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

Optimal. Leaf size=115 \[ \frac {a (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{7/2}}-\frac {a \sqrt {c+d x^2} (b c-a d)}{b^3}-\frac {a \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b d} \]

[Out]

-1/3*a*(d*x^2+c)^(3/2)/b^2+1/5*(d*x^2+c)^(5/2)/b/d+a*(-a*d+b*c)^(3/2)*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*d+b*
c)^(1/2))/b^(7/2)-a*(-a*d+b*c)*(d*x^2+c)^(1/2)/b^3

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Rubi [A]  time = 0.11, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {446, 80, 50, 63, 208} \[ -\frac {a \left (c+d x^2\right )^{3/2}}{3 b^2}-\frac {a \sqrt {c+d x^2} (b c-a d)}{b^3}+\frac {a (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{7/2}}+\frac {\left (c+d x^2\right )^{5/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(b*c - a*d)*Sqrt[c + d*x^2])/b^3) - (a*(c + d*x^2)^(3/2))/(3*b^2) + (c + d*x^2)^(5/2)/(5*b*d) + (a*(b*c -
a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/b^(7/2)

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 80

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

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

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Mathematica [A]  time = 0.12, size = 108, normalized size = 0.94 \[ \frac {\sqrt {c+d x^2} \left (15 a^2 d^2-5 a b d \left (4 c+d x^2\right )+3 b^2 \left (c+d x^2\right )^2\right )}{15 b^3 d}+\frac {a (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^2]*(15*a^2*d^2 + 3*b^2*(c + d*x^2)^2 - 5*a*b*d*(4*c + d*x^2)))/(15*b^3*d) + (a*(b*c - a*d)^(3/2)
*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]])/b^(7/2)

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fricas [A]  time = 0.68, size = 397, normalized size = 3.45 \[ \left [-\frac {15 \, {\left (a b c d - a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, {\left (3 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} + {\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{60 \, b^{3} d}, \frac {15 \, {\left (a b c d - a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (3 \, b^{2} d^{2} x^{4} + 3 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} + {\left (6 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, b^{3} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/60*(15*(a*b*c*d - a^2*d^2)*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*(3*b^2*d^2*x^4 + 3*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2 + (6*b^2*c*d - 5*a*b*d^2)*x^2)*sqrt(d*x
^2 + c))/(b^3*d), 1/30*(15*(a*b*c*d - a^2*d^2)*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*(3*b^2*d^2*x^4 + 3*b^2*c^2 - 20*a*b*
c*d + 15*a^2*d^2 + (6*b^2*c*d - 5*a*b*d^2)*x^2)*sqrt(d*x^2 + c))/(b^3*d)]

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giac [A]  time = 0.37, size = 151, normalized size = 1.31 \[ -\frac {{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{3}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{4} d^{4} - 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b^{3} d^{5} - 15 \, \sqrt {d x^{2} + c} a b^{3} c d^{5} + 15 \, \sqrt {d x^{2} + c} a^{2} b^{2} d^{6}}{15 \, b^{5} d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 1897, normalized size = 16.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^3*(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 = 0.65, size = 179, normalized size = 1.56 \[ \frac {{\left (d\,x^2+c\right )}^{5/2}}{5\,b\,d}-{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {c}{3\,b\,d}+\frac {a\,d^2-b\,c\,d}{3\,b^2\,d^2}\right )-\frac {a\,\mathrm {atan}\left (\frac {a\,\sqrt {b}\,\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^{3/2}}{a^3\,d^2-2\,a^2\,b\,c\,d+a\,b^2\,c^2}\right )\,{\left (a\,d-b\,c\right )}^{3/2}}{b^{7/2}}+\frac {\sqrt {d\,x^2+c}\,\left (a\,d^2-b\,c\,d\right )\,\left (\frac {c}{b\,d}+\frac {a\,d^2-b\,c\,d}{b^2\,d^2}\right )}{b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + d*x^2)^(5/2)/(5*b*d) - (c + d*x^2)^(3/2)*(c/(3*b*d) + (a*d^2 - b*c*d)/(3*b^2*d^2)) - (a*atan((a*b^(1/2)*(
c + d*x^2)^(1/2)*(a*d - b*c)^(3/2))/(a^3*d^2 + a*b^2*c^2 - 2*a^2*b*c*d))*(a*d - b*c)^(3/2))/b^(7/2) + ((c + d*
x^2)^(1/2)*(a*d^2 - b*c*d)*(c/(b*d) + (a*d^2 - b*c*d)/(b^2*d^2)))/(b*d)

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sympy [A]  time = 46.42, size = 104, normalized size = 0.90 \[ - \frac {a \left (c + d x^{2}\right )^{\frac {3}{2}}}{3 b^{2}} - \frac {a \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{4} \sqrt {\frac {a d - b c}{b}}} + \frac {\left (c + d x^{2}\right )^{\frac {5}{2}}}{5 b d} + \frac {\sqrt {c + d x^{2}} \left (a^{2} d - a b c\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(c + d*x**2)**(3/2)/(3*b**2) - a*(a*d - b*c)**2*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(b**4*sqrt((a*d
- b*c)/b)) + (c + d*x**2)**(5/2)/(5*b*d) + sqrt(c + d*x**2)*(a**2*d - a*b*c)/b**3

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