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

Optimal. Leaf size=88 \[ \frac {a \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2}}-\frac {a \sqrt {c+d x^2}}{b^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ -\frac {a \sqrt {c+d x^2}}{b^2}+\frac {a \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2}}+\frac {\left (c+d x^2\right )^{3/2}}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Sqrt[c + d*x^2])/b^2) + (c + d*x^2)^(3/2)/(3*b*d) + (a*Sqrt[b*c - a*d]*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 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 \sqrt {c+d x^2}}{a+b x^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x \sqrt {c+d x}}{a+b x} \, dx,x,x^2\right )\\ &=\frac {\left (c+d x^2\right )^{3/2}}{3 b d}-\frac {a \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^2\right )}{2 b}\\ &=-\frac {a \sqrt {c+d x^2}}{b^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b d}-\frac {(a (b c-a d)) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 b^2}\\ &=-\frac {a \sqrt {c+d x^2}}{b^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b d}-\frac {(a (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 )}{b^2 d}\\ &=-\frac {a \sqrt {c+d x^2}}{b^2}+\frac {\left (c+d x^2\right )^{3/2}}{3 b d}+\frac {a \sqrt {b c-a d} \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 = 0.07, size = 85, normalized size = 0.97 \[ \frac {a \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{5/2}}+\frac {\sqrt {c+d x^2} \left (b \left (c+d x^2\right )-3 a d\right )}{3 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.72, size = 295, normalized size = 3.35 \[ \left [\frac {3 \, a d \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} + 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (b d x^{2} + b c - 3 \, a d\right )} \sqrt {d x^{2} + c}}{12 \, b^{2} d}, \frac {3 \, a d \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (b d x^{2} + b c - 3 \, a d\right )} \sqrt {d x^{2} + c}}{6 \, b^{2} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(d*x^2+c)^(1/2)/(b*x^2+a),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.32, size = 96, normalized size = 1.09 \[ -\frac {{\left (a b c - a^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d^{2} - 3 \, \sqrt {d x^{2} + c} a b d^{3}}{3 \, b^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(d*x^2+c)^(1/2)/(b*x^2+a),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.01, size = 963, normalized size = 10.94 \[ -\frac {a^{2} d \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, b^{3}}-\frac {a^{2} d \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, b^{3}}+\frac {a c \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, b^{2}}+\frac {a c \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-\frac {a d -b c}{b}}\, b^{2}}+\frac {\sqrt {-a b}\, a \sqrt {d}\, \ln \left (\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) d -\frac {\sqrt {-a b}\, d}{b}}{\sqrt {d}}+\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\right )}{2 b^{3}}-\frac {\sqrt {-a b}\, a \sqrt {d}\, \ln \left (\frac {\left (x -\frac {\sqrt {-a b}}{b}\right ) d +\frac {\sqrt {-a b}\, d}{b}}{\sqrt {d}}+\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\right )}{2 b^{3}}-\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a}{2 b^{2}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a}{2 b^{2}}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3 b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(1/2)/(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 = 0.59, size = 86, normalized size = 0.98 \[ \frac {{\left (d\,x^2+c\right )}^{3/2}}{3\,b\,d}-\frac {a\,\sqrt {d\,x^2+c}}{b^2}+\frac {a\,\mathrm {atan}\left (\frac {a\,\sqrt {b}\,\sqrt {d\,x^2+c}\,\sqrt {a\,d-b\,c}}{a^2\,d-a\,b\,c}\right )\,\sqrt {a\,d-b\,c}}{b^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 7.23, size = 87, normalized size = 0.99 \[ \frac {2 \left (- \frac {a d^{2} \sqrt {c + d x^{2}}}{2 b^{2}} + \frac {a d^{2} \left (a d - b c\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{2 b^{3} \sqrt {\frac {a d - b c}{b}}} + \frac {d \left (c + d x^{2}\right )^{\frac {3}{2}}}{6 b}\right )}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(d*x**2+c)**(1/2)/(b*x**2+a),x)

[Out]

2*(-a*d**2*sqrt(c + d*x**2)/(2*b**2) + a*d**2*(a*d - b*c)*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(2*b**3*s
qrt((a*d - b*c)/b)) + d*(c + d*x**2)**(3/2)/(6*b))/d**2

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