3.7.58 \(\int \frac {x^4 \sqrt {c+d x^2}}{a+b x^2} \, dx\)

Optimal. Leaf size=157 \[ \frac {a^{3/2} \sqrt {b c-a d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^3}-\frac {\left (-8 a^2 d^2+4 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 b^3 d^{3/2}}+\frac {x \sqrt {c+d x^2} (b c-4 a d)}{8 b^2 d}+\frac {x^3 \sqrt {c+d x^2}}{4 b} \]

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Rubi [A]  time = 0.23, antiderivative size = 157, 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 = {478, 582, 523, 217, 206, 377, 205} \begin {gather*} -\frac {\left (-8 a^2 d^2+4 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 b^3 d^{3/2}}+\frac {a^{3/2} \sqrt {b c-a d} \tan ^{-1}\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} (b c-4 a d)}{8 b^2 d}+\frac {x^3 \sqrt {c+d x^2}}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - 4*a*d)*x*Sqrt[c + d*x^2])/(8*b^2*d) + (x^3*Sqrt[c + d*x^2])/(4*b) + (a^(3/2)*Sqrt[b*c - a*d]*ArcTan[(S
qrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/b^3 - ((b^2*c^2 + 4*a*b*c*d - 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqr
t[c + d*x^2]])/(8*b^3*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 478

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(m + n*(p + q) + 1)), x] - Dist[e^n/(b*(m + n*(p +
q) + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[a*c*(m - n + 1) + (a*d*(m - n + 1) - n*q*(b
*c - a*d))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 0] &&
GtQ[m - n + 1, 0] && 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 \sqrt {c+d x^2}}{a+b x^2} \, dx &=\frac {x^3 \sqrt {c+d x^2}}{4 b}-\frac {\int \frac {x^2 \left (3 a c+(-b c+4 a d) x^2\right )}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{4 b}\\ &=\frac {(b c-4 a d) x \sqrt {c+d x^2}}{8 b^2 d}+\frac {x^3 \sqrt {c+d x^2}}{4 b}+\frac {\int \frac {-a c (b c-4 a d)+\left (-b^2 c^2-4 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 {(b c-4 a d) x \sqrt {c+d x^2}}{8 b^2 d}+\frac {x^3 \sqrt {c+d x^2}}{4 b}+\frac {\left (a^2 (b c-a d)\right ) \int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx}{b^3}-\frac {\left (b^2 c^2+4 a b c d-8 a^2 d^2\right ) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{8 b^3 d}\\ &=\frac {(b c-4 a d) x \sqrt {c+d x^2}}{8 b^2 d}+\frac {x^3 \sqrt {c+d x^2}}{4 b}+\frac {\left (a^2 (b c-a d)\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^3}-\frac {\left (b^2 c^2+4 a b c d-8 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{8 b^3 d}\\ &=\frac {(b c-4 a d) x \sqrt {c+d x^2}}{8 b^2 d}+\frac {x^3 \sqrt {c+d x^2}}{4 b}+\frac {a^{3/2} \sqrt {b c-a d} \tan ^{-1}\left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^3}-\frac {\left (b^2 c^2+4 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 d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 148, normalized size = 0.94 \begin {gather*} \frac {8 a^{3/2} d^{3/2} \sqrt {b c-a d} \tan ^{-1}\left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )-\left (-8 a^2 d^2+4 a b c d+b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )+b \sqrt {d} x \sqrt {c+d x^2} \left (-4 a d+b c+2 b d x^2\right )}{8 b^3 d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.40, size = 200, normalized size = 1.27 \begin {gather*} -\frac {a^{3/2} \sqrt {b c-a d} \tan ^{-1}\left (\frac {b \sqrt {d} x^2}{\sqrt {a} \sqrt {b c-a d}}-\frac {b x \sqrt {c+d x^2}}{\sqrt {a} \sqrt {b c-a d}}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c-a d}}\right )}{b^3}+\frac {\left (-8 a^2 d^2+4 a b c d+b^2 c^2\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{8 b^3 d^{3/2}}+\frac {\sqrt {c+d x^2} \left (-4 a d x+b c x+2 b d x^3\right )}{8 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.88, size = 857, normalized size = 5.46 \begin {gather*} \left [\frac {4 \, \sqrt {-a b c + a^{2} d} a d^{2} \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 (b^{2} c^{2} + 4 \, 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 (b^{2} c d - 4 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{16 \, b^{3} d^{2}}, \frac {2 \, \sqrt {-a b c + a^{2} d} a d^{2} \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 (b^{2} c^{2} + 4 \, 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 (b^{2} c d - 4 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{8 \, b^{3} d^{2}}, \frac {8 \, \sqrt {a b c - a^{2} d} a d^{2} \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 (b^{2} c^{2} + 4 \, 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 (b^{2} c d - 4 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{16 \, b^{3} d^{2}}, \frac {4 \, \sqrt {a b c - a^{2} d} a d^{2} \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 (b^{2} c^{2} + 4 \, 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 (b^{2} c d - 4 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{8 \, b^{3} d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(d*x^2+c)^(1/2)/(b*x^2+a),x, algorithm="fricas")

[Out]

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

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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^4*(d*x^2+c)^(1/2)/(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 = 0.07, size = 1088, normalized size = 6.93 \begin {gather*} \frac {a^{3} d \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b^{3}}-\frac {a^{3} d \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b^{3}}-\frac {a^{2} c \ln \left (\frac {\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x -\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b^{2}}+\frac {a^{2} c \ln \left (\frac {-\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {2 \left (a d -b c \right )}{b}+2 \sqrt {-\frac {a d -b c}{b}}\, \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}}{x +\frac {\sqrt {-a b}}{b}}\right )}{2 \sqrt {-a b}\, \sqrt {-\frac {a d -b c}{b}}\, b^{2}}+\frac {a^{2} \sqrt {d}\, \ln \left (\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) d -\frac {\sqrt {-a b}\, d}{b}}{\sqrt {d}}+\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\right )}{2 b^{3}}+\frac {a^{2} \sqrt {d}\, \ln \left (\frac {\left (x -\frac {\sqrt {-a b}}{b}\right ) d +\frac {\sqrt {-a b}\, d}{b}}{\sqrt {d}}+\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\right )}{2 b^{3}}-\frac {a c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{2 b^{2} \sqrt {d}}-\frac {c^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{8 b \,d^{\frac {3}{2}}}-\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} d -\frac {2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a^{2}}{2 \sqrt {-a b}\, b^{2}}+\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} d +\frac {2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right ) d}{b}-\frac {a d -b c}{b}}\, a^{2}}{2 \sqrt {-a b}\, b^{2}}-\frac {\sqrt {d \,x^{2}+c}\, a x}{2 b^{2}}-\frac {\sqrt {d \,x^{2}+c}\, c x}{8 b d}+\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}} x}{4 b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(d*x^2+c)^(1/2)/(b*x^2+a),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {d x^{2} + c} x^{4}}{b x^{2} + a}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(d*x^2+c)^(1/2)/(b*x^2+a),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,\sqrt {d\,x^2+c}}{b\,x^2+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(c + d*x^2)^(1/2))/(a + b*x^2),x)

[Out]

int((x^4*(c + d*x^2)^(1/2))/(a + b*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \sqrt {c + d x^{2}}}{a + b x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(d*x**2+c)**(1/2)/(b*x**2+a),x)

[Out]

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

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