3.6.14 \(\int \frac {1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}} \, dx\) [514]
Optimal. Leaf size=326 \[ -\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}-\frac {\left (c^2-5 c d-12 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{3 a^2 (c-d)^3 (c+d) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c-5 d) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{3 a^2 (c-d)^2 f \sqrt {c+d \sin (e+f x)}} \]
[Out]
-1/3*d*(c^2-5*c*d-12*d^2)*cos(f*x+e)/a^2/(c-d)^3/(c+d)/f/(c+d*sin(f*x+e))^(1/2)-1/3*(c-5*d)*cos(f*x+e)/a^2/(c-
d)^2/f/(1+sin(f*x+e))/(c+d*sin(f*x+e))^(1/2)-1/3*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(1/2)+
1/3*(c^2-5*c*d-12*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*P
i+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/a^2/(c-d)^3/(c+d)/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-
1/3*(c-5*d)*(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)/a^2/(c-d)^2/f/(c+d*sin(f*x+e))^(1/2)
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Rubi [A]
time = 0.43, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2845, 3057,
2833, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 f (c-d)^3 (c+d) \sqrt {c+d \sin (e+f x)}}-\frac {\left (c^2-5 c d-12 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 )}{3 a^2 f (c-d)^3 (c+d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) \sqrt {c+d \sin (e+f x)}}+\frac {(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 )}{3 a^2 f (c-d)^2 \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 \sqrt {c+d \sin (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^(3/2)),x]
[Out]
-1/3*(d*(c^2 - 5*c*d - 12*d^2)*Cos[e + f*x])/(a^2*(c - d)^3*(c + d)*f*Sqrt[c + d*Sin[e + f*x]]) - ((c - 5*d)*C
os[e + f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]) - Cos[e + f*x]/(3*(c - d)*f*(a +
a*Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]]) - ((c^2 - 5*c*d - 12*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c +
d)]*Sqrt[c + d*Sin[e + f*x]])/(3*a^2*(c - d)^3*(c + d)*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((c - 5*d)*Ell
ipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*a^2*(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]
Rule 2845
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))
Rule 3057
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])
Rubi steps
\begin {align*} \int \frac {1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}} \, dx &=-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}-\frac {\int \frac {-\frac {1}{2} a (2 c-7 d)-\frac {3}{2} a d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx}{3 a^2 (c-d)}\\ &=-\frac {(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}+\frac {\int \frac {6 a^2 d^2+\frac {1}{2} a^2 (c-5 d) d \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 a^4 (c-d)^2}\\ &=-\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}-\frac {2 \int \frac {-\frac {1}{4} a^2 d^2 (11 c+5 d)+\frac {1}{4} a^2 d \left (c^2-5 c d-12 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 a^4 (c-d)^3 (c+d)}\\ &=-\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}+\frac {(c-5 d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{6 a^2 (c-d)^2}-\frac {\left (c^2-5 c d-12 d^2\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{6 a^2 (c-d)^3 (c+d)}\\ &=-\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}-\frac {\left (\left (c^2-5 c d-12 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}{6 a^2 (c-d)^3 (c+d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left ((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}{6 a^2 (c-d)^2 \sqrt {c+d \sin (e+f x)}}\\ &=-\frac {d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt {c+d \sin (e+f x)}}-\frac {(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt {c+d \sin (e+f x)}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}}-\frac {\left (c^2-5 c d-12 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)}}{3 a^2 (c-d)^3 (c+d) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(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}}}{3 a^2 (c-d)^2 f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 3.28, size = 405, normalized size = 1.24 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left ((c+d \sin (e+f x)) \left (-\frac {2 \left (c^2-5 c d-9 d^2\right )}{c+d}+\frac {2 (c-d) \sin \left (\frac {1}{2} (e+f x)\right )}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}+\frac {-c+d}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {2 (c-6 d) \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}+\frac {6 d^3 \cos (e+f x)}{(c+d) (c+d \sin (e+f x))}\right )+\frac {\left (c^2-5 c d-12 d^2\right ) (c+d \sin (e+f x))-d^2 (11 c+5 d) F\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}}+\left (c^2-5 c d-12 d^2\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c F\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}}}{c+d}\right )}{3 a^2 (c-d)^3 f (1+\sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^(3/2)),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*((c + d*Sin[e + f*x])*((-2*(c^2 - 5*c*d - 9*d^2))/(c + d) + (2*(c - d
)*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (-c + d)/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2
+ (2*(c - 6*d)*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (6*d^3*Cos[e + f*x])/((c + d)*(c + d
*Sin[e + f*x]))) + ((c^2 - 5*c*d - 12*d^2)*(c + d*Sin[e + f*x]) - d^2*(11*c + 5*d)*EllipticF[(-2*e + Pi - 2*f*
x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + (c^2 - 5*c*d - 12*d^2)*((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)])/(c + d)))/(3*a^2*(c - d)^3*f*(1 + Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]])
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1298\) vs.
