\(\int \frac {x^3}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx\) [970]

Optimal result

Integrand size = 26, antiderivative size = 88 \[ \int \frac {x^3}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}-\frac {(b c+a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 b^{3/2} d^{3/2}} \]

[Out]

-1/2*(a*d+b*c)*arctanh(d^(1/2)*(b*x^2+a)^(1/2)/b^(1/2)/(d*x^2+c)^(1/2))/b^(3/2)/d^(3/2)+1/2*(b*x^2+a)^(1/2)*(d
*x^2+c)^(1/2)/b/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {457, 81, 65, 223, 212} \[ \int \frac {x^3}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}-\frac {(a d+b c) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 b^{3/2} d^{3/2}} \]

[In]

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

[Out]

(Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(2*b*d) - ((b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d
*x^2])])/(2*b^(3/2)*d^(3/2))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

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

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}-\frac {(b c+a d) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{4 b d} \\ & = \frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}-\frac {(b c+a d) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x^2}\right )}{2 b^2 d} \\ & = \frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}-\frac {(b c+a d) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )}{2 b^2 d} \\ & = \frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}-\frac {(b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 b^{3/2} d^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{2 b d}-\frac {(b c+a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{2 b^{3/2} d^{3/2}} \]

[In]

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

[Out]

(Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(2*b*d) - ((b*c + a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x^2])/(Sqrt[b]*Sqrt[c + d
*x^2])])/(2*b^(3/2)*d^(3/2))

Maple [A] (verified)

Time = 3.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.47

method	result	size
risch	\(\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{2 b d}-\frac {\left (a d +b c \right ) \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}{4 b d \sqrt {b d}\, \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(129\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (\frac {\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{2 b d}-\frac {\left (a d +b c \right ) \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right )}{4 b d \sqrt {b d}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\)	\(135\)
default	\(-\frac {\left (a \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) d +b \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) c -2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}\right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{4 d \sqrt {b d}\, b \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}\)	\(172\)

[In]

int(x^3/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/b/d-1/4*(a*d+b*c)/b/d*ln((1/2*a*d+1/2*b*c+b*d*x^2)/(b*d)^(1/2)+(b*d*x^4+(a
*d+b*c)*x^2+a*c)^(1/2))/(b*d)^(1/2)*((b*x^2+a)*(d*x^2+c))^(1/2)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.91 \[ \int \frac {x^3}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\left [\frac {4 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} b d + {\left (b c + a d\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} - 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right )}{8 \, b^{2} d^{2}}, \frac {2 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} b d + {\left (b c + a d\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right )}{4 \, b^{2} d^{2}}\right ] \]

[In]

integrate(x^3/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(4*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*b*d + (b*c + a*d)*sqrt(b*d)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d +
a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 - 4*(2*b*d*x^2 + b*c + a*d)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(b*d)))/(b
^2*d^2), 1/4*(2*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*b*d + (b*c + a*d)*sqrt(-b*d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d
)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-b*d)/(b^2*d^2*x^4 + a*b*c*d + (b^2*c*d + a*b*d^2)*x^2)))/(b^2*d^2)]

Sympy [F]

\[ \int \frac {x^3}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\int \frac {x^{3}}{\sqrt {a + b x^{2}} \sqrt {c + d x^{2}}}\, dx \]

[In]

integrate(x**3/(b*x**2+a)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x^3/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),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(a*d-b*c>0)', see `assume?` for
more detail

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.18 \[ \int \frac {x^3}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\frac {{\left (b c + a d\right )} \log \left ({\left | -\sqrt {b x^{2} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d} + \frac {\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a}}{b d}}{2 \, {\left | b \right |}} \]

[In]

integrate(x^3/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

1/2*((b*c + a*d)*log(abs(-sqrt(b*x^2 + a)*sqrt(b*d) + sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)))/(sqrt(b*d)*d) +
sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d)*sqrt(b*x^2 + a)/(b*d))/abs(b)

Mupad [B] (verification not implemented)

Time = 10.12 (sec) , antiderivative size = 279, normalized size of antiderivative = 3.17 \[ \int \frac {x^3}{\sqrt {a+b x^2} \sqrt {c+d x^2}} \, dx=\frac {\frac {\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )\,\left (a\,d+b\,c\right )}{d^3\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^3\,\left (a\,d+b\,c\right )}{b\,d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^3}-\frac {4\,\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2}{d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}}{\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^4}{{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^4}+\frac {b^2}{d^2}-\frac {2\,b\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2}{d\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}\right )\,\left (a\,d+b\,c\right )}{b^{3/2}\,d^{3/2}} \]

[In]

int(x^3/((a + b*x^2)^(1/2)*(c + d*x^2)^(1/2)),x)

[Out]

((((a + b*x^2)^(1/2) - a^(1/2))*(a*d + b*c))/(d^3*((c + d*x^2)^(1/2) - c^(1/2))) + (((a + b*x^2)^(1/2) - a^(1/
2))^3*(a*d + b*c))/(b*d^2*((c + d*x^2)^(1/2) - c^(1/2))^3) - (4*a^(1/2)*c^(1/2)*((a + b*x^2)^(1/2) - a^(1/2))^
2)/(d^2*((c + d*x^2)^(1/2) - c^(1/2))^2))/(((a + b*x^2)^(1/2) - a^(1/2))^4/((c + d*x^2)^(1/2) - c^(1/2))^4 + b
^2/d^2 - (2*b*((a + b*x^2)^(1/2) - a^(1/2))^2)/(d*((c + d*x^2)^(1/2) - c^(1/2))^2)) - (atanh((d^(1/2)*((a + b*
x^2)^(1/2) - a^(1/2)))/(b^(1/2)*((c + d*x^2)^(1/2) - c^(1/2))))*(a*d + b*c))/(b^(3/2)*d^(3/2))