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

Optimal. Leaf size=255 \[ -\frac{d \left (-a^2 e^4+4 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{3/2} e^6}+\frac{\left (a+c x^2\right )^{3/2} \left (47 c d^2-8 a e^2\right )}{60 c^2 e^3}+\frac{d \sqrt{a+c x^2} \left (8 c d^3-e x \left (4 c d^2-a e^2\right )\right )}{8 c e^5}-\frac{d^4 \sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^6}-\frac{13 d \left (a+c x^2\right )^{3/2} (d+e x)}{20 c e^3}+\frac{\left (a+c x^2\right )^{3/2} (d+e x)^2}{5 c e^3} \]

[Out]

(d*(8*c*d^3 - e*(4*c*d^2 - a*e^2)*x)*Sqrt[a + c*x^2])/(8*c*e^5) + ((47*c*d^2 - 8*a*e^2)*(a + c*x^2)^(3/2))/(60
*c^2*e^3) - (13*d*(d + e*x)*(a + c*x^2)^(3/2))/(20*c*e^3) + ((d + e*x)^2*(a + c*x^2)^(3/2))/(5*c*e^3) - (d*(8*
c^2*d^4 + 4*a*c*d^2*e^2 - a^2*e^4)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*c^(3/2)*e^6) - (d^4*Sqrt[c*d^2 + a
*e^2]*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/e^6

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Rubi [A]  time = 0.629224, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {1654, 815, 844, 217, 206, 725} \[ -\frac{d \left (-a^2 e^4+4 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{3/2} e^6}+\frac{\left (a+c x^2\right )^{3/2} \left (47 c d^2-8 a e^2\right )}{60 c^2 e^3}+\frac{d \sqrt{a+c x^2} \left (8 c d^3-e x \left (4 c d^2-a e^2\right )\right )}{8 c e^5}-\frac{d^4 \sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^6}-\frac{13 d \left (a+c x^2\right )^{3/2} (d+e x)}{20 c e^3}+\frac{\left (a+c x^2\right )^{3/2} (d+e x)^2}{5 c e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(8*c*d^3 - e*(4*c*d^2 - a*e^2)*x)*Sqrt[a + c*x^2])/(8*c*e^5) + ((47*c*d^2 - 8*a*e^2)*(a + c*x^2)^(3/2))/(60
*c^2*e^3) - (13*d*(d + e*x)*(a + c*x^2)^(3/2))/(20*c*e^3) + ((d + e*x)^2*(a + c*x^2)^(3/2))/(5*c*e^3) - (d*(8*
c^2*d^4 + 4*a*c*d^2*e^2 - a^2*e^4)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*c^(3/2)*e^6) - (d^4*Sqrt[c*d^2 + a
*e^2]*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/e^6

Rule 1654

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 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 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 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rubi steps

