3.8.44 \(\int \frac {(c+d x^2)^{3/2}}{(a+b x^2)^2} \, dx\) [744]
Optimal. Leaf size=131 \[ \frac {(b c-a d) x \sqrt {c+d x^2}}{2 a b \left (a+b x^2\right )}+\frac {\sqrt {b c-a d} (b c+2 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} b^2}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b^2} \]
[Out]
d^(3/2)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/b^2+1/2*(2*a*d+b*c)*arctan(x*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^2+c)^(1/
2))*(-a*d+b*c)^(1/2)/a^(3/2)/b^2+1/2*(-a*d+b*c)*x*(d*x^2+c)^(1/2)/a/b/(b*x^2+a)
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Rubi [A]
time = 0.06, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {424, 537, 223,
212, 385, 211} \begin {gather*} \frac {\sqrt {b c-a d} (2 a d+b c) \text {ArcTan}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} b^2}+\frac {x \sqrt {c+d x^2} (b c-a d)}{2 a b \left (a+b x^2\right )}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(c + d*x^2)^(3/2)/(a + b*x^2)^2,x]
[Out]
((b*c - a*d)*x*Sqrt[c + d*x^2])/(2*a*b*(a + b*x^2)) + (Sqrt[b*c - a*d]*(b*c + 2*a*d)*ArcTan[(Sqrt[b*c - a*d]*x
)/(Sqrt[a]*Sqrt[c + d*x^2])])/(2*a^(3/2)*b^2) + (d^(3/2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/b^2
Rule 211
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 223
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
Rule 385
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Rule 424
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a*d - c*b)*x*(a + b*x^n)^(
p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 1))), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]
Rule 537
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]
Rubi steps
\begin {align*} \int \frac {\left (c+d x^2\right )^{3/2}}{\left (a+b x^2\right )^2} \, dx &=\frac {(b c-a d) x \sqrt {c+d x^2}}{2 a b \left (a+b x^2\right )}+\frac {\int \frac {c (b c+a d)+2 a d^2 x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 a b}\\ &=\frac {(b c-a d) x \sqrt {c+d x^2}}{2 a b \left (a+b x^2\right )}+\frac {d^2 \int \frac {1}{\sqrt {c+d x^2}} \, dx}{b^2}+\frac {((b c-a d) (b c+2 a d)) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{2 a b^2}\\ &=\frac {(b c-a d) x \sqrt {c+d x^2}}{2 a b \left (a+b x^2\right )}+\frac {d^2 \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{b^2}+\frac {((b c-a d) (b c+2 a d)) \text {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{2 a b^2}\\ &=\frac {(b c-a d) x \sqrt {c+d x^2}}{2 a b \left (a+b x^2\right )}+\frac {\sqrt {b c-a d} (b c+2 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 a^{3/2} b^2}+\frac {d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b^2}\\ \end {align*}
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Mathematica [A]
time = 0.52, size = 148, normalized size = 1.13 \begin {gather*} \frac {\frac {b (b c-a d) x \sqrt {c+d x^2}}{a \left (a+b x^2\right )}-\frac {\sqrt {b c-a d} (b c+2 a d) \tan ^{-1}\left (\frac {a \sqrt {d}+b x \left (\sqrt {d} x-\sqrt {c+d x^2}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{a^{3/2}}-2 d^{3/2} \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{2 b^2} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x^2)^(3/2)/(a + b*x^2)^2,x]
[Out]
((b*(b*c - a*d)*x*Sqrt[c + d*x^2])/(a*(a + b*x^2)) - (Sqrt[b*c - a*d]*(b*c + 2*a*d)*ArcTan[(a*Sqrt[d] + b*x*(S
qrt[d]*x - Sqrt[c + d*x^2]))/(Sqrt[a]*Sqrt[b*c - a*d])])/a^(3/2) - 2*d^(3/2)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2
]])/(2*b^2)
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(3386\) vs.
