c)*cos(f*x+e)/(c^2-d^2)/f/(c+d*sin(f*x+e))^(5/2)-2/15*(-8*a*c*d+3*b*c^2+5*b*d^2)*cos(f*x+e)/(c^2-
d^2)^2/f/(c+d*sin(f*x+e))^(3/2)-2/15*(-23*a*c^2*d-9*a*d^3+3*b*c^3+29*b*c*d^2)*cos(f*x+e)/(c^2-d^2)^3/f/(c+d*si
n(f*x+e))^(1/2)+2/15*(-23*a*c^2*d-9*a*d^3+3*b*c^3+29*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/(c^2-d^2)^
3/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-2/15*(-8*a*c*d+3*b*c^2+5*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/(c^2-d^2)^2/f/(c+d*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 369, 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} \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx=-\frac {2 \left (-8 a c d+3 b c^2+5 b d^2\right ) \cos (e+f x)}{15 f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))^{3/2}}-\frac {2 (b c-a d) \cos (e+f x)}{5 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{5/2}}+\frac {2 \left (-8 a c d+3 b c^2+5 b d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{15 d f \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}}-\frac {2 \left (-23 a c^2 d-9 a d^3+3 b c^3+29 b c d^2\right ) \cos (e+f x)}{15 f \left (c^2-d^2\right )^3 \sqrt {c+d \sin (e+f x)}}-\frac {2 \left (-23 a c^2 d-9 a d^3+3 b c^3+29 b c d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d f \left (c^2-d^2\right )^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}} \]

[In]

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

[Out]

(-2*(b*c - a*d)*Cos[e + f*x])/(5*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^(5/2)) - (2*(3*b*c^2 - 8*a*c*d + 5*b*d^2)*
Cos[e + f*x])/(15*(c^2 - d^2)^2*f*(c + d*Sin[e + f*x])^(3/2)) - (2*(3*b*c^3 - 23*a*c^2*d + 29*b*c*d^2 - 9*a*d^
3)*Cos[e + f*x])/(15*(c^2 - d^2)^3*f*Sqrt[c + d*Sin[e + f*x]]) - (2*(3*b*c^3 - 23*a*c^2*d + 29*b*c*d^2 - 9*a*d
^3)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(15*d*(c^2 - d^2)^3*f*Sqrt[(c + d*S
in[e + f*x])/(c + d)]) + (2*(3*b*c^2 - 8*a*c*d + 5*b*d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c
+ d*Sin[e + f*x])/(c + d)])/(15*d*(c^2 - d^2)^2*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 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 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \int \frac {-\frac {5}{2} (a c-b d)-\frac {3}{2} (b c-a d) \sin (e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 \left (c^2-d^2\right )} \\ & = -\frac {2 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \left (3 b c^2-8 a c d+5 b d^2\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}+\frac {4 \int \frac {\frac {3}{4} \left (5 a c^2-8 b c d+3 a d^2\right )+\frac {1}{4} \left (3 b c^2-8 a c d+5 b d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 \left (c^2-d^2\right )^2} \\ & = -\frac {2 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \left (3 b c^2-8 a c d+5 b d^2\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 \left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}-\frac {8 \int \frac {\frac {1}{8} \left (-15 a c^3+27 b c^2 d-17 a c d^2+5 b d^3\right )+\frac {1}{8} \left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 \left (c^2-d^2\right )^3} \\ & = -\frac {2 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \left (3 b c^2-8 a c d+5 b d^2\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 \left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (3 b c^2-8 a c d+5 b d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d \left (c^2-d^2\right )^2}-\frac {\left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{15 d \left (c^2-d^2\right )^3} \\ & = -\frac {2 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \left (3 b c^2-8 a c d+5 b d^2\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 \left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (\left (3 b c^3-23 a c^2 d+29 b c d^2-9 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}{15 d \left (c^2-d^2\right )^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (\left (3 b c^2-8 a c d+5 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}{15 d \left (c^2-d^2\right )^2 \sqrt {c+d \sin (e+f x)}} \\ & = -\frac {2 (b c-a d) \cos (e+f x)}{5 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{5/2}}-\frac {2 \left (3 b c^2-8 a c d+5 b d^2\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))^{3/2}}-\frac {2 \left (3 b c^3-23 a c^2 d+29 b c d^2-9 a d^3\right ) \cos (e+f x)}{15 \left (c^2-d^2\right )^3 f \sqrt {c+d \sin (e+f x)}}-\frac {2 \left (3 b c^3-23 a c^2 d+29 b c d^2-9 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)}}{15 d \left (c^2-d^2\right )^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 \left (3 b c^2-8 a c d+5 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}}}{15 d \left (c^2-d^2\right )^2 f \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.65 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.79 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx=\frac {2 \left (\frac {\left (\left (3 b c^3-69 c^2 d+29 b c d^2-27 d^3\right ) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-(c-d) \left (3 b c^2-24 c d+5 b d^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right ) \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{5/2}}{(c-d)^3 d}+\frac {\cos (e+f x) \left (-9 b c^5+102 c^4 d-25 b c^3 d^2-15 c^2 d^3+2 b c d^4+9 d^5+d \left (-9 b c^4+162 c^3 d-60 b c^2 d^2+30 c d^3+5 b d^4\right ) \sin (e+f x)+d^2 \left (-3 b c^3+69 c^2 d-29 b c d^2+27 d^3\right ) \sin ^2(e+f x)\right )}{\left (c^2-d^2\right )^3}\right )}{15 f (c+d \sin (e+f x))^{5/2}} \]

