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

Optimal. Leaf size=99 \[ -\frac {3 d \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{5/2}}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}+\frac {3 d \sqrt {c+d x^2}}{2 b^2} \]

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Rubi [A]  time = 0.08, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {444, 47, 50, 63, 208} \begin {gather*} -\frac {3 d \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 b^{5/2}}-\frac {\left (c+d x^2\right )^{3/2}}{2 b \left (a+b x^2\right )}+\frac {3 d \sqrt {c+d x^2}}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

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

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

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.55 \begin {gather*} \frac {d \left (c+d x^2\right )^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\frac {b \left (d x^2+c\right )}{a d-b c}\right )}{5 (a d-b c)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(c + d*x^2)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, -((b*(c + d*x^2))/(-(b*c) + a*d))])/(5*(-(b*c) + a*d)^2)

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IntegrateAlgebraic [A]  time = 0.28, size = 106, normalized size = 1.07 \begin {gather*} \frac {3 d \sqrt {a d-b c} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2} \sqrt {a d-b c}}{b c-a d}\right )}{2 b^{5/2}}+\frac {\sqrt {c+d x^2} \left (3 a d-b c+2 b d x^2\right )}{2 b^2 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.31, size = 333, normalized size = 3.36 \begin {gather*} \left [\frac {3 \, {\left (b d x^{2} + 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 (2 \, b d x^{2} - b c + 3 \, a d\right )} \sqrt {d x^{2} + c}}{8 \, {\left (b^{3} x^{2} + a b^{2}\right )}}, -\frac {3 \, {\left (b d x^{2} + 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 (2 \, b d x^{2} - b c + 3 \, a d\right )} \sqrt {d x^{2} + c}}{4 \, {\left (b^{3} x^{2} + a b^{2}\right )}}\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)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.43, size = 122, normalized size = 1.23 \begin {gather*} \frac {\sqrt {d x^{2} + c} d}{b^{2}} + \frac {3 \, {\left (b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} b^{2}} - \frac {\sqrt {d x^{2} + c} b c d - \sqrt {d x^{2} + c} a d^{2}}{2 \, {\left ({\left (d x^{2} + c\right )} b - b c + a d\right )} b^{2}} \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)^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.01, size = 2821, normalized size = 28.49 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*a/b^2*d^2/(a*d-b*c)/(-(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-3/8*(-a*b)^(1/2)/a/b*d^(1/2)/(a*d-b*c)*c^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))-3/2*a/b^2*d^2/(a*d-b*c)/(-(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-1/4*(-a*b)^(1/2)/a/b*d/(a*
d-b*c)*((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)*x+9/8*(-a*b)^(1/2)/b^2
*d^(3/2)/(a*d-b*c)*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-3/4*(-a*b)^(1/2)*a/b^3*d^(5/2)/(a*d-b*c)*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))+3/
4*a^2/b^3*d^3/(a*d-b*c)/(-(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))+3/4/b*d/(a*d-b*c)/(-(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+1/4*(-a*b)^(1/2)/a/b/(a*d-b*c)/(x+(-a*b)^(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)^(5/2)+3/8*(-a*b)^(1/2)/b^2*d^2/(a*d-b*c)*((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*d/(a*d-b*c)/(-(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+3/4*a^2/b^3*d^3/(a*d-b*c)/(-(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))-3/8*(-a*b)^(1/2)/b^2*d^2/(a*d-b*c)*((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-9/8*(-a*b)^(1/2)/b^2*d^(3/2)/(a*d-b*c)*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+3/4*(-a*b)^(1/2)*a/b^3*d^(5/2)/(a*d-b*c)*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))-1/4*(-a*b)^(1/2)/a/b/(a*d-b*c)/
(x-(-a*b)^(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)^(5/2)-1/4/b*d/(a
*d-b*c)*((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*d/(a*d-b*c)*((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)+3/4*a/b^2*d^2/(a*d-b*c)*((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)+3/4*a/b^2*d^2/(a*d-b*c)*((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)-3/4/b*d/(a*d-b*c)*((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+3/8*(-a*b)^(1/2)/a/b*d^(1/2)/(a*d-b*c)*c^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))+1/4*(-a*b)^(1/2)/a/b*d/(a*d-b*c)*((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)*x-3/4/b*d/(a*d-b*c)*((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+3/8*(-a*b)^(1/2)/a/b*d/(a*d-b*c)*c*((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-3/8*(-a*b)^(1/2)/a/b*d/(a*d-b*c)*c*((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

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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)^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 positive or negative?

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mupad [B]  time = 0.90, size = 117, normalized size = 1.18 \begin {gather*} \frac {\sqrt {d\,x^2+c}\,\left (\frac {a\,d^2}{2}-\frac {b\,c\,d}{2}\right )}{b^3\,\left (d\,x^2+c\right )-b^3\,c+a\,b^2\,d}+\frac {d\,\sqrt {d\,x^2+c}}{b^2}-\frac {3\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,d\,\sqrt {d\,x^2+c}\,\sqrt {a\,d-b\,c}}{a\,d^2-b\,c\,d}\right )\,\sqrt {a\,d-b\,c}}{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)^2,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**2,x)

[Out]

Timed out

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