3.317 \(\int \frac{x^3 \sqrt{a+c x^2}}{d+e x} \, dx\)

Optimal. Leaf size=211 \[ \frac{\left (-a^2 e^4+4 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{3/2} e^5}-\frac{\sqrt{a+c x^2} \left (8 c d^3-e x \left (4 c d^2-a e^2\right )\right )}{8 c e^4}+\frac{d^3 \sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^5}-\frac{7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac{\left (a+c x^2\right )^{3/2} (d+e x)}{4 c e^2} \]

[Out]

-((8*c*d^3 - e*(4*c*d^2 - a*e^2)*x)*Sqrt[a + c*x^2])/(8*c*e^4) - (7*d*(a + c*x^2)^(3/2))/(12*c*e^2) + ((d + e*
x)*(a + c*x^2)^(3/2))/(4*c*e^2) + ((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^5) + (d^3*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^5

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Rubi [A]  time = 0.390428, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{\left (-a^2 e^4+4 a c d^2 e^2+8 c^2 d^4\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{8 c^{3/2} e^5}-\frac{\sqrt{a+c x^2} \left (8 c d^3-e x \left (4 c d^2-a e^2\right )\right )}{8 c e^4}+\frac{d^3 \sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{e^5}-\frac{7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac{\left (a+c x^2\right )^{3/2} (d+e x)}{4 c e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((8*c*d^3 - e*(4*c*d^2 - a*e^2)*x)*Sqrt[a + c*x^2])/(8*c*e^4) - (7*d*(a + c*x^2)^(3/2))/(12*c*e^2) + ((d + e*
x)*(a + c*x^2)^(3/2))/(4*c*e^2) + ((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^5) + (d^3*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^5

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^3 \sqrt{a+c x^2}}{d+e x} \, dx &=\frac{(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}+\frac{\int \frac{\sqrt{a+c x^2} \left (-a d e^2-e \left (3 c d^2+a e^2\right ) x-7 c d e^2 x^2\right )}{d+e x} \, dx}{4 c e^3}\\ &=-\frac{7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac{(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}+\frac{\int \frac{\left (-3 a c d e^4+3 c e^3 \left (4 c d^2-a e^2\right ) x\right ) \sqrt{a+c x^2}}{d+e x} \, dx}{12 c^2 e^5}\\ &=-\frac{\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^4}-\frac{7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac{(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}+\frac{\int \frac{-3 a c^2 d e^4 \left (4 c d^2+a e^2\right )+3 c^2 e^3 \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}{24 c^3 e^7}\\ &=-\frac{\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^4}-\frac{7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac{(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}-\frac{\left (d^3 \left (c d^2+a e^2\right )\right ) \int \frac{1}{(d+e x) \sqrt{a+c x^2}} \, dx}{e^5}+\frac{\left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{8 c e^5}\\ &=-\frac{\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^4}-\frac{7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac{(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}+\frac{\left (d^3 \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^5}+\frac{\left (8 c^2 d^4+4 a c d^2 e^2-a^2 e^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{8 c e^5}\\ &=-\frac{\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^4}-\frac{7 d \left (a+c x^2\right )^{3/2}}{12 c e^2}+\frac{(d+e x) \left (a+c x^2\right )^{3/2}}{4 c e^2}+\frac{\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^5}+\frac{d^3 \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^5}\\ \end{align*}

Mathematica [A]  time = 0.465301, size = 225, normalized size = 1.07 \[ \frac{24 c^{3/2} d^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )+e \sqrt{a+c x^2} \left (a e^2 (3 e x-8 d)+c \left (12 d^2 e x-24 d^3-8 d e^2 x^2+6 e^3 x^3\right )\right )+24 c d^3 \sqrt{a e^2+c d^2} \tanh ^{-1}\left (\frac{a e-c d x}{\sqrt{a+c x^2} \sqrt{a e^2+c d^2}}\right )}{24 c e^5}-\frac{\sqrt{a} \sqrt{a+c x^2} \left (a e^2-4 c d^2\right ) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 c^{3/2} e^3 \sqrt{\frac{c x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a]*(-4*c*d^2 + a*e^2)*Sqrt[a + c*x^2]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/(8*c^(3/2)*e^3*Sqrt[1 + (c*x^2)/a])
+ (e*Sqrt[a + c*x^2]*(a*e^2*(-8*d + 3*e*x) + c*(-24*d^3 + 12*d^2*e*x - 8*d*e^2*x^2 + 6*e^3*x^3)) + 24*c^(3/2)
*d^4*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]] + 24*c*d^3*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])])/(24*c*e^5)

