3.6.49 \(\int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^3} \, dx\) [549]
Optimal. Leaf size=247 \[ -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d)^3 f}+\frac {\sqrt {d} \left (15 c^2+10 c d+7 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 \sqrt {a} (c-d)^3 (c+d)^{5/2} f}+\frac {d \cos (e+f x)}{2 \left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}+\frac {d (7 c+d) \cos (e+f x)}{4 \left (c^2-d^2\right )^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \]
[Out]
-arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/2)/(c-d)^3/f/a^(1/2)+1/4*(15*c^2+10*c*d+7
*d^2)*arctanh(cos(f*x+e)*a^(1/2)*d^(1/2)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))*d^(1/2)/(c-d)^3/(c+d)^(5/2)/f/a^(
1/2)+1/2*d*cos(f*x+e)/(c^2-d^2)/f/(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(1/2)+1/4*d*(7*c+d)*cos(f*x+e)/(c^2-d^2)
^2/f/(c+d*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)
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Rubi [A]
time = 0.51, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2858, 3063,
3064, 2728, 212, 2852, 214} \begin {gather*} \frac {d (7 c+d) \cos (e+f x)}{4 f \left (c^2-d^2\right )^2 \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}+\frac {d \cos (e+f x)}{2 f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^2}+\frac {\sqrt {d} \left (15 c^2+10 c d+7 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{4 \sqrt {a} f (c-d)^3 (c+d)^{5/2}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f (c-d)^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^3),x]
[Out]
-((Sqrt[2]*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*(c - d)^3*f)) + (Sqrt[
d]*(15*c^2 + 10*c*d + 7*d^2)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(
4*Sqrt[a]*(c - d)^3*(c + d)^(5/2)*f) + (d*Cos[e + f*x])/(2*(c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e
+ f*x])^2) + (d*(7*c + d)*Cos[e + f*x])/(4*(c^2 - d^2)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 214
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]
Rule 2728
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Rule 2852
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[-2*(
b/f), Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 2858
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[(-d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] - Dist[
1/(2*b*(n + 1)*(c^2 - d^2)), Int[(c + d*Sin[e + f*x])^(n + 1)*(Simp[a*d - 2*b*c*(n + 1) + b*d*(2*n + 3)*Sin[e
+ f*x], x]/Sqrt[a + b*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Rule 3063
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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Rule 3064
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rubi steps
\begin {align*} \int \frac {1}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^3} \, dx &=\frac {d \cos (e+f x)}{2 \left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}+\frac {\int \frac {a (4 c+d)-3 a d \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx}{4 a \left (c^2-d^2\right )}\\ &=\frac {d \cos (e+f x)}{2 \left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}+\frac {d (7 c+d) \cos (e+f x)}{4 \left (c^2-d^2\right )^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac {\int \frac {-\frac {1}{2} a^2 \left (8 c^2+9 c d+7 d^2\right )+\frac {1}{2} a^2 d (7 c+d) \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx}{4 a^2 \left (c^2-d^2\right )^2}\\ &=\frac {d \cos (e+f x)}{2 \left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}+\frac {d (7 c+d) \cos (e+f x)}{4 \left (c^2-d^2\right )^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac {\int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{(c-d)^3}-\frac {\left (d \left (15 c^2+10 c d+7 d^2\right )\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{8 a (c-d)^3 (c+d)^2}\\ &=\frac {d \cos (e+f x)}{2 \left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}+\frac {d (7 c+d) \cos (e+f x)}{4 \left (c^2-d^2\right )^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac {2 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{(c-d)^3 f}+\frac {\left (d \left (15 c^2+10 c d+7 d^2\right )\right ) \text {Subst}\left (\int \frac {1}{a c+a d-d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{4 (c-d)^3 (c+d)^2 f}\\ &=-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d)^3 f}+\frac {\sqrt {d} \left (15 c^2+10 c d+7 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 \sqrt {a} (c-d)^3 (c+d)^{5/2} f}+\frac {d \cos (e+f x)}{2 \left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}+\frac {d (7 c+d) \cos (e+f x)}{4 \left (c^2-d^2\right )^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 3.41, size = 414, normalized size = 1.68 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\frac {(32+32 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right )}{(c-d)^3}+\frac {\sqrt {d} \left (15 c^2+10 c d+7 d^2\right ) \left (e+f x-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (\sqrt {c+d}+\sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )\right )\right )\right )}{(c-d)^3 (c+d)^{5/2}}+\frac {\sqrt {d} \left (15 c^2+10 c d+7 d^2\right ) \left (e+f x-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (\sqrt {c+d}-\sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )+\sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )\right )\right )\right )}{(-c+d)^3 (c+d)^{5/2}}+\frac {8 d \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c-d) (c+d) (c+d \sin (e+f x))^2}+\frac {4 d (7 c+d) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c-d)^2 (c+d)^2 (c+d \sin (e+f x))}\right )}{16 f \sqrt {a (1+\sin (e+f x))}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[1/(Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^3),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(((32 + 32*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e +
f*x)/4])])/(c - d)^3 + (Sqrt[d]*(15*c^2 + 10*c*d + 7*d^2)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + 2*Log[Sec[(e
+ f*x)/4]^2*(Sqrt[c + d] + Sqrt[d]*Cos[(e + f*x)/2] - Sqrt[d]*Sin[(e + f*x)/2])]))/((c - d)^3*(c + d)^(5/2)) +
(Sqrt[d]*(15*c^2 + 10*c*d + 7*d^2)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + 2*Log[Sec[(e + f*x)/4]^2*(Sqrt[c +
d] - Sqrt[d]*Cos[(e + f*x)/2] + Sqrt[d]*Sin[(e + f*x)/2])]))/((-c + d)^3*(c + d)^(5/2)) + (8*d*(Cos[(e + f*x)/
2] - Sin[(e + f*x)/2]))/((c - d)*(c + d)*(c + d*Sin[e + f*x])^2) + (4*d*(7*c + d)*(Cos[(e + f*x)/2] - Sin[(e +
f*x)/2]))/((c - d)^2*(c + d)^2*(c + d*Sin[e + f*x]))))/(16*f*Sqrt[a*(1 + Sin[e + f*x])])
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1064\) vs.
\(2(214)=428\).
time = 7.18, size = 1065, normalized size = 4.31
| | |
| method |
result |
size |
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| default |
\(\text {Expression too large to display}\) |
\(1065\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^3,x,method=_RETURNVERBOSE)
[Out]
1/4*((-a*(sin(f*x+e)-1))^(3/2)*(a*(c+d)*d)^(1/2)*a^(7/2)*d^4+7*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)
^(1/2))*a^(11/2)*sin(f*x+e)^2*d^5+15*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(11/2)*c^4*d+10*
arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(11/2)*c^3*d^2+7*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/