\(2(368)=736\).
time = 23.01, size = 1299, normalized size = 3.98
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\(\text {Expression too large to display}\) |
\(1299\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
[Out]
(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/a^2*(1/(c-d)*(-1/3/(c-d)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(1+si
n(f*x+e))^2-1/3*(-sin(f*x+e)^2*d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)^2*(c-3*d)/((-d*sin(f*x+e)-c)*(sin(f*x+e)-1
)*(1+sin(f*x+e)))^(1/2)+2*d^2/(3*c^2-6*c*d+3*d^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)*EllipticF(((c+d*sin(f*x+e))/
(c-d))^(1/2),((c-d)/(c+d))^(1/2))-1/3*d*(c-3*d)/(c-d)^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)
)))-d/(c-d)^2*(-(-sin(f*x+e)^2*d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)/((-d*sin(f*x+e)-c)*(sin(f*x+e)-1)*(1+sin(f
*x+e)))^(1/2)-2*d/(2*c-2*d)*(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))-d/(c-d)*(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))))+d^2/(c-d)^2*(2*d*cos(f*x+e)^2/(c
^2-d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*c/(c^2-d^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)*EllipticF(((c+d*
sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+2/(c^2-d^2)*d*(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)*Elliptic
E(((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
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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(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate(1/((a*sin(f*x + e) + a)^2*(d*sin(f*x + e) + c)^(3/2)), x)
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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 0.23, size = 2093, normalized size = 6.42 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
1/18*((sqrt(2)*(2*c^3*d - 10*c^2*d^2 + 9*c*d^3 + 15*d^4)*cos(f*x + e)^3 + sqrt(2)*(2*c^4 - 6*c^3*d - 11*c^2*d^
2 + 33*c*d^3 + 30*d^4)*cos(f*x + e)^2 - sqrt(2)*(2*c^4 - 8*c^3*d - c^2*d^2 + 24*c*d^3 + 15*d^4)*cos(f*x + e) +
(sqrt(2)*(2*c^3*d - 10*c^2*d^2 + 9*c*d^3 + 15*d^4)*cos(f*x + e)^2 - sqrt(2)*(2*c^4 - 8*c^3*d - c^2*d^2 + 24*c
*d^3 + 15*d^4)*cos(f*x + e) - 2*sqrt(2)*(2*c^4 - 8*c^3*d - c^2*d^2 + 24*c*d^3 + 15*d^4))*sin(f*x + e) - 2*sqrt
(2)*(2*c^4 - 8*c^3*d - c^2*d^2 + 24*c*d^3 + 15*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)*(2*c^3*d - 1
0*c^2*d^2 + 9*c*d^3 + 15*d^4)*cos(f*x + e)^3 + sqrt(2)*(2*c^4 - 6*c^3*d - 11*c^2*d^2 + 33*c*d^3 + 30*d^4)*cos(
f*x + e)^2 - sqrt(2)*(2*c^4 - 8*c^3*d - c^2*d^2 + 24*c*d^3 + 15*d^4)*cos(f*x + e) + (sqrt(2)*(2*c^3*d - 10*c^2
*d^2 + 9*c*d^3 + 15*d^4)*cos(f*x + e)^2 - sqrt(2)*(2*c^4 - 8*c^3*d - c^2*d^2 + 24*c*d^3 + 15*d^4)*cos(f*x + e)
- 2*sqrt(2)*(2*c^4 - 8*c^3*d - c^2*d^2 + 24*c*d^3 + 15*d^4))*sin(f*x + e) - 2*sqrt(2)*(2*c^4 - 8*c^3*d - c^2*
d^2 + 24*c*d^3 + 15*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)*(I*c^2*d^2 - 5*I*c*d^3 - 12*I*d^4)*
cos(f*x + e)^3 + sqrt(2)*(I*c^3*d - 3*I*c^2*d^2 - 22*I*c*d^3 - 24*I*d^4)*cos(f*x + e)^2 + sqrt(2)*(-I*c^3*d +
4*I*c^2*d^2 + 17*I*c*d^3 + 12*I*d^4)*cos(f*x + e) + (sqrt(2)*(I*c^2*d^2 - 5*I*c*d^3 - 12*I*d^4)*cos(f*x + e)^2