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Maple [B]  time = 0.237, size = 515, normalized size = 2.4 \begin{align*}{\frac{x}{4\,ce} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{ax}{8\,ce}\sqrt{c{x}^{2}+a}}-{\frac{{a}^{2}}{8\,e}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}-{\frac{d}{3\,c{e}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{{d}^{2}x}{2\,{e}^{3}}\sqrt{c{x}^{2}+a}}+{\frac{a{d}^{2}}{2\,{e}^{3}}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{d}^{3}}{{e}^{4}}\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}^{4}}{{e}^{5}}\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}^{3}a}{{e}^{4}}\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}^{5}c}{{e}^{6}}\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^3*(c*x^2+a)^(1/2)/(e*x+d),x)

[Out]

1/4/e*x*(c*x^2+a)^(3/2)/c-1/8/e*a/c*x*(c*x^2+a)^(1/2)-1/8/e*a^2/c^(3/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))-1/3*d*(c
*x^2+a)^(3/2)/c/e^2+1/2*d^2/e^3*x*(c*x^2+a)^(1/2)+1/2*d^2/e^3*a/c^(1/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))-d^3/e^4*
((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*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^3/e^4/((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^5/e^6
/((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^3*(c*x^2+a)^(1/2)/(e*x+d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 37.417, size = 2076, normalized size = 9.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*x^2+a)^(1/2)/(e*x+d),x, algorithm="fricas")

[Out]

[1/48*(24*sqrt(c*d^2 + a*e^2)*c^2*d^3*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*s
qrt(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)) - 3*(8*c^2*d^4 + 4*a*c*d^2*e^2 -
a^2*e^4)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(6*c^2*e^4*x^3 - 8*c^2*d*e^3*x^2 - 24*c^2
*d^3*e - 8*a*c*d*e^3 + 3*(4*c^2*d^2*e^2 + a*c*e^4)*x)*sqrt(c*x^2 + a))/(c^2*e^5), 1/48*(48*sqrt(-c*d^2 - a*e^2
)*c^2*d^3*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)) - 3*(8*c^2*d^4 + 4*a*c*d^2*e^2 - a^2*e^4)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(6*
c^2*e^4*x^3 - 8*c^2*d*e^3*x^2 - 24*c^2*d^3*e - 8*a*c*d*e^3 + 3*(4*c^2*d^2*e^2 + a*c*e^4)*x)*sqrt(c*x^2 + a))/(
c^2*e^5), 1/24*(12*sqrt(c*d^2 + a*e^2)*c^2*d^3*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)) - 3*(8*c^2*d^4 + 4*a*c*d
^2*e^2 - a^2*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (6*c^2*e^4*x^3 - 8*c^2*d*e^3*x^2 - 24*c^2*d^3*
e - 8*a*c*d*e^3 + 3*(4*c^2*d^2*e^2 + a*c*e^4)*x)*sqrt(c*x^2 + a))/(c^2*e^5), 1/24*(24*sqrt(-c*d^2 - a*e^2)*c^2
*d^3*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))
- 3*(8*c^2*d^4 + 4*a*c*d^2*e^2 - a^2*e^4)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (6*c^2*e^4*x^3 - 8*c^2
*d*e^3*x^2 - 24*c^2*d^3*e - 8*a*c*d*e^3 + 3*(4*c^2*d^2*e^2 + a*c*e^4)*x)*sqrt(c*x^2 + a))/(c^2*e^5)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \sqrt{a + c x^{2}}}{d + e x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(c*x**2+a)**(1/2)/(e*x+d),x)

[Out]

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

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Giac [A]  time = 1.25477, size = 271, normalized size = 1.28 \begin{align*} -\frac{2 \,{\left (c d^{5} + a d^{3} 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 (-5\right )}}{\sqrt{-c d^{2} - a e^{2}}} + \frac{1}{24} \, \sqrt{c x^{2} + a}{\left ({\left (2 \,{\left (3 \, x e^{\left (-1\right )} - 4 \, d e^{\left (-2\right )}\right )} x + \frac{3 \,{\left (4 \, c^{2} d^{2} e^{12} + a c e^{14}\right )} e^{\left (-15\right )}}{c^{2}}\right )} x - \frac{8 \,{\left (3 \, c^{2} d^{3} e^{11} + a c d e^{13}\right )} e^{\left (-15\right )}}{c^{2}}\right )} - \frac{{\left (8 \, c^{2} d^{4} + 4 \, a c d^{2} e^{2} - a^{2} e^{4}\right )} e^{\left (-5\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{8 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(c*x^2+a)^(1/2)/(e*x+d),x, algorithm="giac")

[Out]

-2*(c*d^5 + a*d^3*e^2)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-5)/sqrt
(-c*d^2 - a*e^2) + 1/24*sqrt(c*x^2 + a)*((2*(3*x*e^(-1) - 4*d*e^(-2))*x + 3*(4*c^2*d^2*e^12 + a*c*e^14)*e^(-15
)/c^2)*x - 8*(3*c^2*d^3*e^11 + a*c*d*e^13)*e^(-15)/c^2) - 1/8*(8*c^2*d^4 + 4*a*c*d^2*e^2 - a^2*e^4)*e^(-5)*log
(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(3/2)