\(2(109)=218\).
time = 0.10, size = 3387, normalized size = 25.85
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\(\text {Expression too large to display}\) |
\(3387\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x^2+c)^(3/2)/(b*x^2+a)^2,x,method=_RETURNVERBOSE)
[Out]
-1/4/b/a*(1/(a*d-b*c)*b/(x+1/b*(-a*b)^(1/2))*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))
-(a*d-b*c)/b)^(5/2)+3*d*(-a*b)^(1/2)/(a*d-b*c)*(1/3*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)
^(1/2))-(a*d-b*c)/b)^(3/2)-d*(-a*b)^(1/2)/b*(1/4*(2*d*(x+1/b*(-a*b)^(1/2))-2*d*(-a*b)^(1/2)/b)/d*(d*(x+1/b*(-a
*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1/8*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d^(3/
2)*ln((-d*(-a*b)^(1/2)/b+d*(x+1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(
-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)))-(a*d-b*c)/b*((d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2
))-(a*d-b*c)/b)^(1/2)-d^(1/2)*(-a*b)^(1/2)/b*ln((-d*(-a*b)^(1/2)/b+d*(x+1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x+1/b*(
-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+(a*d-b*c)/b/(-(a*d-b*c)/b)^(1/2)*ln
((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*
(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x+1/b*(-a*b)^(1/2)))))-4*d/(a*d-b*c)*b*(1/8*(2*d*(x+1
/b*(-a*b)^(1/2))-2*d*(-a*b)^(1/2)/b)/d*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-
b*c)/b)^(3/2)+3/16*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d*(1/4*(2*d*(x+1/b*(-a*b)^(1/2))-2*d*(-a*b)^(1/2)/b)/d*(d*(x+1
/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1/8*(-4*d*(a*d-b*c)/b+4*d^2*a/b)
/d^(3/2)*ln((-d*(-a*b)^(1/2)/b+d*(x+1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x
+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)))))+1/4/a/(-a*b)^(1/2)*(1/3*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b
*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+d*(-a*b)^(1/2)/b*(1/4*(2*d*(x-1/b*(-a*b)^(1/2))+2*d*(-a*b)^(1/2)/b)/d
*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1/8*(-4*d*(a*d-b*c)/b+4*
d^2*a/b)/d^(3/2)*ln((d*(-a*b)^(1/2)/b+d*(x-1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/
2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)))-(a*d-b*c)/b*((d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/
b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+d^(1/2)*(-a*b)^(1/2)/b*ln((d*(-a*b)^(1/2)/b+d*(x-1/b*(-a*b)^(1/2)))/d^(1/2)
+(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+(a*d-b*c)/b/(-(a*d-b*c)
/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*(d*(x-1/b*(-a*b)^(
1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))))-1/4/a/(-a*b)^(1/2)
*(1/3*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)-d*(-a*b)^(1/2)/b*(1
/4*(2*d*(x+1/b*(-a*b)^(1/2))-2*d*(-a*b)^(1/2)/b)/d*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^
(1/2))-(a*d-b*c)/b)^(1/2)+1/8*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d^(3/2)*ln((-d*(-a*b)^(1/2)/b+d*(x+1/b*(-a*b)^(1/2)
))/d^(1/2)+(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)))-(a*d-b*c)/b*
((d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)-d^(1/2)*(-a*b)^(1/2)/b*l
n((-d*(-a*b)^(1/2)/b+d*(x+1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b
)^(1/2))-(a*d-b*c)/b)^(1/2))+(a*d-b*c)/b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a
*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)*(d*(x+1/b*(-a*b)^(1/2))^2-2*d*(-a*b)^(1/2)/b*(x+1/b*(-a*b)^(1/2))-(a*d-b*c)/
b)^(1/2))/(x+1/b*(-a*b)^(1/2)))))-1/4/b/a*(1/(a*d-b*c)*b/(x-1/b*(-a*b)^(1/2))*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-