(a*(c+d)*d)^(1/2))*a^(11/2)*c^2*d^3+(-a*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a^(9/2)*d^4-4*(a*(c+d)*d)^(1/2
)*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)^2*a^5*c^2*d^2-8*(a*(c+d)*d)^(1/2)*
2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)^2*a^5*c*d^3-8*(a*(c+d)*d)^(1/2)*2^(1
/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)*a^5*c^3*d-16*(a*(c+d)*d)^(1/2)*2^(1/2)*a
rctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)*a^5*c^2*d^2-8*(a*(c+d)*d)^(1/2)*2^(1/2)*arcta
nh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)*a^5*c*d^3+10*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/
(a*(c+d)*d)^(1/2))*a^(11/2)*sin(f*x+e)^2*c*d^4+30*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(11
/2)*sin(f*x+e)*c^3*d^2+20*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(11/2)*sin(f*x+e)*c^2*d^3+1
4*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(11/2)*sin(f*x+e)*c*d^4+9*(-a*(sin(f*x+e)-1))^(1/2)
*(a*(c+d)*d)^(1/2)*a^(9/2)*c^3*d-(-a*(sin(f*x+e)-1))^(1/2)*(a*(c+d)*d)^(1/2)*a^(9/2)*c^2*d^2-9*(-a*(sin(f*x+e)
-1))^(1/2)*(a*(c+d)*d)^(1/2)*a^(9/2)*c*d^3-7*(-a*(sin(f*x+e)-1))^(3/2)*(a*(c+d)*d)^(1/2)*a^(7/2)*c^2*d^2+6*(-a
*(sin(f*x+e)-1))^(3/2)*(a*(c+d)*d)^(1/2)*a^(7/2)*c*d^3-4*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)
-1))^(1/2)*2^(1/2)/a^(1/2))*a^5*c^4+15*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(11/2)*sin(f*x
+e)^2*c^2*d^3-8*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*a^5*c^3*d-4*(
a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*a^5*c^2*d^2-4*(a*(c+d)*d)^(1/2
)*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)^2*a^5*d^4)*(-a*(sin(f*x+e)-1))^(1/
2)*(1+sin(f*x+e))/a^(11/2)/(a*(c+d)*d)^(1/2)/(c+d*sin(f*x+e))^2/(c+d)^2/(c-d)^3/cos(f*x+e)/(a+a*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))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
integrate(1/(sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^3), x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1346 vs.
\(2 (224) = 448\).
time = 0.79, size = 2991, normalized size = 12.11 \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))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
[1/16*((15*a*c^4 + 40*a*c^3*d + 42*a*c^2*d^2 + 24*a*c*d^3 + 7*a*d^4 - (15*a*c^2*d^2 + 10*a*c*d^3 + 7*a*d^4)*co
s(f*x + e)^3 - (30*a*c^3*d + 35*a*c^2*d^2 + 24*a*c*d^3 + 7*a*d^4)*cos(f*x + e)^2 + (15*a*c^4 + 10*a*c^3*d + 22
*a*c^2*d^2 + 10*a*c*d^3 + 7*a*d^4)*cos(f*x + e) + (15*a*c^4 + 40*a*c^3*d + 42*a*c^2*d^2 + 24*a*c*d^3 + 7*a*d^4
- (15*a*c^2*d^2 + 10*a*c*d^3 + 7*a*d^4)*cos(f*x + e)^2 + 2*(15*a*c^3*d + 10*a*c^2*d^2 + 7*a*c*d^3)*cos(f*x +
e))*sin(f*x + e))*sqrt(d/(a*c + a*d))*log((d^2*cos(f*x + e)^3 - (6*c*d + 7*d^2)*cos(f*x + e)^2 - c^2 - 2*c*d -
d^2 - 4*((c*d + d^2)*cos(f*x + e)^2 - c^2 - 4*c*d - 3*d^2 - (c^2 + 3*c*d + 2*d^2)*cos(f*x + e) + (c^2 + 4*c*d
+ 3*d^2 + (c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d/(a*c + a*d)) - (c^2 + 8*c*d
+ 9*d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 + 2*(3*c*d + 4*d^2)*cos(f*x + e))*sin(f*x + e
))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*co
s(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 8*sqrt(2)*(a*c^4 + 4*a*c^3*d + 6*a*c^2
*d^2 + 4*a*c*d^3 + a*d^4 - (a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e)^3 - (2*a*c^3*d + 5*a*c^2*d^2 + 4*a*c*d
^3 + a*d^4)*cos(f*x + e)^2 + (a*c^4 + 2*a*c^3*d + 2*a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e) + (a*c^4 + 4*a
*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 - (a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e)^2 + 2*(a*c^3*d + 2*a*c