[In]

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

[Out]

(2*((((3*b*c^3 - 69*c^2*d + 29*b*c*d^2 - 27*d^3)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - (c - d)*(3*
b*c^2 - 24*c*d + 5*b*d^2)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*((c + d*Sin[e + f*x])/(c + d))^(5/2
))/((c - d)^3*d) + (Cos[e + f*x]*(-9*b*c^5 + 102*c^4*d - 25*b*c^3*d^2 - 15*c^2*d^3 + 2*b*c*d^4 + 9*d^5 + d*(-9
*b*c^4 + 162*c^3*d - 60*b*c^2*d^2 + 30*c*d^3 + 5*b*d^4)*Sin[e + f*x] + d^2*(-3*b*c^3 + 69*c^2*d - 29*b*c*d^2 +
27*d^3)*Sin[e + f*x]^2))/(c^2 - d^2)^3))/(15*f*(c + d*Sin[e + f*x])^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1048\) vs. \(2(411)=822\).

Time = 24.20 (sec) , antiderivative size = 1049, normalized size of antiderivative = 2.90

method	result	size
default	\(\text {Expression too large to display}\)	\(1049\)
parts	\(\text {Expression too large to display}\)	\(1628\)

[In]

int((a+b*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)*(b/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))))+(a*d-b*c)/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-s
in(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 = 0.20 (sec) , antiderivative size = 1573, normalized size of antiderivative = 4.35 \[ \int \frac {3+b \sin (e+f x)}{(c+d \sin (e+f x))^{7/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/45*((3*sqrt(2)*(6*b*c^5*d^2 - a*c^4*d^3 - 23*b*c^3*d^4 + 33*a*c^2*d^5 - 15*b*c*d^6)*cos(f*x + e)^2 + (sqrt(2
)*(6*b*c^4*d^3 - a*c^3*d^4 - 23*b*c^2*d^5 + 33*a*c*d^6 - 15*b*d^7)*cos(f*x + e)^2 - sqrt(2)*(18*b*c^6*d - 3*a*
c^5*d^2 - 63*b*c^4*d^3 + 98*a*c^3*d^4 - 68*b*c^2*d^5 + 33*a*c*d^6 - 15*b*d^7))*sin(f*x + e) - sqrt(2)*(6*b*c^7
- a*c^6*d - 5*b*c^5*d^2 + 30*a*c^4*d^3 - 84*b*c^3*d^4 + 99*a*c^2*d^5 - 45*b*c*d^6))*sqrt(I*d)*weierstrassPInv
erse(-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*b*c^5*d^2 - a*c^4*d^3 - 23*b*c^3*d^4 + 33*a*c^2*d^5 - 15*b*c*d^6)*cos(f*x + e)^2 + (s
