\(\int (d+e x)^3 (a+c x^2)^{3/2} \, dx\) [535]

Optimal result

Integrand size = 19, antiderivative size = 180 \[ \int (d+e x)^3 \left (a+c x^2\right )^{3/2} \, dx=\frac {3 a d \left (2 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{16 c}+\frac {d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac {3 a^2 d \left (2 c d^2-a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}} \]

[Out]

1/8*d*(-a*e^2+2*c*d^2)*x*(c*x^2+a)^(3/2)/c+1/7*e*(e*x+d)^2*(c*x^2+a)^(5/2)/c+1/70*e*(15*c*d*e*x-4*a*e^2+32*c*d
^2)*(c*x^2+a)^(5/2)/c^2+3/16*a^2*d*(-a*e^2+2*c*d^2)*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/c^(3/2)+3/16*a*d*(-a*e^
2+2*c*d^2)*x*(c*x^2+a)^(1/2)/c

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {757, 794, 201, 223, 212} \[ \int (d+e x)^3 \left (a+c x^2\right )^{3/2} \, dx=\frac {3 a^2 d \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (2 c d^2-a e^2\right )}{16 c^{3/2}}+\frac {e \left (a+c x^2\right )^{5/2} \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right )}{70 c^2}+\frac {d x \left (a+c x^2\right )^{3/2} \left (2 c d^2-a e^2\right )}{8 c}+\frac {3 a d x \sqrt {a+c x^2} \left (2 c d^2-a e^2\right )}{16 c}+\frac {e \left (a+c x^2\right )^{5/2} (d+e x)^2}{7 c} \]

[In]

Int[(d + e*x)^3*(a + c*x^2)^(3/2),x]

[Out]

(3*a*d*(2*c*d^2 - a*e^2)*x*Sqrt[a + c*x^2])/(16*c) + (d*(2*c*d^2 - a*e^2)*x*(a + c*x^2)^(3/2))/(8*c) + (e*(d +
e*x)^2*(a + c*x^2)^(5/2))/(7*c) + (e*(4*(8*c*d^2 - a*e^2) + 15*c*d*e*x)*(a + c*x^2)^(5/2))/(70*c^2) + (3*a^2*
d*(2*c*d^2 - a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(16*c^(3/2))

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

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

Rule 223

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 757

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

Rule 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {\int (d+e x) \left (7 c d^2-2 a e^2+9 c d e x\right ) \left (a+c x^2\right )^{3/2} \, dx}{7 c} \\ & = \frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac {\left (d \left (2 c d^2-a e^2\right )\right ) \int \left (a+c x^2\right )^{3/2} \, dx}{2 c} \\ & = \frac {d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac {\left (3 a d \left (2 c d^2-a e^2\right )\right ) \int \sqrt {a+c x^2} \, dx}{8 c} \\ & = \frac {3 a d \left (2 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{16 c}+\frac {d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac {\left (3 a^2 d \left (2 c d^2-a e^2\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{16 c} \\ & = \frac {3 a d \left (2 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{16 c}+\frac {d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac {\left (3 a^2 d \left (2 c d^2-a e^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{16 c} \\ & = \frac {3 a d \left (2 c d^2-a e^2\right ) x \sqrt {a+c x^2}}{16 c}+\frac {d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac {e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac {e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac {3 a^2 d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.96 \[ \int (d+e x)^3 \left (a+c x^2\right )^{3/2} \, dx=\frac {\sqrt {a+c x^2} \left (-32 a^3 e^3+a^2 c e \left (336 d^2+105 d e x+16 e^2 x^2\right )+4 c^3 x^3 \left (35 d^3+84 d^2 e x+70 d e^2 x^2+20 e^3 x^3\right )+2 a c^2 x \left (175 d^3+336 d^2 e x+245 d e^2 x^2+64 e^3 x^3\right )\right )+105 a^2 \sqrt {c} d \left (-2 c d^2+a e^2\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{560 c^2} \]

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-32*a^3*e^3 + a^2*c*e*(336*d^2 + 105*d*e*x + 16*e^2*x^2) + 4*c^3*x^3*(35*d^3 + 84*d^2*e*x +
70*d*e^2*x^2 + 20*e^3*x^3) + 2*a*c^2*x*(175*d^3 + 336*d^2*e*x + 245*d*e^2*x^2 + 64*e^3*x^3)) + 105*a^2*Sqrt[c]
*d*(-2*c*d^2 + a*e^2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(560*c^2)

