3.8.94 \(\int \frac {x^3}{(a+b x^2)^{3/2} \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=83 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{b^{3/2} \sqrt {d}}+\frac {a \sqrt {c+d x^2}}{b \sqrt {a+b x^2} (b c-a d)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {446, 78, 63, 217, 206} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{b^{3/2} \sqrt {d}}+\frac {a \sqrt {c+d x^2}}{b \sqrt {a+b x^2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

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 78

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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 446

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

________________________________________________________________________________________

Mathematica [A]  time = 0.52, size = 118, normalized size = 1.42 \begin {gather*} \frac {\frac {a b \left (c+d x^2\right )}{\sqrt {a+b x^2} (b c-a d)}+\frac {\sqrt {b c-a d} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )}{\sqrt {d}}}{b^2 \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 1.75, size = 83, normalized size = 1.00 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {d} \sqrt {a+b x^2}}\right )}{b^{3/2} \sqrt {d}}+\frac {a \sqrt {c+d x^2}}{b \sqrt {a+b x^2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.44, size = 367, normalized size = 4.42 \begin {gather*} \left [\frac {4 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} a b d + {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x^{2}\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 )}{4 \, {\left (a b^{3} c d - a^{2} b^{2} d^{2} + {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{2}\right )}}, \frac {2 \, \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} a b d - {\left (a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x^{2}\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 )}{2 \, {\left (a b^{3} c d - a^{2} b^{2} d^{2} + {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*a*b*d + (a*b*c - a^2*d + (b^2*c - a*b*d)*x^2)*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)*sq
rt(d*x^2 + c)*sqrt(b*d)))/(a*b^3*c*d - a^2*b^2*d^2 + (b^4*c*d - a*b^3*d^2)*x^2), 1/2*(2*sqrt(b*x^2 + a)*sqrt(d
*x^2 + c)*a*b*d - (a*b*c - a^2*d + (b^2*c - a*b*d)*x^2)*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)))/(a*b^3*c*d - a^2*b^2*d^
2 + (b^4*c*d - a*b^3*d^2)*x^2)]

________________________________________________________________________________________

giac [B]  time = 0.48, size = 135, normalized size = 1.63 \begin {gather*} \frac {\frac {4 \, \sqrt {b d} a b}{{\left (b^{2} c - a b d - {\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 )}^{2}\right )} {\left | b \right |}} - \frac {\sqrt {b d} \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 )}^{2}\right )}{d {\left | b \right |}}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*sqrt(b*d)*a*b/((b^2*c - a*b*d - (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)*
abs(b)) - sqrt(b*d)*log((sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)/(d*abs(b)))/b

________________________________________________________________________________________

maple [B]  time = 0.03, size = 320, normalized size = 3.86 \begin {gather*} \frac {\left (a b d \,x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-b^{2} c \,x^{2} \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+a^{2} d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-a b c \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {x^{4} b d +a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, a \right ) \sqrt {d \,x^{2}+c}}{2 \sqrt {b \,x^{2}+a}\, \sqrt {b d}\, \left (a d -b c \right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b*x^2+a)^(3/2)/(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 details)Is a*d-b*c zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{{\left (b\,x^2+a\right )}^{3/2}\,\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________