qrt(2)*(6*b*c^4*d^3 - a*c^3*d^4 - 23*b*c^2*d^5 + 33*a*c*d^6 - 15*b*d^7)*cos(f*x + e)^2 - sqrt(2)*(18*b*c^6*d -
3*a*c^5*d^2 - 63*b*c^4*d^3 + 98*a*c^3*d^4 - 68*b*c^2*d^5 + 33*a*c*d^6 - 15*b*d^7))*sin(f*x + e) - sqrt(2)*(6*
b*c^7 - a*c^6*d - 5*b*c^5*d^2 + 30*a*c^4*d^3 - 84*b*c^3*d^4 + 99*a*c^2*d^5 - 45*b*c*d^6))*sqrt(-I*d)*weierstra
ssPInverse(-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*b*c^4*d^3 - 23*I*a*c^3*d^4 + 29*I*b*c^2*d^5 - 9*I*a*c*d^6)*cos(f*x + e)^2
+ (sqrt(2)*(3*I*b*c^3*d^4 - 23*I*a*c^2*d^5 + 29*I*b*c*d^6 - 9*I*a*d^7)*cos(f*x + e)^2 + sqrt(2)*(-9*I*b*c^5*d^
2 + 69*I*a*c^4*d^3 - 90*I*b*c^3*d^4 + 50*I*a*c^2*d^5 - 29*I*b*c*d^6 + 9*I*a*d^7))*sin(f*x + e) + sqrt(2)*(-3*I
*b*c^6*d + 23*I*a*c^5*d^2 - 38*I*b*c^4*d^3 + 78*I*a*c^3*d^4 - 87*I*b*c^2*d^5 + 27*I*a*c*d^6))*sqrt(I*d)*weiers
trassZeta(-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*b*c^4*d^3 + 23*I*a*c^3*d^4 - 29*I*b*c^2*d^5 + 9*I*a*c*d^6)*cos(f*x + e)^2 + (sqrt(2)*(-3*I*b*c^3*d^4 + 23
*I*a*c^2*d^5 - 29*I*b*c*d^6 + 9*I*a*d^7)*cos(f*x + e)^2 + sqrt(2)*(9*I*b*c^5*d^2 - 69*I*a*c^4*d^3 + 90*I*b*c^3
*d^4 - 50*I*a*c^2*d^5 + 29*I*b*c*d^6 - 9*I*a*d^7))*sin(f*x + e) + sqrt(2)*(3*I*b*c^6*d - 23*I*a*c^5*d^2 + 38*I
*b*c^4*d^3 - 78*I*a*c^3*d^4 + 87*I*b*c^2*d^5 - 27*I*a*c*d^6))*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)) - 6*((3*b*c^3*d^4 - 23*a*c^2*d^5 + 29*b*c*d^6
- 9*a*d^7)*cos(f*x + e)^3 - (9*b*c^4*d^3 - 54*a*c^3*d^4 + 60*b*c^2*d^5 - 10*a*c*d^6 - 5*b*d^7)*cos(f*x + e)*s
in(f*x + e) - (9*b*c^5*d^2 - 34*a*c^4*d^3 + 28*b*c^3*d^4 - 18*a*c^2*d^5 + 27*b*c*d^6 - 12*a*d^7)*cos(f*x + e))
*sqrt(d*sin(f*x + e) + c))/(3*(c^7*d^4 - 3*c^5*d^6 + 3*c^3*d^8 - c*d^10)*f*cos(f*x + e)^2 - (c^9*d^2 - 6*c^5*d
^6 + 8*c^3*d^8 - 3*c*d^10)*f + ((c^6*d^5 - 3*c^4*d^7 + 3*c^2*d^9 - d^11)*f*cos(f*x + e)^2 - (3*c^8*d^3 - 8*c^6
*d^5 + 6*c^4*d^7 - d^11)*f)*sin(f*x + e))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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