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.05

method	result	size
risch	\(-\frac {\left (-80 e^{3} c^{3} x^{6}-280 d \,e^{2} c^{3} x^{5}-128 e^{3} c^{2} a \,x^{4}-336 d^{2} e \,c^{3} x^{4}-490 a \,c^{2} d \,e^{2} x^{3}-140 c^{3} d^{3} x^{3}-16 a^{2} c \,e^{3} x^{2}-672 c^{2} d^{2} a e \,x^{2}-105 d \,e^{2} a^{2} c x -350 d^{3} c^{2} a x +32 a^{3} e^{3}-336 d^{2} e \,a^{2} c \right ) \sqrt {c \,x^{2}+a}}{560 c^{2}}-\frac {3 a^{2} d \left (e^{2} a -2 c \,d^{2}\right ) \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{16 c^{\frac {3}{2}}}\)	\(189\)
default	\(d^{3} \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )+e^{3} \left (\frac {x^{2} \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{7 c}-\frac {2 a \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{35 c^{2}}\right )+3 d \,e^{2} \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{6 c}-\frac {a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4}\right )}{6 c}\right )+\frac {3 d^{2} e \left (c \,x^{2}+a \right )^{\frac {5}{2}}}{5 c}\)	\(191\)

[In]

int((e*x+d)^3*(c*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/560/c^2*(-80*c^3*e^3*x^6-280*c^3*d*e^2*x^5-128*a*c^2*e^3*x^4-336*c^3*d^2*e*x^4-490*a*c^2*d*e^2*x^3-140*c^3*
d^3*x^3-16*a^2*c*e^3*x^2-672*a*c^2*d^2*e*x^2-105*a^2*c*d*e^2*x-350*a*c^2*d^3*x+32*a^3*e^3-336*a^2*c*d^2*e)*(c*
x^2+a)^(1/2)-3/16*a^2/c^(3/2)*d*(a*e^2-2*c*d^2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 402, normalized size of antiderivative = 2.23 \[ \int (d+e x)^3 \left (a+c x^2\right )^{3/2} \, dx=\left [\frac {105 \, {\left (2 \, a^{2} c d^{3} - a^{3} d e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (80 \, c^{3} e^{3} x^{6} + 280 \, c^{3} d e^{2} x^{5} + 336 \, a^{2} c d^{2} e - 32 \, a^{3} e^{3} + 16 \, {\left (21 \, c^{3} d^{2} e + 8 \, a c^{2} e^{3}\right )} x^{4} + 70 \, {\left (2 \, c^{3} d^{3} + 7 \, a c^{2} d e^{2}\right )} x^{3} + 16 \, {\left (42 \, a c^{2} d^{2} e + a^{2} c e^{3}\right )} x^{2} + 35 \, {\left (10 \, a c^{2} d^{3} + 3 \, a^{2} c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{1120 \, c^{2}}, -\frac {105 \, {\left (2 \, a^{2} c d^{3} - a^{3} d e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (80 \, c^{3} e^{3} x^{6} + 280 \, c^{3} d e^{2} x^{5} + 336 \, a^{2} c d^{2} e - 32 \, a^{3} e^{3} + 16 \, {\left (21 \, c^{3} d^{2} e + 8 \, a c^{2} e^{3}\right )} x^{4} + 70 \, {\left (2 \, c^{3} d^{3} + 7 \, a c^{2} d e^{2}\right )} x^{3} + 16 \, {\left (42 \, a c^{2} d^{2} e + a^{2} c e^{3}\right )} x^{2} + 35 \, {\left (10 \, a c^{2} d^{3} + 3 \, a^{2} c d e^{2}\right )} x\right )} \sqrt {c x^{2} + a}}{560 \, c^{2}}\right ] \]

[In]

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

[Out]

