3.7.87 \(\int \frac {x^2 (c+d x^2)^{3/2}}{a+b x^2} \, dx\) [687]

Optimal. Leaf size=158 \[ \frac {(5 b c-4 a d) x \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \sqrt {c+d x^2}}{4 b}-\frac {\sqrt {a} (b c-a d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^3}+\frac {\left (3 b^2 c^2-12 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 b^3 \sqrt {d}} \]

[Out]

-(-a*d+b*c)^(3/2)*arctan(x*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^2+c)^(1/2))*a^(1/2)/b^3+1/8*(8*a^2*d^2-12*a*b*c*d+3*b
^2*c^2)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/b^3/d^(1/2)+1/8*(-4*a*d+5*b*c)*x*(d*x^2+c)^(1/2)/b^2+1/4*d*x^3*(d*x
^2+c)^(1/2)/b

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Rubi [A]
time = 0.16, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {488, 596, 537, 223, 212, 385, 211} \begin {gather*} \frac {\left (8 a^2 d^2-12 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 b^3 \sqrt {d}}-\frac {\sqrt {a} (b c-a d)^{3/2} \text {ArcTan}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^3}+\frac {x \sqrt {c+d x^2} (5 b c-4 a d)}{8 b^2}+\frac {d x^3 \sqrt {c+d x^2}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5*b*c - 4*a*d)*x*Sqrt[c + d*x^2])/(8*b^2) + (d*x^3*Sqrt[c + d*x^2])/(4*b) - (Sqrt[a]*(b*c - a*d)^(3/2)*ArcTa
n[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/b^3 + ((3*b^2*c^2 - 12*a*b*c*d + 8*a^2*d^2)*ArcTanh[(Sqrt[d]
*x)/Sqrt[c + d*x^2]])/(8*b^3*Sqrt[d])

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 488

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[d*(e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*e*(m + n*(p + q) + 1))), x] + Dist[1/(b*(m + n*(p + q) + 1
)), Int[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*((c*b - a*d)*(m + 1) + c*b*n*(p + q)) + (d*(c*b - a*d
)*(m + 1) + d*n*(q - 1)*(b*c - a*d) + c*b*d*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && N
eQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, 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]