^2*d^2 + a*c*d^3)*cos(f*x + e))*sin(f*x + e))*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(
2)*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 -
(cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(a) - 4*(9*c^3*d - 15*c^2*d^2 + 3*c*d^3 + 3*d^4 + (7*
c^2*d^2 - 6*c*d^3 - d^4)*cos(f*x + e)^2 + (9*c^3*d - 8*c^2*d^2 - 3*c*d^3 + 2*d^4)*cos(f*x + e) - (9*c^3*d - 15
*c^2*d^2 + 3*c*d^3 + 3*d^4 - (7*c^2*d^2 - 6*c*d^3 - d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))
/((a*c^5*d^2 - a*c^4*d^3 - 2*a*c^3*d^4 + 2*a*c^2*d^5 + a*c*d^6 - a*d^7)*f*cos(f*x + e)^3 + (2*a*c^6*d - a*c^5*
d^2 - 5*a*c^4*d^3 + 2*a*c^3*d^4 + 4*a*c^2*d^5 - a*c*d^6 - a*d^7)*f*cos(f*x + e)^2 - (a*c^7 - a*c^6*d - a*c^5*d
^2 + a*c^4*d^3 - a*c^3*d^4 + a*c^2*d^5 + a*c*d^6 - a*d^7)*f*cos(f*x + e) - (a*c^7 + a*c^6*d - 3*a*c^5*d^2 - 3*
a*c^4*d^3 + 3*a*c^3*d^4 + 3*a*c^2*d^5 - a*c*d^6 - a*d^7)*f + ((a*c^5*d^2 - a*c^4*d^3 - 2*a*c^3*d^4 + 2*a*c^2*d
^5 + a*c*d^6 - a*d^7)*f*cos(f*x + e)^2 - 2*(a*c^6*d - a*c^5*d^2 - 2*a*c^4*d^3 + 2*a*c^3*d^4 + a*c^2*d^5 - a*c*
d^6)*f*cos(f*x + e) - (a*c^7 + a*c^6*d - 3*a*c^5*d^2 - 3*a*c^4*d^3 + 3*a*c^3*d^4 + 3*a*c^2*d^5 - a*c*d^6 - a*d
^7)*f)*sin(f*x + e)), -1/8*((15*a*c^4 + 40*a*c^3*d + 42*a*c^2*d^2 + 24*a*c*d^3 + 7*a*d^4 - (15*a*c^2*d^2 + 10*
a*c*d^3 + 7*a*d^4)*cos(f*x + e)^3 - (30*a*c^3*d + 35*a*c^2*d^2 + 24*a*c*d^3 + 7*a*d^4)*cos(f*x + e)^2 + (15*a*
c^4 + 10*a*c^3*d + 22*a*c^2*d^2 + 10*a*c*d^3 + 7*a*d^4)*cos(f*x + e) + (15*a*c^4 + 40*a*c^3*d + 42*a*c^2*d^2 +
24*a*c*d^3 + 7*a*d^4 - (15*a*c^2*d^2 + 10*a*c*d^3 + 7*a*d^4)*cos(f*x + e)^2 + 2*(15*a*c^3*d + 10*a*c^2*d^2 +
7*a*c*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(-d/(a*c + a*d))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e
) - c - 2*d)*sqrt(-d/(a*c + a*d))/(d*cos(f*x + e))) - 4*sqrt(2)*(a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 +
a*d^4 - (a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e)^3 - (2*a*c^3*d + 5*a*c^2*d^2 + 4*a*c*d^3 + a*d^4)*cos(f*
x + e)^2 + (a*c^4 + 2*a*c^3*d + 2*a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e) + (a*c^4 + 4*a*c^3*d + 6*a*c^2*d
^2 + 4*a*c*d^3 + a*d^4 - (a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e)^2 + 2*(a*c^3*d + 2*a*c^2*d^2 + a*c*d^3)*
cos(f*x + e))*sin(f*x + e))*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(a*sin(f*x
+ e) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)
*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(a) + 2*(9*c^3*d - 15*c^2*d^2 + 3*c*d^3 + 3*d^4 + (7*c^2*d^2 - 6*c*d^3
- d^4)*cos(f*x + e)^2 + (9*c^3*d - 8*c^2*d^2 - 3*c*d^3 + 2*d^4)*cos(f*x + e) - (9*c^3*d - 15*c^2*d^2 + 3*c*d^3
+ 3*d^4 - (7*c^2*d^2 - 6*c*d^3 - d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((a*c^5*d^2 - a*c
^4*d^3 - 2*a*c^3*d^4 + 2*a*c^2*d^5 + a*c*d^6 - a*d^7)*f*cos(f*x + e)^3 + (2*a*c^6*d - a*c^5*d^2 - 5*a*c^4*d^3
+ 2*a*c^3*d^4 + 4*a*c^2*d^5 - a*c*d^6 - a*d^7)*f*cos(f*x + e)^2 - (a*c^7 - a*c^6*d - a*c^5*d^2 + a*c^4*d^3 - a
*c^3*d^4 + a*c^2*d^5 + a*c*d^6 - a*d^7)*f*cos(f*x + e) - (a*c^7 + a*c^6*d - 3*a*c^5*d^2 - 3*a*c^4*d^3 + 3*a*c^
3*d^4 + 3*a*c^2*d^5 - a*c*d^6 - a*d^7)*f + ((a*c^5*d^2 - a*c^4*d^3 - 2*a*c^3*d^4 + 2*a*c^2*d^5 + a*c*d^6 - a*d
^7)*f*cos(f*x + e)^2 - 2*(a*c^6*d - a*c^5*d^2 - 2*a*c^4*d^3 + 2*a*c^3*d^4 + a*c^2*d^5 - a*c*d^6)*f*cos(f*x + e
) - (a*c^7 + a*c^6*d - 3*a*c^5*d^2 - 3*a*c^4*d^3 + 3*a*c^3*d^4 + 3*a*c^2*d^5 - a*c*d^6 - a*d^7)*f)*sin(f*x + e
))]
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))**(1/2)/(c+d*sin(f*x+e))**3,x)
[Out]
Timed out