+ sqrt(2)*(-I*c^3*d + 4*I*c^2*d^2 + 17*I*c*d^3 + 12*I*d^4)*cos(f*x + e) + 2*sqrt(2)*(-I*c^3*d + 4*I*c^2*d^2 +
17*I*c*d^3 + 12*I*d^4))*sin(f*x + e) + 2*sqrt(2)*(-I*c^3*d + 4*I*c^2*d^2 + 17*I*c*d^3 + 12*I*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)*(-I*c^2*d^2 + 5*I*c*d^3 + 12*I*d^4)*cos(f*x + e)^3 + sqrt(2)*(-I*c^3*d + 3*I*c^2*d^2 + 22*I*c*d^3 + 24*I*d^
4)*cos(f*x + e)^2 + sqrt(2)*(I*c^3*d - 4*I*c^2*d^2 - 17*I*c*d^3 - 12*I*d^4)*cos(f*x + e) + (sqrt(2)*(-I*c^2*d^
2 + 5*I*c*d^3 + 12*I*d^4)*cos(f*x + e)^2 + sqrt(2)*(I*c^3*d - 4*I*c^2*d^2 - 17*I*c*d^3 - 12*I*d^4)*cos(f*x + e
) + 2*sqrt(2)*(I*c^3*d - 4*I*c^2*d^2 - 17*I*c*d^3 - 12*I*d^4))*sin(f*x + e) + 2*sqrt(2)*(I*c^3*d - 4*I*c^2*d^2
- 17*I*c*d^3 - 12*I*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)) + 6*(c^3*d - c^2*d^2 - c*d^3 + d^4 - (c^2*d^2 - 5*c*d^3 - 12*d^4)*cos(f*x + e)^3
+ (c^3*d - 4*c^2*d^2 - 6*c*d^3 - 7*d^4)*cos(f*x + e)^2 + 2*(c^3*d - 2*c^2*d^2 - 6*c*d^3 - 9*d^4)*cos(f*x + e)
- (c^3*d - c^2*d^2 - c*d^3 + d^4 - (c^2*d^2 - 5*c*d^3 - 12*d^4)*cos(f*x + e)^2 - (c^3*d - 3*c^2*d^2 - 11*c*d^
3 - 19*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(d*sin(f*x + e) + c))/((a^2*c^4*d^2 - 2*a^2*c^3*d^3 + 2*a^2*c*d^5
- a^2*d^6)*f*cos(f*x + e)^3 + (a^2*c^5*d - 4*a^2*c^3*d^3 + 2*a^2*c^2*d^4 + 3*a^2*c*d^5 - 2*a^2*d^6)*f*cos(f*x
+ e)^2 - (a^2*c^5*d - a^2*c^4*d^2 - 2*a^2*c^3*d^3 + 2*a^2*c^2*d^4 + a^2*c*d^5 - a^2*d^6)*f*cos(f*x + e) - 2*(a
^2*c^5*d - a^2*c^4*d^2 - 2*a^2*c^3*d^3 + 2*a^2*c^2*d^4 + a^2*c*d^5 - a^2*d^6)*f + ((a^2*c^4*d^2 - 2*a^2*c^3*d^
3 + 2*a^2*c*d^5 - a^2*d^6)*f*cos(f*x + e)^2 - (a^2*c^5*d - a^2*c^4*d^2 - 2*a^2*c^3*d^3 + 2*a^2*c^2*d^4 + a^2*c
*d^5 - a^2*d^6)*f*cos(f*x + e) - 2*(a^2*c^5*d - a^2*c^4*d^2 - 2*a^2*c^3*d^3 + 2*a^2*c^2*d^4 + a^2*c*d^5 - a^2*
d^6)*f)*sin(f*x + e))
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{c \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )} + 2 c \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )} + c \sqrt {c + d \sin {\left (e + f x \right )}} + d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{3}{\left (e + f x \right )} + 2 d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )} + d \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )}}\, dx}{a^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))**2/(c+d*sin(f*x+e))**(3/2),x)
[Out]
Integral(1/(c*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2 + 2*c*sqrt(c + d*sin(e + f*x))*sin(e + f*x) + c*sqrt(c
+ d*sin(e + f*x)) + d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**3 + 2*d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2
+ d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)), x)/a**2
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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(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(3/2),x, algorithm="giac")
[Out]
integrate(1/((a*sin(f*x + e) + a)^2*(d*sin(f*x + e) + c)^(3/2)), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/((a + a*sin(e + f*x))^2*(c + d*sin(e + f*x))^(3/2)),x)
[Out]
int(1/((a + a*sin(e + f*x))^2*(c + d*sin(e + f*x))^(3/2)), x)
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