\begin{align*} \int \frac{x^4 \sqrt{a+c x^2}}{d+e x} \, dx &=\frac{(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}+\frac{\int \frac{\sqrt{a+c x^2} \left (-2 a d^2 e^2-d e \left (3 c d^2+4 a e^2\right ) x-e^2 \left (11 c d^2+2 a e^2\right ) x^2-13 c d e^3 x^3\right )}{d+e x} \, dx}{5 c e^4}\\ &=-\frac{13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac{(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}+\frac{\int \frac{\sqrt{a+c x^2} \left (5 a c d^2 e^5+3 c d e^4 \left (9 c d^2-a e^2\right ) x+c e^5 \left (47 c d^2-8 a e^2\right ) x^2\right )}{d+e x} \, dx}{20 c^2 e^7}\\ &=\frac{\left (47 c d^2-8 a e^2\right ) \left (a+c x^2\right )^{3/2}}{60 c^2 e^3}-\frac{13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac{(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}+\frac{\int \frac{\left (15 a c^2 d^2 e^7-15 c^2 d e^6 \left (4 c d^2-a e^2\right ) x\right ) \sqrt{a+c x^2}}{d+e x} \, dx}{60 c^3 e^9}\\ &=\frac{d \left (8 c d^3-e \left (4 c d^2-a e^2\right ) x\right ) \sqrt{a+c x^2}}{8 c e^5}+\frac{\left (47 c d^2-8 a e^2\right ) \left (a+c x^2\right )^{3/2}}{60 c^2 e^3}-\frac{13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac{(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}+\frac{\int \frac{15 a c^3 d^2 e^7 \left (4 c d^2+a e^2\right )-15 c^3 d e^6 \left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right ) x}{(d+e x) \sqrt{a+c x^2}} \, dx}{120 c^4 e^{11}}\\ &=\frac{d \left (8 c d^3-e \left (4 c d^2-a e^2\right ) x\right ) \sqrt{a+c x^2}}{8 c e^5}+\frac{\left (47 c d^2-8 a e^2\right ) \left (a+c x^2\right )^{3/2}}{60 c^2 e^3}-\frac{13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac{(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}+\frac{\left (d^4 \left (c d^2+a e^2\right )\right ) \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{e^6}-\frac{\left (d \left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right )\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{8 c e^6}\\ &=\frac{d \left (8 c d^3-e \left (4 c d^2-a e^2\right ) x\right ) \sqrt{a+c x^2}}{8 c e^5}+\frac{\left (47 c d^2-8 a e^2\right ) \left (a+c x^2\right )^{3/2}}{60 c^2 e^3}-\frac{13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac{(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}-\frac{\left (d^4 \left (c d^2+a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c d^2+a e^2-x^2} \, dx,x,\frac{a e-c d x}{\sqrt{a+c x^2}}\right )}{e^6}-\frac{\left (d \left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{8 c e^6}\\ &=\frac{d \left (8 c d^3-e \left (4 c d^2-a e^2\right ) x\right ) \sqrt{a+c x^2}}{8 c e^5}+\frac{\left (47 c d^2-8 a e^2\right ) \left (a+c x^2\right )^{3/2}}{60 c^2 e^3}-\frac{13 d (d+e x) \left (a+c x^2\right )^{3/2}}{20 c e^3}+\frac{(d+e x)^2 \left (a+c x^2\right )^{3/2}}{5 c e^3}-\frac{d \left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{3/2} e^6}-\frac{d^4 \sqrt{c d^2+a e^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{c d^2+a e^2} \sqrt{a+c x^2}}\right )}{e^6}\\ \end{align*}

Mathematica [A]  time = 0.668256, size = 259, normalized size = 1.02 \[ \frac{e \sqrt{a+c x^2} \left (-16 a^2 e^4+a c e^2 \left (40 d^2-15 d e x+8 e^2 x^2\right )+2 c^2 \left (20 d^2 e^2 x^2-30 d^3 e x+60 d^4-15 d e^3 x^3+12 e^4 x^4\right )\right )-120 c^2 d^4 \sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )-120 c^{5/2} d^5 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )+\frac{15 \sqrt{a} \sqrt{c} d e^2 \sqrt{a+c x^2} \left (a e^2-4 c d^2\right ) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{\frac{c x^2}{a}+1}}}{120 c^2 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[a + c*x^2]*(-16*a^2*e^4 + a*c*e^2*(40*d^2 - 15*d*e*x + 8*e^2*x^2) + 2*c^2*(60*d^4 - 30*d^3*e*x + 20*d^
2*e^2*x^2 - 15*d*e^3*x^3 + 12*e^4*x^4)) + (15*Sqrt[a]*Sqrt[c]*d*e^2*(-4*c*d^2 + a*e^2)*Sqrt[a + c*x^2]*ArcSinh
[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[1 + (c*x^2)/a] - 120*c^(5/2)*d^5*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]] - 120*c^2*d^
4*Sqrt[c*d^2 + a*e^2]*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(120*c^2*e^6)

