3.988 \(\int \frac {x}{(a+b x^2)^{7/2} \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=113 \[ -\frac {8 d^2 \sqrt {c+d x^2}}{15 \sqrt {a+b x^2} (b c-a d)^3}+\frac {4 d \sqrt {c+d x^2}}{15 \left (a+b x^2\right )^{3/2} (b c-a d)^2}-\frac {\sqrt {c+d x^2}}{5 \left (a+b x^2\right )^{5/2} (b c-a d)} \]

[Out]

-1/5*(d*x^2+c)^(1/2)/(-a*d+b*c)/(b*x^2+a)^(5/2)+4/15*d*(d*x^2+c)^(1/2)/(-a*d+b*c)^2/(b*x^2+a)^(3/2)-8/15*d^2*(
d*x^2+c)^(1/2)/(-a*d+b*c)^3/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {444, 45, 37} \[ -\frac {8 d^2 \sqrt {c+d x^2}}{15 \sqrt {a+b x^2} (b c-a d)^3}+\frac {4 d \sqrt {c+d x^2}}{15 \left (a+b x^2\right )^{3/2} (b c-a d)^2}-\frac {\sqrt {c+d x^2}}{5 \left (a+b x^2\right )^{5/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[x/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

-Sqrt[c + d*x^2]/(5*(b*c - a*d)*(a + b*x^2)^(5/2)) + (4*d*Sqrt[c + d*x^2])/(15*(b*c - a*d)^2*(a + b*x^2)^(3/2)
) - (8*d^2*Sqrt[c + d*x^2])/(15*(b*c - a*d)^3*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 )^{7/2} \sqrt {c+d x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{7/2} \sqrt {c+d x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {c+d x^2}}{5 (b c-a d) \left (a+b x^2\right )^{5/2}}-\frac {(2 d) \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{5/2} \sqrt {c+d x}} \, dx,x,x^2\right )}{5 (b c-a d)}\\ &=-\frac {\sqrt {c+d x^2}}{5 (b c-a d) \left (a+b x^2\right )^{5/2}}+\frac {4 d \sqrt {c+d x^2}}{15 (b c-a d)^2 \left (a+b x^2\right )^{3/2}}+\frac {\left (4 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx,x,x^2\right )}{15 (b c-a d)^2}\\ &=-\frac {\sqrt {c+d x^2}}{5 (b c-a d) \left (a+b x^2\right )^{5/2}}+\frac {4 d \sqrt {c+d x^2}}{15 (b c-a d)^2 \left (a+b x^2\right )^{3/2}}-\frac {8 d^2 \sqrt {c+d x^2}}{15 (b c-a d)^3 \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 83, normalized size = 0.73 \[ -\frac {\sqrt {c+d x^2} \left (15 a^2 d^2-10 a b d \left (c-2 d x^2\right )+b^2 \left (3 c^2-4 c d x^2+8 d^2 x^4\right )\right )}{15 \left (a+b x^2\right )^{5/2} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x/((a + b*x^2)^(7/2)*Sqrt[c + d*x^2]),x]

[Out]

-1/15*(Sqrt[c + d*x^2]*(15*a^2*d^2 - 10*a*b*d*(c - 2*d*x^2) + b^2*(3*c^2 - 4*c*d*x^2 + 8*d^2*x^4)))/((b*c - a*
d)^3*(a + b*x^2)^(5/2))

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fricas [B]  time = 1.24, size = 259, normalized size = 2.29 \[ -\frac {{\left (8 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} - 10 \, a b c d + 15 \, a^{2} d^{2} - 4 \, {\left (b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{15 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{6} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{4} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

