3.685 \(\int \frac {x^4 (c+d x^2)^{3/2}}{a+b x^2} \, dx\)

Optimal. Leaf size=210 \[ \frac {a^{3/2} (b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^4}-\frac {(b c-2 a d) \left (-8 a^2 d^2+8 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^4 d^{3/2}}+\frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-10 a b c d+b^2 c^2\right )}{16 b^3 d}+\frac {x^3 \sqrt {c+d x^2} (7 b c-6 a d)}{24 b^2}+\frac {d x^5 \sqrt {c+d x^2}}{6 b} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {477, 582, 523, 217, 206, 377, 205} \[ \frac {x \sqrt {c+d x^2} \left (8 a^2 d^2-10 a b c d+b^2 c^2\right )}{16 b^3 d}-\frac {(b c-2 a d) \left (-8 a^2 d^2+8 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^4 d^{3/2}}+\frac {a^{3/2} (b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^4}+\frac {x^3 \sqrt {c+d x^2} (7 b c-6 a d)}{24 b^2}+\frac {d x^5 \sqrt {c+d x^2}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(c + d*x^2)^(3/2))/(a + b*x^2),x]

[Out]

((b^2*c^2 - 10*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(16*b^3*d) + ((7*b*c - 6*a*d)*x^3*Sqrt[c + d*x^2])/(24*
b^2) + (d*x^5*Sqrt[c + d*x^2])/(6*b) + (a^(3/2)*(b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c +
d*x^2])])/b^4 - ((b*c - 2*a*d)*(b^2*c^2 + 8*a*b*c*d - 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(16*b^
4*d^(3/2))

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 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^4 \left (c+d x^2\right )^{3/2}}{a+b x^2} \, dx &=\frac {d x^5 \sqrt {c+d x^2}}{6 b}+\frac {\int \frac {x^4 \left (c (6 b c-5 a d)+d (7 b c-6 a d) x^2\right )}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{6 b}\\ &=\frac {(7 b c-6 a d) x^3 \sqrt {c+d x^2}}{24 b^2}+\frac {d x^5 \sqrt {c+d x^2}}{6 b}-\frac {\int \frac {x^2 \left (3 a c d (7 b c-6 a d)-3 d \left (b^2 c^2-10 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 d}\\ &=\frac {\left (b^2 c^2-10 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3 d}+\frac {(7 b c-6 a d) x^3 \sqrt {c+d x^2}}{24 b^2}+\frac {d x^5 \sqrt {c+d x^2}}{6 b}+\frac {\int \frac {-3 a c d \left (b^2 c^2-10 a b c d+8 a^2 d^2\right )-3 d (b c-2 a d) \left (b^2 c^2+8 a b c d-8 a^2 d^2\right ) x^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{48 b^3 d^2}\\ &=\frac {\left (b^2 c^2-10 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3 d}+\frac {(7 b c-6 a d) x^3 \sqrt {c+d x^2}}{24 b^2}+\frac {d x^5 \sqrt {c+d x^2}}{6 b}+\frac {\left (a^2 (b c-a d)^2\right ) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{b^4}-\frac {\left ((b c-2 a d) \left (b^2 c^2+8 a b c d-8 a^2 d^2\right )\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{16 b^4 d}\\ &=\frac {\left (b^2 c^2-10 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3 d}+\frac {(7 b c-6 a d) x^3 \sqrt {c+d x^2}}{24 b^2}+\frac {d x^5 \sqrt {c+d x^2}}{6 b}+\frac {\left (a^2 (b c-a d)^2\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 ((b c-2 a d) \left (b^2 c^2+8 a b c d-8 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{16 b^4 d}\\ &=\frac {\left (b^2 c^2-10 a b c d+8 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^3 d}+\frac {(7 b c-6 a d) x^3 \sqrt {c+d x^2}}{24 b^2}+\frac {d x^5 \sqrt {c+d x^2}}{6 b}+\frac {a^{3/2} (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^4}-\frac {(b c-2 a d) \left (b^2 c^2+8 a b c d-8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^4 d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 196, normalized size = 0.93 \[ \frac {48 a^{3/2} d^{3/2} (b c-a d)^{3/2} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )+b \sqrt {d} x \sqrt {c+d x^2} \left (24 a^2 d^2-6 a b d \left (5 c+2 d x^2\right )+b^2 \left (3 c^2+14 c d x^2+8 d^2 x^4\right )\right )-3 \left (16 a^3 d^3-24 a^2 b c d^2+6 a b^2 c^2 d+b^3 c^3\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )}{48 b^4 d^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(c + d*x^2)^(3/2))/(a + b*x^2),x]

[Out]

(b*Sqrt[d]*x*Sqrt[c + d*x^2]*(24*a^2*d^2 - 6*a*b*d*(5*c + 2*d*x^2) + b^2*(3*c^2 + 14*c*d*x^2 + 8*d^2*x^4)) + 4
8*a^(3/2)*d^(3/2)*(b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])] - 3*(b^3*c^3 + 6*a*b
^2*c^2*d - 24*a^2*b*c*d^2 + 16*a^3*d^3)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(48*b^4*d^(3/2))

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fricas [A]  time = 3.96, size = 1119, normalized size = 5.33 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(d*x^2+c)^(3/2)/(b*x^2+a),x, algorithm="fricas")

[Out]

[1/96*(3*(b^3*c^3 + 6*a*b^2*c^2*d - 24*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*(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*(7*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(b^3*c^2*d - 10*a*b^2*c*d^2
+ 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d^2), 1/48*(3*(b^3*c^3 + 6*a*b^2*c^2*d - 24*a^2*b*c*d^2 + 16*a^3*d^3)*
sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - 12*(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)) + (8*b^3*d^3*x^5 + 2*(7*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 +
3*(b^3*c^2*d - 10*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d^2), 1/96*(48*(a*b*c*d^2 - a^2*d^3)*sq
rt(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*(b^3*c^3 + 6*a*b^2*c^2*d - 24*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*(7*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(b^3*c^2*d -
10*a*b^2*c*d^2 + 8*a^2*b*d^3)*x)*sqrt(d*x^2 + c))/(b^4*d^2), 1/48*(24*(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*(b^3*c^3 + 6*a*b^2*c^2*d - 24*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*(7*b^3*c*d^2 - 6*a*b^2*d^3)*x^3 + 3*(b^3*c^2*d - 10*a*b^2*c*d^2 + 8*a^2*b*d^3)*
x)*sqrt(d*x^2 + c))/(b^4*d^2)]

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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^4*(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,x):;OUTPUT:

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maple [B]  time = 0.02, size = 2081, normalized size = 9.91 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(d*x^2+c)^(3/2)/(b*x^2+a),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} x^{4}}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(d*x^2+c)^(3/2)/(b*x^2+a),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)^(3/2)*x^4/(b*x^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\left (d\,x^2+c\right )}^{3/2}}{b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(c + d*x^2)^(3/2))/(a + b*x^2),x)

[Out]

int((x^4*(c + d*x^2)^(3/2))/(a + b*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (c + d x^{2}\right )^{\frac {3}{2}}}{a + b x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(d*x**2+c)**(3/2)/(b*x**2+a),x)

[Out]

Integral(x**4*(c + d*x**2)**(3/2)/(a + b*x**2), x)

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