3.44 \(\int \frac {x^3}{(a+b \sin (c+d x^2))^2} \, dx\)

Optimal. Leaf size=324 \[ -\frac {a \text {Li}_2\left (\frac {i b e^{i \left (d x^2+c\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 d^2 \left (a^2-b^2\right )^{3/2}}+\frac {a \text {Li}_2\left (\frac {i b e^{i \left (d x^2+c\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 d^2 \left (a^2-b^2\right )^{3/2}}-\frac {\log \left (a+b \sin \left (c+d x^2\right )\right )}{2 d^2 \left (a^2-b^2\right )}-\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 d \left (a^2-b^2\right )^{3/2}}+\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )}{2 d \left (a^2-b^2\right )^{3/2}}+\frac {b x^2 \cos \left (c+d x^2\right )}{2 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^2\right )\right )} \]

[Out]

-1/2*ln(a+b*sin(d*x^2+c))/(a^2-b^2)/d^2-1/2*I*a*x^2*ln(1-I*b*exp(I*(d*x^2+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(
3/2)/d+1/2*I*a*x^2*ln(1-I*b*exp(I*(d*x^2+c))/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d-1/2*a*polylog(2,I*b*exp(I*
(d*x^2+c))/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/d^2+1/2*a*polylog(2,I*b*exp(I*(d*x^2+c))/(a+(a^2-b^2)^(1/2)))/
(a^2-b^2)^(3/2)/d^2+1/2*b*x^2*cos(d*x^2+c)/(a^2-b^2)/d/(a+b*sin(d*x^2+c))

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Rubi [A]  time = 0.60, antiderivative size = 324, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3379, 3324, 3323, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac {a \text {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 d^2 \left (a^2-b^2\right )^{3/2}}+\frac {a \text {PolyLog}\left (2,\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )}{2 d^2 \left (a^2-b^2\right )^{3/2}}-\frac {\log \left (a+b \sin \left (c+d x^2\right )\right )}{2 d^2 \left (a^2-b^2\right )}-\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 d \left (a^2-b^2\right )^{3/2}}+\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )}{2 d \left (a^2-b^2\right )^{3/2}}+\frac {b x^2 \cos \left (c+d x^2\right )}{2 d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^2\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*Sin[c + d*x^2])^2,x]

[Out]

((-I/2)*a*x^2*Log[1 - (I*b*E^(I*(c + d*x^2)))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) + ((I/2)*a*x^2*Log
[1 - (I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) - Log[a + b*Sin[c + d*x^2]]/(2*(a^2
- b^2)*d^2) - (a*PolyLog[2, (I*b*E^(I*(c + d*x^2)))/(a - Sqrt[a^2 - b^2])])/(2*(a^2 - b^2)^(3/2)*d^2) + (a*Po
lyLog[2, (I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/(2*(a^2 - b^2)^(3/2)*d^2) + (b*x^2*Cos[c + d*x^2])/(2
*(a^2 - b^2)*d*(a + b*Sin[c + d*x^2]))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3324

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(c + d*x)^m*Cos[
e + f*x])/(f*(a^2 - b^2)*(a + b*Sin[e + f*x])), x] + (Dist[a/(a^2 - b^2), Int[(c + d*x)^m/(a + b*Sin[e + f*x])
, x], x] - Dist[(b*d*m)/(f*(a^2 - b^2)), Int[((c + d*x)^(m - 1)*Cos[e + f*x])/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rubi steps

