3.697 \(\int \frac {x (c+d x^2)^{5/2}}{a+b x^2} \, dx\)
Optimal. Leaf size=119 \[ -\frac {(b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{7/2}}+\frac {\sqrt {c+d x^2} (b c-a d)^2}{b^3}+\frac {\left (c+d x^2\right )^{3/2} (b c-a d)}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b} \]
[Out]
1/3*(-a*d+b*c)*(d*x^2+c)^(3/2)/b^2+1/5*(d*x^2+c)^(5/2)/b-(-a*d+b*c)^(5/2)*arctanh(b^(1/2)*(d*x^2+c)^(1/2)/(-a*
d+b*c)^(1/2))/b^(7/2)+(-a*d+b*c)^2*(d*x^2+c)^(1/2)/b^3
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Rubi [A] time = 0.11, antiderivative size = 119, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used =
{444, 50, 63, 208} \[ \frac {\sqrt {c+d x^2} (b c-a d)^2}{b^3}+\frac {\left (c+d x^2\right )^{3/2} (b c-a d)}{3 b^2}-\frac {(b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{b^{7/2}}+\frac {\left (c+d x^2\right )^{5/2}}{5 b} \]
Antiderivative was successfully verified.
[In]
Int[(x*(c + d*x^2)^(5/2))/(a + b*x^2),x]
[Out]
((b*c - a*d)^2*Sqrt[c + d*x^2])/b^3 + ((b*c - a*d)*(c + d*x^2)^(3/2))/(3*b^2) + (c + d*x^2)^(5/2)/(5*b) - ((b*
c - a*d)^(5/2)*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 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 )^{5/2}}{a+b x^2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(c+d x)^{5/2}}{a+b x} \, dx,x,x^2\right )\\ &=\frac {\left (c+d x^2\right )^{5/2}}{5 b}+\frac {(b c-a d) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{a+b x} \, dx,x,x^2\right )}{2 b}\\ &=\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b}+\frac {(b c-a d)^2 \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^2\right )}{2 b^2}\\ &=\frac {(b c-a d)^2 \sqrt {c+d x^2}}{b^3}+\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b}+\frac {(b c-a d)^3 \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 b^3}\\ &=\frac {(b c-a d)^2 \sqrt {c+d x^2}}{b^3}+\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b}+\frac {(b c-a d)^3 \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 {(b c-a d)^2 \sqrt {c+d x^2}}{b^3}+\frac {(b c-a d) \left (c+d x^2\right )^{3/2}}{3 b^2}+\frac {\left (c+d x^2\right )^{5/2}}{5 b}-\frac {(b c-a d)^{5/2} \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 = 0.16, size = 113, normalized size = 0.95 \[ \frac {(b c-a d) \left (\sqrt {b} \sqrt {c+d x^2} \left (-3 a d+4 b c+b d x^2\right )-3 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )\right )}{3 b^{7/2}}+\frac {\left (c+d x^2\right )^{5/2}}{5 b} \]
Antiderivative was successfully verified.
[In]
Integrate[(x*(c + d*x^2)^(5/2))/(a + b*x^2),x]
[Out]
(c + d*x^2)^(5/2)/(5*b) + ((b*c - a*d)*(Sqrt[b]*Sqrt[c + d*x^2]*(4*b*c - 3*a*d + b*d*x^2) - 3*(b*c - a*d)^(3/2
)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^2])/Sqrt[b*c - a*d]]))/(3*b^(7/2))
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fricas [A] time = 0.94, size = 405, normalized size = 3.40 \[ \left [\frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b^{2} d^{2} x^{4} + 8 \, b^{2} c^{2} - 8 \, a b c d + a^{2} d^{2} + 2 \, {\left (4 \, b^{2} c d - 3 \, a b d^{2}\right )} x^{2} - 4 \, {\left (b^{2} d x^{2} + 2 \, b^{2} c - a b d\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{b}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 4 \, {\left (3 \, b^{2} d^{2} x^{4} + 23 \, b^{2} c^{2} - 35 \, a b c d + 15 \, a^{2} d^{2} + {\left (11 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{60 \, b^{3}}, -\frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {{\left (b d x^{2} + 2 \, b c - a d\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{b}}}{2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (3 \, b^{2} d^{2} x^{4} + 23 \, b^{2} c^{2} - 35 \, a b c d + 15 \, a^{2} d^{2} + {\left (11 \, b^{2} c d - 5 \, a b d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{30 \, b^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(d*x^2+c)^(5/2)/(b*x^2+a),x, algorithm="fricas")
[Out]
[1/60*(15*(b^2*c^2 - 2*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 + 23*b^2*c^2 - 35*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d - 5*a*b*d^2)*
x^2)*sqrt(d*x^2 + c))/b^3, -1/30*(15*(b^2*c^2 - 2*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
+ 23*b^2*c^2 - 35*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d - 5*a*b*d^2)*x^2)*sqrt(d*x^2 + c))/b^3]