[1/1120*(105*(2*a^2*c*d^3 - a^3*d*e^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(80*c^3*e^3
*x^6 + 280*c^3*d*e^2*x^5 + 336*a^2*c*d^2*e - 32*a^3*e^3 + 16*(21*c^3*d^2*e + 8*a*c^2*e^3)*x^4 + 70*(2*c^3*d^3
+ 7*a*c^2*d*e^2)*x^3 + 16*(42*a*c^2*d^2*e + a^2*c*e^3)*x^2 + 35*(10*a*c^2*d^3 + 3*a^2*c*d*e^2)*x)*sqrt(c*x^2 +
a))/c^2, -1/560*(105*(2*a^2*c*d^3 - a^3*d*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (80*c^3*e^3*x^6
+ 280*c^3*d*e^2*x^5 + 336*a^2*c*d^2*e - 32*a^3*e^3 + 16*(21*c^3*d^2*e + 8*a*c^2*e^3)*x^4 + 70*(2*c^3*d^3 + 7*a
*c^2*d*e^2)*x^3 + 16*(42*a*c^2*d^2*e + a^2*c*e^3)*x^2 + 35*(10*a*c^2*d^3 + 3*a^2*c*d*e^2)*x)*sqrt(c*x^2 + a))/
c^2]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (165) = 330\).

Time = 0.59 (sec) , antiderivative size = 366, normalized size of antiderivative = 2.03 \[ \int (d+e x)^3 \left (a+c x^2\right )^{3/2} \, dx=\begin {cases} \sqrt {a + c x^{2}} \left (\frac {c d e^{2} x^{5}}{2} + \frac {c e^{3} x^{6}}{7} + \frac {x^{4} \cdot \left (\frac {8 a c e^{3}}{7} + 3 c^{2} d^{2} e\right )}{5 c} + \frac {x^{3} \cdot \left (\frac {7 a c d e^{2}}{2} + c^{2} d^{3}\right )}{4 c} + \frac {x^{2} \left (a^{2} e^{3} + 6 a c d^{2} e - \frac {4 a \left (\frac {8 a c e^{3}}{7} + 3 c^{2} d^{2} e\right )}{5 c}\right )}{3 c} + \frac {x \left (3 a^{2} d e^{2} + 2 a c d^{3} - \frac {3 a \left (\frac {7 a c d e^{2}}{2} + c^{2} d^{3}\right )}{4 c}\right )}{2 c} + \frac {3 a^{2} d^{2} e - \frac {2 a \left (a^{2} e^{3} + 6 a c d^{2} e - \frac {4 a \left (\frac {8 a c e^{3}}{7} + 3 c^{2} d^{2} e\right )}{5 c}\right )}{3 c}}{c}\right ) + \left (a^{2} d^{3} - \frac {a \left (3 a^{2} d e^{2} + 2 a c d^{3} - \frac {3 a \left (\frac {7 a c d e^{2}}{2} + c^{2} d^{3}\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {c} \sqrt {a + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {c x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\a^{\frac {3}{2}} \left (\begin {cases} d^{3} x & \text {for}\: e = 0 \\\frac {\left (d + e x\right )^{4}}{4 e} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a + c*x**2)*(c*d*e**2*x**5/2 + c*e**3*x**6/7 + x**4*(8*a*c*e**3/7 + 3*c**2*d**2*e)/(5*c) + x**
3*(7*a*c*d*e**2/2 + c**2*d**3)/(4*c) + x**2*(a**2*e**3 + 6*a*c*d**2*e - 4*a*(8*a*c*e**3/7 + 3*c**2*d**2*e)/(5*
c))/(3*c) + x*(3*a**2*d*e**2 + 2*a*c*d**3 - 3*a*(7*a*c*d*e**2/2 + c**2*d**3)/(4*c))/(2*c) + (3*a**2*d**2*e - 2
*a*(a**2*e**3 + 6*a*c*d**2*e - 4*a*(8*a*c*e**3/7 + 3*c**2*d**2*e)/(5*c))/(3*c))/c) + (a**2*d**3 - a*(3*a**2*d*
e**2 + 2*a*c*d**3 - 3*a*(7*a*c*d*e**2/2 + c**2*d**3)/(4*c))/(2*c))*Piecewise((log(2*sqrt(c)*sqrt(a + c*x**2) +
2*c*x)/sqrt(c), Ne(a, 0)), (x*log(x)/sqrt(c*x**2), True)), Ne(c, 0)), (a**(3/2)*Piecewise((d**3*x, Eq(e, 0)),
((d + e*x)**4/(4*e), True)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.06 \[ \int (d+e x)^3 \left (a+c x^2\right )^{3/2} \, dx=\frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} e^{3} x^{2}}{7 \, c} + \frac {1}{4} \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} d^{3} x + \frac {3}{8} \, \sqrt {c x^{2} + a} a d^{3} x + \frac {{\left (c x^{2} + a\right )}^{\frac {5}{2}} d e^{2} x}{2 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} a d e^{2} x}{8 \, c} - \frac {3 \, \sqrt {c x^{2} + a} a^{2} d e^{2} x}{16 \, c} + \frac {3 \, a^{2} d^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {c}} - \frac {3 \, a^{3} d e^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, c^{\frac {3}{2}}} + \frac {3 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} d^{2} e}{5 \, c} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {5}{2}} a e^{3}}{35 \, c^{2}} \]

