3.696 \(\int \frac {x^2 (c+d x^2)^{5/2}}{a+b x^2} \, dx\)

Optimal. Leaf size=217 \[ \frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-18 a b c d+11 b^2 c^2\right )}{16 b^3}+\frac {\left (-16 a^3 d^3+40 a^2 b c d^2-30 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^4 \sqrt {d}}-\frac {\sqrt {a} (b c-a d)^{5/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^4}+\frac {d x^3 \sqrt {c+d x^2} (3 b c-2 a d)}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b} \]

[Out]

1/6*d*x^3*(d*x^2+c)^(3/2)/b-(-a*d+b*c)^(5/2)*arctan(x*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^2+c)^(1/2))*a^(1/2)/b^4+1/
16*(-16*a^3*d^3+40*a^2*b*c*d^2-30*a*b^2*c^2*d+5*b^3*c^3)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/b^4/d^(1/2)+1/16*(
8*a^2*d^2-18*a*b*c*d+11*b^2*c^2)*x*(d*x^2+c)^(1/2)/b^3+1/8*d*(-2*a*d+3*b*c)*x^3*(d*x^2+c)^(1/2)/b^2

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Rubi [A]  time = 0.39, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {477, 581, 582, 523, 217, 206, 377, 205} \[ \frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-18 a b c d+11 b^2 c^2\right )}{16 b^3}+\frac {\left (40 a^2 b c d^2-16 a^3 d^3-30 a b^2 c^2 d+5 b^3 c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^4 \sqrt {d}}+\frac {d x^3 \sqrt {c+d x^2} (3 b c-2 a d)}{8 b^2}-\frac {\sqrt {a} (b c-a d)^{5/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^4}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(c + d*x^2)^(5/2))/(a + b*x^2),x]

[Out]

((11*b^2*c^2 - 18*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*b^3) + (d*(3*b*c - 2*a*d)*x^3*Sqrt[c + d*x^2])/(
8*b^2) + (d*x^3*(c + d*x^2)^(3/2))/(6*b) - (Sqrt[a]*(b*c - a*d)^(5/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt
[c + d*x^2])])/b^4 + ((5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*ArcTanh[(Sqrt[d]*x)/Sqrt[c +
d*x^2]])/(16*b^4*Sqrt[d])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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 377

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 477

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 523

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 581

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

Rule 582

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 )^{5/2}}{a+b x^2} \, dx &=\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}+\frac {\int \frac {x^2 \sqrt {c+d x^2} \left (3 c (2 b c-a d)+3 d (3 b c-2 a d) x^2\right )}{a+b x^2} \, dx}{6 b}\\ &=\frac {d (3 b c-2 a d) x^3 \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}+\frac {\int \frac {x^2 \left (3 c \left (8 b^2 c^2-13 a b c d+6 a^2 d^2\right )+3 d \left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x^2\right )}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{24 b^2}\\ &=\frac {\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3}+\frac {d (3 b c-2 a d) x^3 \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac {\int \frac {3 a c d \left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right )-3 d \left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{48 b^3 d}\\ &=\frac {\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3}+\frac {d (3 b c-2 a d) x^3 \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac {\left (a (b c-a d)^3\right ) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{b^4}+\frac {\left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{16 b^4}\\ &=\frac {\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3}+\frac {d (3 b c-2 a d) x^3 \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac {\left (a (b c-a d)^3\right ) \operatorname {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{b^4}+\frac {\left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{16 b^4}\\ &=\frac {\left (11 b^2 c^2-18 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3}+\frac {d (3 b c-2 a d) x^3 \sqrt {c+d x^2}}{8 b^2}+\frac {d x^3 \left (c+d x^2\right )^{3/2}}{6 b}-\frac {\sqrt {a} (b c-a d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^4}+\frac {\left (5 b^3 c^3-30 a b^2 c^2 d+40 a^2 b c d^2-16 a^3 d^3\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^4 \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 187, normalized size = 0.86 \[ \frac {b x \sqrt {c+d x^2} \left (24 a^2 d^2-6 a b d \left (9 c+2 d x^2\right )+b^2 \left (33 c^2+26 c d x^2+8 d^2 x^4\right )\right )+\frac {3 \left (-16 a^3 d^3+40 a^2 b c d^2-30 a b^2 c^2 d+5 b^3 c^3\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{\sqrt {d}}-48 \sqrt {a} (b c-a d)^{5/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{48 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(c + d*x^2)^(5/2))/(a + b*x^2),x]