a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(5/2)-3*d*(-a*b)^(1/2)/(a*d-b*c)*(1/3*(d*(x-1/b*(-a*b)^(1/2))^2
+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+d*(-a*b)^(1/2)/b*(1/4*(2*d*(x-1/b*(-a*b)^(1/2))+2*
d*(-a*b)^(1/2)/b)/d*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1/8*(
-4*d*(a*d-b*c)/b+4*d^2*a/b)/d^(3/2)*ln((d*(-a*b)^(1/2)/b+d*(x-1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x-1/b*(-a*b)^(1/2
))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)))-(a*d-b*c)/b*((d*(x-1/b*(-a*b)^(1/2))^2+2*d*(
-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+d^(1/2)*(-a*b)^(1/2)/b*ln((d*(-a*b)^(1/2)/b+d*(x-1/b*(-a
*b)^(1/2)))/d^(1/2)+(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))+(a*d
-b*c)/b/(-(a*d-b*c)/b)^(1/2)*ln((-2*(a*d-b*c)/b+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))+2*(-(a*d-b*c)/b)^(1/2)
*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2))/(x-1/b*(-a*b)^(1/2)))))
-4*d/(a*d-b*c)*b*(1/8*(2*d*(x-1/b*(-a*b)^(1/2))+2*d*(-a*b)^(1/2)/b)/d*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/
2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(3/2)+3/16*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d*(1/4*(2*d*(x-1/b*(-a*b)^(1/2)
)+2*d*(-a*b)^(1/2)/b)/d*(d*(x-1/b*(-a*b)^(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2))-(a*d-b*c)/b)^(1/2)+1
/8*(-4*d*(a*d-b*c)/b+4*d^2*a/b)/d^(3/2)*ln((d*(-a*b)^(1/2)/b+d*(x-1/b*(-a*b)^(1/2)))/d^(1/2)+(d*(x-1/b*(-a*b)^
(1/2))^2+2*d*(-a*b)^(1/2)/b*(x-1/b*(-a*b)^(1/2)...
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^(3/2)/(b*x^2+a)^2,x, algorithm="maxima")
[Out]
integrate((d*x^2 + c)^(3/2)/(b*x^2 + a)^2, x)
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Fricas [A]
time = 1.76, size = 903, normalized size = 6.89 \begin {gather*} \left [\frac {4 \, {\left (b^{2} c - a b d\right )} \sqrt {d x^{2} + c} x + 4 \, {\left (a b d x^{2} + a^{2} d\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + {\left (a b c + 2 \, a^{2} d + {\left (b^{2} c + 2 \, a b d\right )} x^{2}\right )} \sqrt {-\frac {b c - a d}{a}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left (a^{2} c x - {\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{8 \, {\left (a b^{3} x^{2} + a^{2} b^{2}\right )}}, \frac {4 \, {\left (b^{2} c - a b d\right )} \sqrt {d x^{2} + c} x - 8 \, {\left (a b d x^{2} + a^{2} d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (a b c + 2 \, a^{2} d + {\left (b^{2} c + 2 \, a b d\right )} x^{2}\right )} \sqrt {-\frac {b c - a d}{a}} \log \left (\frac {{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{2} - 4 \, {\left (a^{2} c x - {\left (a b c - 2 \, a^{2} d\right )} x^{3}\right )} \sqrt {d x^{2} + c} \sqrt {-\frac {b c - a d}{a}}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right )}{8 \, {\left (a b^{3} x^{2} + a^{2} b^{2}\right )}}, \frac {2 \, {\left (b^{2} c - a b d\right )} \sqrt {d x^{2} + c} x + {\left (a b c + 2 \, a^{2} d + {\left (b^{2} c + 2 \, a b d\right )} x^{2}\right )} \sqrt {\frac {b c - a d}{a}} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{a}}}{2 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + {\left (b c^{2} - a c d\right )} x\right )}}\right ) + 2 \, {\left (a b d x^{2} + a^{2} d\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right )}{4 \, {\left (a b^{3} x^{2} + a^{2} b^{2}\right )}}, \frac {2 \, {\left (b^{2} c - a b d\right )} \sqrt {d x^{2} + c} x - 4 \, {\left (a b d x^{2} + a^{2} d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (a b c + 2 \, a^{2} d + {\left (b^{2} c + 2 \, a b d\right )} x^{2}\right )} \sqrt {\frac {b c - a d}{a}} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c} \sqrt {\frac {b c - a d}{a}}}{2 \, {\left ({\left (b c d - a d^{2}\right )} x^{3} + {\left (b c^{2} - a c d\right )} x\right )}}\right )}{4 \, {\left (a b^{3} x^{2} + a^{2} b^{2}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^(3/2)/(b*x^2+a)^2,x, algorithm="fricas")