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giac [A] time = 0.44, size = 184, normalized size = 1.55 \[ \frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} b^{3}} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{4} + 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{4} c + 15 \, \sqrt {d x^{2} + c} b^{4} c^{2} - 5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b^{3} d - 30 \, \sqrt {d x^{2} + c} a b^{3} c d + 15 \, \sqrt {d x^{2} + c} a^{2} b^{2} d^{2}}{15 \, b^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(d*x^2+c)^(5/2)/(b*x^2+a),x, algorithm="giac")
[Out]
(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*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 + 5*(d*x^2 + c)^(3/2)*b^4*c + 15*sqrt(d*x^2 + c)*b^4*c^2 - 5*(
d*x^2 + c)^(3/2)*a*b^3*d - 30*sqrt(d*x^2 + c)*a*b^3*c*d + 15*sqrt(d*x^2 + c)*a^2*b^2*d^2)/b^5
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maple [B] time = 0.01, size = 3078, normalized size = 25.87 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*(d*x^2+c)^(5/2)/(b*x^2+a),x)
[Out]
-1/8/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)^(3/2)*x-15/
16/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^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)*c+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^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)*c+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^2-1/4/b^3*(-a*b)^(1/2)*d^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-5/4/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*c-7/16/b^2*(-a*b)^(1/2)*d*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+1/4/b^3*(-a*b)^(1/2)*d^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+5/4/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*c-3/2/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*c+3/2/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^2-3/2/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*c+3/2/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^2+7/16/b^2*(-a*b)^(1/2)*d*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-1/6/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)*a*d+1/2/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)*a^2*d^2-1/2/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^3-1/6/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)*a*d+1/2/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)*a^2*d^2-1/2/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^3+1/10/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)+1/10/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/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*c-1/2/b^4*d^(5/2)*(-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
^2+1/8/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)^(3/2)*x+1
5/16/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^2-1/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*c+1/2/b^4*d^(5/2)*(-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^2+1/2/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))*a^3*d^3+1/2/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))*a^3*d^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*(d*x^2+c)^(5/2)/(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 = 0.61, size = 137, normalized size = 1.15 \[ \frac {{\left (d\,x^2+c\right )}^{5/2}}{5\,b}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^{5/2}}{a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3}\right )\,{\left (a\,d-b\,c\right )}^{5/2}}{b^{7/2}}-\frac {{\left (d\,x^2+c\right )}^{3/2}\,\left (a\,d-b\,c\right )}{3\,b^2}+\frac {\sqrt {d\,x^2+c}\,{\left (a\,d-b\,c\right )}^2}{b^3} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((x*(c + d*x^2)^(5/2))/(a + b*x^2),x)
[Out]
(c + d*x^2)^(5/2)/(5*b) - (atan((b^(1/2)*(c + d*x^2)^(1/2)*(a*d - b*c)^(5/2))/(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2
*d - 3*a^2*b*c*d^2))*(a*d - b*c)^(5/2))/b^(7/2) - ((c + d*x^2)^(3/2)*(a*d - b*c))/(3*b^2) + ((c + d*x^2)^(1/2)
*(a*d - b*c)^2)/b^3
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sympy [A] time = 50.76, size = 117, normalized size = 0.98 \[ \frac {\left (c + d x^{2}\right )^{\frac {5}{2}}}{5 b} + \frac {\left (c + d x^{2}\right )^{\frac {3}{2}} \left (- a d + b c\right )}{3 b^{2}} + \frac {\sqrt {c + d x^{2}} \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{b^{3}} - \frac {\left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{b^{4} \sqrt {\frac {a d - b c}{b}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(d*x**2+c)**(5/2)/(b*x**2+a),x)
[Out]
(c + d*x**2)**(5/2)/(5*b) + (c + d*x**2)**(3/2)*(-a*d + b*c)/(3*b**2) + sqrt(c + d*x**2)*(a**2*d**2 - 2*a*b*c*
d + b**2*c**2)/b**3 - (a*d - b*c)**3*atan(sqrt(c + d*x**2)/sqrt((a*d - b*c)/b))/(b**4*sqrt((a*d - b*c)/b))
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