[In]

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

[Out]

1/7*(c*x^2 + a)^(5/2)*e^3*x^2/c + 1/4*(c*x^2 + a)^(3/2)*d^3*x + 3/8*sqrt(c*x^2 + a)*a*d^3*x + 1/2*(c*x^2 + a)^
(5/2)*d*e^2*x/c - 1/8*(c*x^2 + a)^(3/2)*a*d*e^2*x/c - 3/16*sqrt(c*x^2 + a)*a^2*d*e^2*x/c + 3/8*a^2*d^3*arcsinh
(c*x/sqrt(a*c))/sqrt(c) - 3/16*a^3*d*e^2*arcsinh(c*x/sqrt(a*c))/c^(3/2) + 3/5*(c*x^2 + a)^(5/2)*d^2*e/c - 2/35
*(c*x^2 + a)^(5/2)*a*e^3/c^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.21 \[ \int (d+e x)^3 \left (a+c x^2\right )^{3/2} \, dx=\frac {1}{560} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, {\left (2 \, c e^{3} x + 7 \, c d e^{2}\right )} x + \frac {2 \, {\left (21 \, c^{6} d^{2} e + 8 \, a c^{5} e^{3}\right )}}{c^{5}}\right )} x + \frac {35 \, {\left (2 \, c^{6} d^{3} + 7 \, a c^{5} d e^{2}\right )}}{c^{5}}\right )} x + \frac {8 \, {\left (42 \, a c^{5} d^{2} e + a^{2} c^{4} e^{3}\right )}}{c^{5}}\right )} x + \frac {35 \, {\left (10 \, a c^{5} d^{3} + 3 \, a^{2} c^{4} d e^{2}\right )}}{c^{5}}\right )} x + \frac {16 \, {\left (21 \, a^{2} c^{4} d^{2} e - 2 \, a^{3} c^{3} e^{3}\right )}}{c^{5}}\right )} - \frac {3 \, {\left (2 \, a^{2} c d^{3} - a^{3} d e^{2}\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{16 \, c^{\frac {3}{2}}} \]

[In]

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

[Out]

1/560*sqrt(c*x^2 + a)*((2*((4*(5*(2*c*e^3*x + 7*c*d*e^2)*x + 2*(21*c^6*d^2*e + 8*a*c^5*e^3)/c^5)*x + 35*(2*c^6
*d^3 + 7*a*c^5*d*e^2)/c^5)*x + 8*(42*a*c^5*d^2*e + a^2*c^4*e^3)/c^5)*x + 35*(10*a*c^5*d^3 + 3*a^2*c^4*d*e^2)/c
^5)*x + 16*(21*a^2*c^4*d^2*e - 2*a^3*c^3*e^3)/c^5) - 3/16*(2*a^2*c*d^3 - a^3*d*e^2)*log(abs(-sqrt(c)*x + sqrt(
c*x^2 + a)))/c^(3/2)

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^3 \left (a+c x^2\right )^{3/2} \, dx=\int {\left (c\,x^2+a\right )}^{3/2}\,{\left (d+e\,x\right )}^3 \,d x \]

[In]

int((a + c*x^2)^(3/2)*(d + e*x)^3,x)

[Out]

int((a + c*x^2)^(3/2)*(d + e*x)^3, x)