[Out]

(b*x*Sqrt[c + d*x^2]*(24*a^2*d^2 - 6*a*b*d*(9*c + 2*d*x^2) + b^2*(33*c^2 + 26*c*d*x^2 + 8*d^2*x^4)) - 48*Sqrt[
a]*(b*c - a*d)^(5/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])] + (3*(5*b^3*c^3 - 30*a*b^2*c^2*d +
40*a^2*b*c*d^2 - 16*a^3*d^3)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/Sqrt[d])/(48*b^4)

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fricas [A]  time = 6.05, size = 1161, normalized size = 5.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(d*x^2+c)^(5/2)/(b*x^2+a),x, algorithm="fricas")

[Out]

[-1/96*(3*(5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*
sqrt(d)*x - c) - 24*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*sqrt(-a*b*c + a^2*d)*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*(8*b^3*d^3*x^5 + 2*(13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c
^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d), -1/48*(3*(5*b^3*c^3 - 30*a*b^2*c^2*d + 40*a^
2*b*c*d^2 - 16*a^3*d^3)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - 12*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*s
qrt(-a*b*c + a^2*d)*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)) - (8*b^3*d^3*x^5
+ 2*(13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b
^4*d), -1/96*(48*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*sqrt(a*b*c - a^2*d)*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*(5*b^3*c^3 - 30*a*b
^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(8*b^3*d^3
*x^5 + 2*(13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c)
)/(b^4*d), -1/48*(24*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*sqrt(a*b*c - a^2*d)*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*(5*b^3*c^3 - 30
*a*b^2*c^2*d + 40*a^2*b*c*d^2 - 16*a^3*d^3)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (8*b^3*d^3*x^5 + 2*(
13*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(11*b^3*c^2*d - 18*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(d*x^2+c)^(5/2)/(b*x^2+a),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:

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maple [B]  time = 0.02, size = 3235, normalized size = 14.91 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(d*x^2+c)^(5/2)/(b*x^2+a),x)

[Out]

1/6*a/(-a*b)^(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)^(3/2)*c-1/8*a/
b^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+5/16/b*c^3/d^(1/2)*ln
(d^(1/2)*x+(d*x^2+c)^(1/2))+5/24/b*c*x*(d*x^2+c)^(3/2)+5/16/b*c^2*x*(d*x^2+c)^(1/2)+1/10*a/(-a*b)^(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)^(5/2)-1/2*a^3/b^4*d^(5/2)*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*d^(5/2)*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))-3/2*a^2/(-a*b)^(1/2)/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))*d*c^2-3/2*a^3/(-a*b)^(1/2)/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^2*c+3/2*a^2/(-a*b)^(1/2)/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))*d*c^2+3/2*a^3/(-a*b)^(
1/2)/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^2*c-1/1
0*a/(-a*b)^(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)^(5/2)+1/2*a/(-a*
b)^(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^3-7/1
6*a/b^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+a^2/(-a*b)^(1/2
)/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)*d*c-1/2*a^4/(-a*b)^(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))*d^3+1/2*a^4/
(-a*b)^(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))*
d^3-1/2*a/(-a*b)^(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^3-7/16*a/b^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-
a^2/(-a*b)^(1/2)/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)*d*c-1/2*
a/(-a*b)^(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*a^2/
b^3*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+5/4*a^2/b^3*d^(3/2)
*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^3/(-a*b)^(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)*d^2+1/6/b*x*(d*x^2+c)^(5/2)-1/8*a/b^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-1/6*a^2/(-a*b)^(1/2)/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)*d+1/4*a^2/b^3*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+5/4*a^2/b^3*d^(3/2)*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^3/(-a*b)^(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)*d^2+1/2*a/(-a*b)^(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^2-15/16*a/b^2*d^(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-
15/16*a/b^2*d^(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*a^2/(-a*b)^(1/2)/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)*d-1/6*a/(-a*b)^(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)^(3/2)*c

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} x^{2}}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(d*x^2+c)^(5/2)/(b*x^2+a),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)^(5/2)*x^2/(b*x^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,{\left (d\,x^2+c\right )}^{5/2}}{b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(c + d*x^2)^(5/2))/(a + b*x^2),x)

[Out]

int((x^2*(c + d*x^2)^(5/2))/(a + b*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}{a + b x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(d*x**2+c)**(5/2)/(b*x**2+a),x)

[Out]

Integral(x**2*(c + d*x**2)**(5/2)/(a + b*x**2), x)

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