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Maple [B]  time = 0.278, size = 560, normalized size = 2.2 \begin{align*}{\frac{{x}^{2}}{5\,ce} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{2\,a}{15\,{c}^{2}e} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{dx}{4\,c{e}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{adx}{8\,c{e}^{2}}\sqrt{c{x}^{2}+a}}+{\frac{{a}^{2}d}{8\,{e}^{2}}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{{d}^{2}}{3\,{e}^{3}c} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{{d}^{3}x}{2\,{e}^{4}}\sqrt{c{x}^{2}+a}}-{\frac{{d}^{3}a}{2\,{e}^{4}}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}}+{\frac{{d}^{4}}{{e}^{5}}\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}-{\frac{{d}^{5}}{{e}^{6}}\sqrt{c}\ln \left ({ \left ( -{\frac{cd}{e}}+ \left ({\frac{d}{e}}+x \right ) c \right ){\frac{1}{\sqrt{c}}}}+\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) }-{\frac{{d}^{4}a}{{e}^{5}}\ln \left ({ \left ( 2\,{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}}-{\frac{{d}^{6}c}{{e}^{7}}\ln \left ({ \left ( 2\,{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}\sqrt{ \left ({\frac{d}{e}}+x \right ) ^{2}c-2\,{\frac{cd}{e} \left ({\frac{d}{e}}+x \right ) }+{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{{\frac{a{e}^{2}+c{d}^{2}}{{e}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/e*x^2*(c*x^2+a)^(3/2)/c-2/15/e*a/c^2*(c*x^2+a)^(3/2)-1/4*d/e^2*x*(c*x^2+a)^(3/2)/c+1/8*d/e^2*a/c*x*(c*x^2+
a)^(1/2)+1/8*d/e^2*a^2/c^(3/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))+1/3*d^2/e^3*(c*x^2+a)^(3/2)/c-1/2*d^3/e^4*x*(c*x^
2+a)^(1/2)-1/2*d^3/e^4*a/c^(1/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))+d^4/e^5*((d/e+x)^2*c-2*c*d/e*(d/e+x)+(a*e^2+c*d
^2)/e^2)^(1/2)-d^5/e^6*c^(1/2)*ln((-c*d/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(
1/2))-d^4/e^5/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2+c*d^2)/e^2-2*c*d/e*(d/e+x)+2*((a*e^2+c*d^2)/e^2)^(1/2)*((
d/e+x)^2*c-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))/(d/e+x))*a-d^6/e^7/((a*e^2+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2
+c*d^2)/e^2-2*c*d/e*(d/e+x)+2*((a*e^2+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c-2*c*d/e*(d/e+x)+(a*e^2+c*d^2)/e^2)^(1/2))
/(d/e+x))*c

