3.8.98 \(\int \frac {x}{(a+b x^2)^{5/2} \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=74 \[ \frac {2 d \sqrt {c+d x^2}}{3 \sqrt {a+b x^2} (b c-a d)^2}-\frac {\sqrt {c+d x^2}}{3 \left (a+b x^2\right )^{3/2} (b c-a d)} \]

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Rubi [A]  time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {444, 45, 37} \begin {gather*} \frac {2 d \sqrt {c+d x^2}}{3 \sqrt {a+b x^2} (b c-a d)^2}-\frac {\sqrt {c+d x^2}}{3 \left (a+b x^2\right )^{3/2} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/((a + b*x^2)^(5/2)*Sqrt[c + d*x^2]),x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b x^2\right )^{5/2} \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {c+d x^2}}{3 (b c-a d) \left (a+b x^2\right )^{3/2}}-\frac {d \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx,x,x^2\right )}{3 (b c-a d)}\\ &=-\frac {\sqrt {c+d x^2}}{3 (b c-a d) \left (a+b x^2\right )^{3/2}}+\frac {2 d \sqrt {c+d x^2}}{3 (b c-a d)^2 \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 52, normalized size = 0.70 \begin {gather*} \frac {\sqrt {c+d x^2} \left (3 a d-b c+2 b d x^2\right )}{3 \left (a+b x^2\right )^{3/2} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/((a + b*x^2)^(5/2)*Sqrt[c + d*x^2]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 1.81, size = 65, normalized size = 0.88 \begin {gather*} \frac {\frac {3 d \sqrt {c+d x^2}}{\sqrt {a+b x^2}}-\frac {b \left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^{3/2}}}{3 (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/((a + b*x^2)^(5/2)*Sqrt[c + d*x^2]),x]

[Out]

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

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fricas [B]  time = 1.43, size = 126, normalized size = 1.70 \begin {gather*} \frac {{\left (2 \, b d x^{2} - b c + 3 \, a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{3 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2} + {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{4} + 2 \, {\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.45, size = 129, normalized size = 1.74 \begin {gather*} \frac {4 \, {\left (b^{2} c - a b d - 3 \, {\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 )} \sqrt {b d} b^{2} d}{3 \, {\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 )}^{3} {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

4/3*(b^2*c - a*b*d - 3*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2)*sqrt(b*d)*b^2*d/
((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)^3*abs(b))

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maple [A]  time = 0.01, size = 60, normalized size = 0.81 \begin {gather*} \frac {\sqrt {d \,x^{2}+c}\, \left (2 b d \,x^{2}+3 a d -b c \right )}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*x^2+a)^(5/2)/(d*x^2+c)^(1/2),x)

[Out]

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

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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/(b*x^2+a)^(5/2)/(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?

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mupad [B]  time = 1.34, size = 137, normalized size = 1.85 \begin {gather*} \frac {\sqrt {b\,x^2+a}\,\left (\frac {x^2\,\left (3\,a\,d^2+b\,c\,d\right )}{3\,b^2\,{\left (a\,d-b\,c\right )}^2}-\frac {b\,c^2-3\,a\,c\,d}{3\,b^2\,{\left (a\,d-b\,c\right )}^2}+\frac {2\,d^2\,x^4}{3\,b\,{\left (a\,d-b\,c\right )}^2}\right )}{x^4\,\sqrt {d\,x^2+c}+\frac {a^2\,\sqrt {d\,x^2+c}}{b^2}+\frac {2\,a\,x^2\,\sqrt {d\,x^2+c}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((a + b*x^2)^(5/2)*(c + d*x^2)^(1/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (a + b x^{2}\right )^{\frac {5}{2}} \sqrt {c + d x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x**2+a)**(5/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x/((a + b*x**2)**(5/2)*sqrt(c + d*x**2)), x)

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