[Out]
[1/8*(4*(b^2*c - a*b*d)*sqrt(d*x^2 + c)*x + 4*(a*b*d*x^2 + a^2*d)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqr
t(d)*x - c) + (a*b*c + 2*a^2*d + (b^2*c + 2*a*b*d)*x^2)*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2
*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*sqrt
(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(a*b^3*x^2 + a^2*b^2), 1/8*(4*(b^2*c - a*b*d)*sqrt(d*x^2 + c)*
x - 8*(a*b*d*x^2 + a^2*d)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (a*b*c + 2*a^2*d + (b^2*c + 2*a*b*d)*x
^2)*sqrt(-(b*c - a*d)/a)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2
- 4*(a^2*c*x - (a*b*c - 2*a^2*d)*x^3)*sqrt(d*x^2 + c)*sqrt(-(b*c - a*d)/a))/(b^2*x^4 + 2*a*b*x^2 + a^2)))/(a*b
^3*x^2 + a^2*b^2), 1/4*(2*(b^2*c - a*b*d)*sqrt(d*x^2 + c)*x + (a*b*c + 2*a^2*d + (b^2*c + 2*a*b*d)*x^2)*sqrt((
b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a*c)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/a)/((b*c*d - a*d^2)*x^3 +
(b*c^2 - a*c*d)*x)) + 2*(a*b*d*x^2 + a^2*d)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c))/(a*b^3*x^
2 + a^2*b^2), 1/4*(2*(b^2*c - a*b*d)*sqrt(d*x^2 + c)*x - 4*(a*b*d*x^2 + a^2*d)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt
(d*x^2 + c)) + (a*b*c + 2*a^2*d + (b^2*c + 2*a*b*d)*x^2)*sqrt((b*c - a*d)/a)*arctan(1/2*((b*c - 2*a*d)*x^2 - a
*c)*sqrt(d*x^2 + c)*sqrt((b*c - a*d)/a)/((b*c*d - a*d^2)*x^3 + (b*c^2 - a*c*d)*x)))/(a*b^3*x^2 + a^2*b^2)]
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c + d x^{2}\right )^{\frac {3}{2}}}{\left (a + b x^{2}\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x**2+c)**(3/2)/(b*x**2+a)**2,x)
[Out]
Integral((c + d*x**2)**(3/2)/(a + b*x**2)**2, x)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 315 vs.
\(2 (109) = 218\).
time = 0.50, size = 315, normalized size = 2.40 \begin {gather*} -\frac {d^{\frac {3}{2}} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{2 \, b^{2}} - \frac {{\left (b^{2} c^{2} \sqrt {d} + a b c d^{\frac {3}{2}} - 2 \, a^{2} d^{\frac {5}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt {a b c d - a^{2} d^{2}} a b^{2}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b^{2} c^{2} \sqrt {d} - 3 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c d^{\frac {3}{2}} + 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} d^{\frac {5}{2}} - b^{2} c^{3} \sqrt {d} + a b c^{2} d^{\frac {3}{2}}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} a b^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x^2+c)^(3/2)/(b*x^2+a)^2,x, algorithm="giac")
[Out]
-1/2*d^(3/2)*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/b^2 - 1/2*(b^2*c^2*sqrt(d) + a*b*c*d^(3/2) - 2*a^2*d^(5/2))*
arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/(sqrt(a*b*c*d - a^2*d^2)
*a*b^2) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*b^2*c^2*sqrt(d) - 3*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*b*c*d^(3/2) +
2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*d^(5/2) - b^2*c^3*sqrt(d) + a*b*c^2*d^(3/2))/(((sqrt(d)*x - sqrt(d*x^2
+ c))^4*b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2)*a*b^2)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,x^2+c\right )}^{3/2}}{{\left (b\,x^2+a\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c + d*x^2)^(3/2)/(a + b*x^2)^2,x)
[Out]
int((c + d*x^2)^(3/2)/(a + b*x^2)^2, x)
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