\begin {align*} \int \frac {x^3}{\left (a+b \sin \left (c+d x^2\right )\right )^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{(a+b \sin (c+d x))^2} \, dx,x,x^2\right )\\ &=\frac {b x^2 \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )}+\frac {a \operatorname {Subst}\left (\int \frac {x}{a+b \sin (c+d x)} \, dx,x,x^2\right )}{2 \left (a^2-b^2\right )}-\frac {b \operatorname {Subst}\left (\int \frac {\cos (c+d x)}{a+b \sin (c+d x)} \, dx,x,x^2\right )}{2 \left (a^2-b^2\right ) d}\\ &=\frac {b x^2 \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )}+\frac {a \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx,x,x^2\right )}{a^2-b^2}-\frac {\operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \sin \left (c+d x^2\right )\right )}{2 \left (a^2-b^2\right ) d^2}\\ &=-\frac {\log \left (a+b \sin \left (c+d x^2\right )\right )}{2 \left (a^2-b^2\right ) d^2}+\frac {b x^2 \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )}-\frac {(i a b) \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx,x,x^2\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {(i a b) \operatorname {Subst}\left (\int \frac {e^{i (c+d x)} x}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx,x,x^2\right )}{\left (a^2-b^2\right )^{3/2}}\\ &=-\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d}+\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d}-\frac {\log \left (a+b \sin \left (c+d x^2\right )\right )}{2 \left (a^2-b^2\right ) d^2}+\frac {b x^2 \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )}+\frac {(i a) \operatorname {Subst}\left (\int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx,x,x^2\right )}{2 \left (a^2-b^2\right )^{3/2} d}-\frac {(i a) \operatorname {Subst}\left (\int \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx,x,x^2\right )}{2 \left (a^2-b^2\right )^{3/2} d}\\ &=-\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d}+\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d}-\frac {\log \left (a+b \sin \left (c+d x^2\right )\right )}{2 \left (a^2-b^2\right ) d^2}+\frac {b x^2 \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )}+\frac {a \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{2 \left (a^2-b^2\right )^{3/2} d^2}-\frac {a \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 i b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{2 \left (a^2-b^2\right )^{3/2} d^2}\\ &=-\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d}+\frac {i a x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d}-\frac {\log \left (a+b \sin \left (c+d x^2\right )\right )}{2 \left (a^2-b^2\right ) d^2}-\frac {a \text {Li}_2\left (\frac {i b e^{i \left (c+d x^2\right )}}{a-\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d^2}+\frac {a \text {Li}_2\left (\frac {i b e^{i \left (c+d x^2\right )}}{a+\sqrt {a^2-b^2}}\right )}{2 \left (a^2-b^2\right )^{3/2} d^2}+\frac {b x^2 \cos \left (c+d x^2\right )}{2 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^2\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 0.96, size = 302, normalized size = 0.93 \[ \frac {-\frac {a \text {Li}_2\left (-\frac {i b e^{i \left (d x^2+c\right )}}{\sqrt {a^2-b^2}-a}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {a \text {Li}_2\left (\frac {i b e^{i \left (d x^2+c\right )}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {i a d x^2 \log \left (1+\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}-a}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {i a d x^2 \log \left (1-\frac {i b e^{i \left (c+d x^2\right )}}{\sqrt {a^2-b^2}+a}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {\log \left (a+b \sin \left (c+d x^2\right )\right )}{a^2-b^2}+\frac {b d x^2 \cos \left (c+d x^2\right )}{\left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^2\right )\right )}}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*a*d*x^2*Log[1 + (I*b*E^(I*(c + d*x^2)))/(-a + Sqrt[a^2 - b^2])])/(a^2 - b^2)^(3/2) + (I*a*d*x^2*Log[1 -
(I*b*E^(I*(c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/(a^2 - b^2)^(3/2) - Log[a + b*Sin[c + d*x^2]]/(a^2 - b^2) - (
a*PolyLog[2, ((-I)*b*E^(I*(c + d*x^2)))/(-a + Sqrt[a^2 - b^2])])/(a^2 - b^2)^(3/2) + (a*PolyLog[2, (I*b*E^(I*(
c + d*x^2)))/(a + Sqrt[a^2 - b^2])])/(a^2 - b^2)^(3/2) + (b*d*x^2*Cos[c + d*x^2])/((a^2 - b^2)*(a + b*Sin[c +
d*x^2])))/(2*d^2)

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fricas [B]  time = 1.17, size = 1517, normalized size = 4.68 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*sin(d*x^2+c))^2,x, algorithm="fricas")