Rule 596

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx &=\frac {d x^3 \sqrt {c+d x^2}}{4 b}+\frac {\int \frac {x^2 \left (c (4 b c-3 a d)+d (5 b c-4 a d) x^2\right )}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{4 b}\\ &=\frac {(5 b c-4 a d) x \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \sqrt {c+d x^2}}{4 b}-\frac {\int \frac {a c d (5 b c-4 a d)-d \left (3 b^2 c^2-12 a b c d+8 a^2 d^2\right ) x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{8 b^2 d}\\ &=\frac {(5 b c-4 a d) x \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \sqrt {c+d x^2}}{4 b}-\frac {\left (a (b c-a d)^2\right ) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{b^3}+\frac {\left (3 b^2 c^2-12 a b c d+8 a^2 d^2\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{8 b^3}\\ &=\frac {(5 b c-4 a d) x \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \sqrt {c+d x^2}}{4 b}-\frac {\left (a (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{b^3}+\frac {\left (3 b^2 c^2-12 a b c d+8 a^2 d^2\right ) \text {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{8 b^3}\\ &=\frac {(5 b c-4 a d) x \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \sqrt {c+d x^2}}{4 b}-\frac {\sqrt {a} (b c-a d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^3}+\frac {\left (3 b^2 c^2-12 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 b^3 \sqrt {d}}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 158, normalized size = 1.00 \begin {gather*} \frac {b x \sqrt {c+d x^2} \left (5 b c-4 a d+2 b d x^2\right )+8 \sqrt {a} (b c-a d)^{3/2} \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 )+\frac {\left (-3 b^2 c^2+12 a b c d-8 a^2 d^2\right ) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{\sqrt {d}}}{8 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x*Sqrt[c + d*x^2]*(5*b*c - 4*a*d + 2*b*d*x^2) + 8*Sqrt[a]*(b*c - a*d)^(3/2)*ArcTan[(a*Sqrt[d] + b*x*(Sqrt[d
]*x - Sqrt[c + d*x^2]))/(Sqrt[a]*Sqrt[b*c - a*d])] + ((-3*b^2*c^2 + 12*a*b*c*d - 8*a^2*d^2)*Log[-(Sqrt[d]*x) +
Sqrt[c + d*x^2]])/Sqrt[d])/(8*b^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1305\) vs. \(2(132)=264\).
time = 0.12, size = 1306, normalized size = 8.27 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]
time = 2.29, size = 894, normalized size = 5.66 \begin {gather*} \left [\frac {{\left (3 \, b^{2} c^{2} - 12 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 4 \, \sqrt {-a b c + a^{2} d} {\left (b c d - a d^{2}\right )} \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 ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt {-a b c + a^{2} d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) + 2 \, {\left (2 \, b^{2} d^{2} x^{3} + {\left (5 \, b^{2} c d - 4 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{16 \, b^{3} d}, -\frac {{\left (3 \, b^{2} c^{2} - 12 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + 2 \, \sqrt {-a b c + a^{2} d} {\left (b c d - a d^{2}\right )} \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 ({\left (b c - 2 \, a d\right )} x^{3} - a c x\right )} \sqrt {-a b c + a^{2} d} \sqrt {d x^{2} + c}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - {\left (2 \, b^{2} d^{2} x^{3} + {\left (5 \, b^{2} c d - 4 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{8 \, b^{3} d}, -\frac {8 \, \sqrt {a b c - a^{2} d} {\left (b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {a b c - a^{2} d} {\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) - {\left (3 \, b^{2} c^{2} - 12 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{3} + {\left (5 \, b^{2} c d - 4 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{16 \, b^{3} d}, -\frac {4 \, \sqrt {a b c - a^{2} d} {\left (b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {a b c - a^{2} d} {\left ({\left (b c - 2 \, a d\right )} x^{2} - a c\right )} \sqrt {d x^{2} + c}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{3} + {\left (a b c^{2} - a^{2} c d\right )} x\right )}}\right ) + {\left (3 \, b^{2} c^{2} - 12 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, b^{2} d^{2} x^{3} + {\left (5 \, b^{2} c d - 4 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{8 \, b^{3} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((3*b^2*c^2 - 12*a*b*c*d + 8*a^2*d^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 4*sqrt(-
a*b*c + a^2*d)*(b*c*d - a*d^2)*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*((b*c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 2*(
2*b^2*d^2*x^3 + (5*b^2*c*d - 4*a*b*d^2)*x)*sqrt(d*x^2 + c))/(b^3*d), -1/8*((3*b^2*c^2 - 12*a*b*c*d + 8*a^2*d^2
)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + 2*sqrt(-a*b*c + a^2*d)*(b*c*d - a*d^2)*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*((b*c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^
2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - (2*b^2*d^2*x^3 + (5*b^2*c*d - 4*a*b*d^2)*x)*sqrt(d*x^2 +
c))/(b^3*d), -1/16*(8*sqrt(a*b*c - a^2*d)*(b*c*d - a*d^2)*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)*x^2 -
a*c)*sqrt(d*x^2 + c)/((a*b*c*d - a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) - (3*b^2*c^2 - 12*a*b*c*d + 8*a^2*d^2)
*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(2*b^2*d^2*x^3 + (5*b^2*c*d - 4*a*b*d^2)*x)*sqrt(
d*x^2 + c))/(b^3*d), -1/8*(4*sqrt(a*b*c - a^2*d)*(b*c*d - a*d^2)*arctan(1/2*sqrt(a*b*c - a^2*d)*((b*c - 2*a*d)
*x^2 - a*c)*sqrt(d*x^2 + c)/((a*b*c*d - a^2*d^2)*x^3 + (a*b*c^2 - a^2*c*d)*x)) + (3*b^2*c^2 - 12*a*b*c*d + 8*a
^2*d^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (2*b^2*d^2*x^3 + (5*b^2*c*d - 4*a*b*d^2)*x)*sqrt(d*x^2 +
c))/(b^3*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}{a + b x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,{\left (d\,x^2+c\right )}^{3/2}}{b\,x^2+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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