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

Optimal. Leaf size=91 \[ -\frac {(b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2}}+\frac {\sqrt {c+d x^2} (b c-a d)}{b^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b} \]

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Rubi [A]  time = 0.08, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {444, 50, 63, 208} \begin {gather*} \frac {\sqrt {c+d x^2} (b c-a d)}{b^2}-\frac {(b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2}}+\frac {\left (c+d x^2\right )^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*Sqrt[c + d*x^2])/b^2 + (c + d*x^2)^(3/2)/(3*b) - ((b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x
^2])/Sqrt[b*c - a*d]])/b^(5/2)

Rule 50

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

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 208

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

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

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Mathematica [A]  time = 0.09, size = 83, normalized size = 0.91 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-3 a d+4 b c+b d x^2\right )}{3 b^2}-\frac {(b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(4*b*c - 3*a*d + b*d*x^2))/(3*b^2) - ((b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqr
t[b*c - a*d]])/b^(5/2)

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IntegrateAlgebraic [A]  time = 0.13, size = 93, normalized size = 1.02 \begin {gather*} \frac {\sqrt {c+d x^2} \left (-3 a d+4 b c+b d x^2\right )}{3 b^2}-\frac {(a d-b c)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(4*b*c - 3*a*d + b*d*x^2))/(3*b^2) - ((-(b*c) + a*d)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]
*Sqrt[c + d*x^2])/(b*c - a*d)])/b^(5/2)

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fricas [A]  time = 1.62, size = 303, normalized size = 3.33 \begin {gather*} \left [-\frac {3 \, {\left (b c - a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (b d x^{2} + 4 \, b c - 3 \, a d\right )} \sqrt {d x^{2} + c}}{12 \, b^{2}}, -\frac {3 \, {\left (b c - a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (b d x^{2} + 4 \, b c - 3 \, a d\right )} \sqrt {d x^{2} + c}}{6 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*(b*c - a*d)*sqrt((b*c - a*d)/b)*log((b^2*d^2*x^4 + 8*b^2*c^2 - 8*a*b*c*d + a^2*d^2 + 2*(4*b^2*c*d -
3*a*b*d^2)*x^2 + 4*(b^2*d*x^2 + 2*b^2*c - a*b*d)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/b))/(b^2*x^4 + 2*a*b*x^2 + a
^2)) - 4*(b*d*x^2 + 4*b*c - 3*a*d)*sqrt(d*x^2 + c))/b^2, -1/6*(3*(b*c - a*d)*sqrt(-(b*c - a*d)/b)*arctan(-1/2*
(b*d*x^2 + 2*b*c - a*d)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/b)/(b*c^2 - a*c*d + (b*c*d - a*d^2)*x^2)) - 2*(b*d*x
^2 + 4*b*c - 3*a*d)*sqrt(d*x^2 + c))/b^2]

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giac [A]  time = 0.38, size = 112, normalized size = 1.23 \begin {gather*} \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} + 3 \, \sqrt {d x^{2} + c} b^{2} c - 3 \, \sqrt {d x^{2} + c} a b d}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 1856, normalized size = 20.40

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/b*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(3/2)-1/4/b^2*(-a*b)^(1/2)*d*
((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*x-3/4/b^2*d^(1/2)*(-a*b)^(1/2
)*ln(((x+(-a*b)^(1/2)/b)*d-(-a*b)^(1/2)/b*d)/d^(1/2)+((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)
/b*d-(a*d-b*c)/b)^(1/2))*c-1/2/b^2*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^
(1/2)*a*d+1/2/b*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*c+1/2/b^3*d^(
3/2)*(-a*b)^(1/2)*ln(((x+(-a*b)^(1/2)/b)*d-(-a*b)^(1/2)/b*d)/d^(1/2)+((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x
+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*a-1/2/b^3/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)
/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b
*c)/b)^(1/2))/(x+(-a*b)^(1/2)/b))*a^2*d^2+1/b^2/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*
d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)
/b)^(1/2))/(x+(-a*b)^(1/2)/b))*a*d*c-1/2/b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-2*(
a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x+(-a*b)^(1/2)/b)^2*d-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(
1/2))/(x+(-a*b)^(1/2)/b))*c^2+1/6/b*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)
^(3/2)+1/4/b^2*(-a*b)^(1/2)*d*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)
*x+3/4/b^2*d^(1/2)*(-a*b)^(1/2)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(1/2)+((x-(-a*b)^(1/2)/b)^2*d+2*(
-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*c-1/2/b^2*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a
*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2)*a*d+1/2/b*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a
*d-b*c)/b)^(1/2)*c-1/2/b^3*d^(3/2)*(-a*b)^(1/2)*ln(((x-(-a*b)^(1/2)/b)*d+(-a*b)^(1/2)/b*d)/d^(1/2)+((x-(-a*b)^
(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))*a-1/2/b^3/(-(a*d-b*c)/b)^(1/2)*ln((2*(-
a*b)^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*
(x-(-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*a^2*d^2+1/b^2/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)
^(1/2)*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(
-a*b)^(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*a*d*c-1/2/b/(-(a*d-b*c)/b)^(1/2)*ln((2*(-a*b)^(1/2)
*(x-(-a*b)^(1/2)/b)/b*d-2*(a*d-b*c)/b+2*(-(a*d-b*c)/b)^(1/2)*((x-(-a*b)^(1/2)/b)^2*d+2*(-a*b)^(1/2)*(x-(-a*b)^
(1/2)/b)/b*d-(a*d-b*c)/b)^(1/2))/(x-(-a*b)^(1/2)/b))*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*(d*x^2+c)^(3/2)/(b*x^2+a),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 positive or negative?

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mupad [B]  time = 0.62, size = 98, normalized size = 1.08 \begin {gather*} \frac {{\left (d\,x^2+c\right )}^{3/2}}{3\,b}-\frac {\sqrt {d\,x^2+c}\,\left (a\,d-b\,c\right )}{b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^{3/2}}{a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}\right )\,{\left (a\,d-b\,c\right )}^{3/2}}{b^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 33.91, size = 80, normalized size = 0.88 \begin {gather*} \frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{3 b} + \frac {\sqrt {c + d x^{2}} \left (- a d + b c\right )}{b^{2}} + \frac {\left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{3} \sqrt {\frac {a d - b c}{b}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + d*x**2)**(3/2)/(3*b) + sqrt(c + d*x**2)*(-a*d + b*c)/b**2 + (a*d - b*c)**2*atan(sqrt(c + d*x**2)/sqrt((a*
d - b*c)/b))/(b**3*sqrt((a*d - b*c)/b))

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