[Out]

1/4*(2*(a^2*b - b^3)*d*x^2*cos(d*x^2 + c) + (I*a*b^2*sin(d*x^2 + c) + I*a^2*b)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1
/2*(2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) + 2*(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^
2) + 2*b)/b + 1) + (-I*a*b^2*sin(d*x^2 + c) - I*a^2*b)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(d*x^2 + c)
+ 2*a*sin(d*x^2 + c) - 2*(b*cos(d*x^2 + c) - I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (-I*a
*b^2*sin(d*x^2 + c) - I*a^2*b)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) +
2*(b*cos(d*x^2 + c) + I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (I*a*b^2*sin(d*x^2 + c) + I*
a^2*b)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) - 2*(b*cos(d*x^2 + c) + I
*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) + (a^2*b*d*x^2 + a^2*b*c + (a*b^2*d*x^2 + a*b^2*c)*sin
(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) + 2*(b*cos(d*x^2 + c) -
I*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - (a^2*b*d*x^2 + a^2*b*c + (a*b^2*d*x^2 + a*b^2*c)*sin(d
*x^2 + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) - 2*(b*cos(d*x^2 + c) - I
*b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + (a^2*b*d*x^2 + a^2*b*c + (a*b^2*d*x^2 + a*b^2*c)*sin(d*x
^2 + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) + 2*(b*cos(d*x^2 + c) + I*
b*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - (a^2*b*d*x^2 + a^2*b*c + (a*b^2*d*x^2 + a*b^2*c)*sin(d*x^
2 + c))*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x^2 + c) + 2*a*sin(d*x^2 + c) - 2*(b*cos(d*x^2 + c) + I*b
*sin(d*x^2 + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - (a^3 - a*b^2 + (a^2*b - b^3)*sin(d*x^2 + c) + (a*b^2*c*sin
(d*x^2 + c) + a^2*b*c)*sqrt(-(a^2 - b^2)/b^2))*log(2*b*cos(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2
- b^2)/b^2) + 2*I*a) - (a^3 - a*b^2 + (a^2*b - b^3)*sin(d*x^2 + c) + (a*b^2*c*sin(d*x^2 + c) + a^2*b*c)*sqrt(-
(a^2 - b^2)/b^2))*log(2*b*cos(d*x^2 + c) - 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) - (a^3 -
a*b^2 + (a^2*b - b^3)*sin(d*x^2 + c) - (a*b^2*c*sin(d*x^2 + c) + a^2*b*c)*sqrt(-(a^2 - b^2)/b^2))*log(-2*b*co
s(d*x^2 + c) + 2*I*b*sin(d*x^2 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) - (a^3 - a*b^2 + (a^2*b - b^3)*sin(d
*x^2 + c) - (a*b^2*c*sin(d*x^2 + c) + a^2*b*c)*sqrt(-(a^2 - b^2)/b^2))*log(-2*b*cos(d*x^2 + c) - 2*I*b*sin(d*x
^2 + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a))/((a^4*b - 2*a^2*b^3 + b^5)*d^2*sin(d*x^2 + c) + (a^5 - 2*a^3*b^
2 + a*b^4)*d^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{{\left (b \sin \left (d x^{2} + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*sin(d*x^2+c))^2,x, algorithm="giac")

[Out]

integrate(x^3/(b*sin(d*x^2 + c) + a)^2, x)

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maple [F]  time = 1.02, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\left (a +b \sin \left (d \,x^{2}+c \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a+b*sin(d*x^2+c))^2,x)

[Out]

int(x^3/(a+b*sin(d*x^2+c))^2,x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(a+b*sin(d*x^2+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3}{{\left (a+b\,\sin \left (d\,x^2+c\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + b*sin(c + d*x^2))^2,x)

[Out]

int(x^3/(a + b*sin(c + d*x^2))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(a+b*sin(d*x**2+c))**2,x)

[Out]

Integral(x**3/(a + b*sin(c + d*x**2))**2, x)

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