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 40.7479, size = 2407, normalized size = 9.44 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/240*(120*sqrt(c*d^2 + a*e^2)*c^2*d^4*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 - 2
*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) - 15*(8*c^2*d^5 + 4*a*c*d^3*e^2
- a^2*d*e^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(24*c^2*e^5*x^4 - 30*c^2*d*e^4*x^3 +
120*c^2*d^4*e + 40*a*c*d^2*e^3 - 16*a^2*e^5 + 8*(5*c^2*d^2*e^3 + a*c*e^5)*x^2 - 15*(4*c^2*d^3*e^2 + a*c*d*e^4
)*x)*sqrt(c*x^2 + a))/(c^2*e^6), -1/240*(240*sqrt(-c*d^2 - a*e^2)*c^2*d^4*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x -
a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) + 15*(8*c^2*d^5 + 4*a*c*d^3*e^2 - a^2*d*e
^4)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) - 2*(24*c^2*e^5*x^4 - 30*c^2*d*e^4*x^3 + 120*c^2*d
^4*e + 40*a*c*d^2*e^3 - 16*a^2*e^5 + 8*(5*c^2*d^2*e^3 + a*c*e^5)*x^2 - 15*(4*c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(
c*x^2 + a))/(c^2*e^6), 1/120*(60*sqrt(c*d^2 + a*e^2)*c^2*d^4*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d
^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + 15*(8*c^
2*d^5 + 4*a*c*d^3*e^2 - a^2*d*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (24*c^2*e^5*x^4 - 30*c^2*d*e^
4*x^3 + 120*c^2*d^4*e + 40*a*c*d^2*e^3 - 16*a^2*e^5 + 8*(5*c^2*d^2*e^3 + a*c*e^5)*x^2 - 15*(4*c^2*d^3*e^2 + a*
c*d*e^4)*x)*sqrt(c*x^2 + a))/(c^2*e^6), -1/120*(120*sqrt(-c*d^2 - a*e^2)*c^2*d^4*arctan(sqrt(-c*d^2 - a*e^2)*(
c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - 15*(8*c^2*d^5 + 4*a*c*d^3*e^2 -
a^2*d*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (24*c^2*e^5*x^4 - 30*c^2*d*e^4*x^3 + 120*c^2*d^4*e +
40*a*c*d^2*e^3 - 16*a^2*e^5 + 8*(5*c^2*d^2*e^3 + a*c*e^5)*x^2 - 15*(4*c^2*d^3*e^2 + a*c*d*e^4)*x)*sqrt(c*x^2 +
a))/(c^2*e^6)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \sqrt{a + c x^{2}}}{d + e x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1966, size = 340, normalized size = 1.33 \begin{align*} \frac{2 \,{\left (c d^{6} + a d^{4} e^{2}\right )} \arctan \left (-\frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )} e + \sqrt{c} d}{\sqrt{-c d^{2} - a e^{2}}}\right ) e^{\left (-6\right )}}{\sqrt{-c d^{2} - a e^{2}}} + \frac{1}{120} \, \sqrt{c x^{2} + a}{\left ({\left (2 \,{\left (3 \,{\left (4 \, x e^{\left (-1\right )} - 5 \, d e^{\left (-2\right )}\right )} x + \frac{4 \,{\left (5 \, c^{3} d^{2} e^{18} + a c^{2} e^{20}\right )} e^{\left (-21\right )}}{c^{3}}\right )} x - \frac{15 \,{\left (4 \, c^{3} d^{3} e^{17} + a c^{2} d e^{19}\right )} e^{\left (-21\right )}}{c^{3}}\right )} x + \frac{8 \,{\left (15 \, c^{3} d^{4} e^{16} + 5 \, a c^{2} d^{2} e^{18} - 2 \, a^{2} c e^{20}\right )} e^{\left (-21\right )}}{c^{3}}\right )} + \frac{{\left (8 \, c^{\frac{5}{2}} d^{5} + 4 \, a c^{\frac{3}{2}} d^{3} e^{2} - a^{2} \sqrt{c} d e^{4}\right )} e^{\left (-6\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{8 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(c*d^6 + a*d^4*e^2)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-6)/sqrt(
-c*d^2 - a*e^2) + 1/120*sqrt(c*x^2 + a)*((2*(3*(4*x*e^(-1) - 5*d*e^(-2))*x + 4*(5*c^3*d^2*e^18 + a*c^2*e^20)*e
^(-21)/c^3)*x - 15*(4*c^3*d^3*e^17 + a*c^2*d*e^19)*e^(-21)/c^3)*x + 8*(15*c^3*d^4*e^16 + 5*a*c^2*d^2*e^18 - 2*
a^2*c*e^20)*e^(-21)/c^3) + 1/8*(8*c^(5/2)*d^5 + 4*a*c^(3/2)*d^3*e^2 - a^2*sqrt(c)*d*e^4)*e^(-6)*log(abs(-sqrt(
c)*x + sqrt(c*x^2 + a)))/c^2