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 585 vs.
\(2 (224) = 448\).
time = 0.67, size = 585, normalized size = 2.37 \begin {gather*} \frac {\sqrt {2} {\left (\frac {\sqrt {2} {\left (15 \, c^{2} d + 10 \, c d^{2} + 7 \, d^{3}\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{{\left (c^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - c^{4} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, c^{3} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, c^{2} d^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + c d^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - d^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {-c d - d^{2}}} + \frac {4 \, \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, c^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, c d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - d^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {4 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, c^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, c d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - d^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, {\left (14 \, c d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, d^{3} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 9 \, c^{2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 8 \, c d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d^{3} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (c^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, c^{2} d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + d^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} {\left (2 \, d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}^{2}}\right )}}{8 \, \sqrt {a} f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^3,x, algorithm="giac")
[Out]
1/8*sqrt(2)*(sqrt(2)*(15*c^2*d + 10*c*d^2 + 7*d^3)*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)/sqrt(-c*d -
d^2))/((c^5*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - c^4*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*c^3*d^2*sgn(c
os(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*c^2*d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + c*d^4*sgn(cos(-1/4*pi + 1/2*f
*x + 1/2*e)) - d^5*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(-c*d - d^2)) + 4*log(sin(-1/4*pi + 1/2*f*x + 1/2*
e) + 1)/(c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*c*d^2*sgn(c
os(-1/4*pi + 1/2*f*x + 1/2*e)) - d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 4*log(-sin(-1/4*pi + 1/2*f*x + 1/2
*e) + 1)/(c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*c^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*c*d^2*sgn(
cos(-1/4*pi + 1/2*f*x + 1/2*e)) - d^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + 2*(14*c*d^2*sin(-1/4*pi + 1/2*f*x
+ 1/2*e)^3 + 2*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 9*c^2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 8*c*d^2*sin(-1
/4*pi + 1/2*f*x + 1/2*e) + d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((c^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*c
^2*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + d^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*(2*d*sin(-1/4*pi + 1/2*f
*x + 1/2*e)^2 - c - d)^2))/(sqrt(a)*f)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{\sqrt {a+a\,\sin \left (e+f\,x\right )}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/((a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^3),x)
[Out]
int(1/((a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^3), x)
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