-1/15*(8*b^2*d^2*x^4 + 3*b^2*c^2 - 10*a*b*c*d + 15*a^2*d^2 - 4*(b^2*c*d - 5*a*b*d^2)*x^2)*sqrt(b*x^2 + a)*sqrt
(d*x^2 + c)/(a^3*b^3*c^3 - 3*a^4*b^2*c^2*d + 3*a^5*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*
d^2 - a^3*b^3*d^3)*x^6 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^4*b^2*d^3)*x^4 + 3*(a^2*b^4*c^3
- 3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b*d^3)*x^2)

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giac [B]  time = 0.48, size = 243, normalized size = 2.15 \[ -\frac {16 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 5 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} b^{2} c + 5 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2} a b d + 10 \, {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{4}\right )} \sqrt {b d} b^{3} d^{2}}{15 \, {\left (b^{2} c - a b d - {\left (\sqrt {b x^{2} + a} \sqrt {b d} - \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d}\right )}^{2}\right )}^{5} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

-16/15*(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 5*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*
b*d))^2*b^2*c + 5*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b*d + 10*(sqrt(b*x^2
+ a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4)*sqrt(b*d)*b^3*d^2/((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)^5*abs(b))

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maple [A]  time = 0.01, size = 113, normalized size = 1.00 \[ \frac {\sqrt {d \,x^{2}+c}\, \left (8 b^{2} d^{2} x^{4}+20 a b \,d^{2} x^{2}-4 b^{2} c d \,x^{2}+15 a^{2} d^{2}-10 a b c d +3 b^{2} c^{2}\right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (a^{3} d^{3}-3 a^{2} c \,d^{2} b +3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*x^2+a)^(7/2)/(d*x^2+c)^(1/2),x)

[Out]

1/15*(d*x^2+c)^(1/2)*(8*b^2*d^2*x^4+20*a*b*d^2*x^2-4*b^2*c*d*x^2+15*a^2*d^2-10*a*b*c*d+3*b^2*c^2)/(b*x^2+a)^(5
/2)/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)

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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/(b*x^2+a)^(7/2)/(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 = 1.47, size = 216, normalized size = 1.91 \[ \frac {\sqrt {b\,x^2+a}\,\left (\frac {15\,a^2\,c\,d^2-10\,a\,b\,c^2\,d+3\,b^2\,c^3}{15\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {8\,d^3\,x^6}{15\,b\,{\left (a\,d-b\,c\right )}^3}+\frac {x^2\,\left (15\,a^2\,d^3+10\,a\,b\,c\,d^2-b^2\,c^2\,d\right )}{15\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {4\,d^2\,x^4\,\left (5\,a\,d+b\,c\right )}{15\,b^2\,{\left (a\,d-b\,c\right )}^3}\right )}{x^6\,\sqrt {d\,x^2+c}+\frac {a^3\,\sqrt {d\,x^2+c}}{b^3}+\frac {3\,a\,x^4\,\sqrt {d\,x^2+c}}{b}+\frac {3\,a^2\,x^2\,\sqrt {d\,x^2+c}}{b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((a + b*x^2)^(7/2)*(c + d*x^2)^(1/2)),x)

[Out]

((a + b*x^2)^(1/2)*((3*b^2*c^3 + 15*a^2*c*d^2 - 10*a*b*c^2*d)/(15*b^3*(a*d - b*c)^3) + (8*d^3*x^6)/(15*b*(a*d
- b*c)^3) + (x^2*(15*a^2*d^3 - b^2*c^2*d + 10*a*b*c*d^2))/(15*b^3*(a*d - b*c)^3) + (4*d^2*x^4*(5*a*d + b*c))/(
15*b^2*(a*d - b*c)^3)))/(x^6*(c + d*x^2)^(1/2) + (a^3*(c + d*x^2)^(1/2))/b^3 + (3*a*x^4*(c + d*x^2)^(1/2))/b +
(3*a^2*x^2*(c + d*x^2)^(1/2))/b^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (a + b x^{2}\right )^{\frac {7}{2}} \sqrt {c + d x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x**2+a)**(7/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(x/((a + b*x**2)**(7/2)*sqrt(c + d*x**2)), x)

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