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

Optimal. Leaf size=283 \[ \frac {3 \left (c d^2-a e^2\right )^5 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{256 c^{7/2} d^{7/2} e^{5/2}}-\frac {3 \left (c d^2-a e^2\right )^3 \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{128 c^3 d^3 e^2}+\frac {\left (c d^2-a e^2\right ) \left (a e^2+c d^2+2 c d e x\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{16 c^2 d^2 e}+\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 c d} \]

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Rubi [A]  time = 0.16, antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {640, 612, 621, 206} \begin {gather*} -\frac {3 \left (c d^2-a e^2\right )^3 \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{128 c^3 d^3 e^2}+\frac {\left (c d^2-a e^2\right ) \left (a e^2+c d^2+2 c d e x\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{16 c^2 d^2 e}+\frac {3 \left (c d^2-a e^2\right )^5 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{256 c^{7/2} d^{7/2} e^{5/2}}+\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{5 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(c*d^2 - a*e^2)^3*(c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(128*c^3*d^3*e^
2) + ((c*d^2 - a*e^2)*(c*d^2 + a*e^2 + 2*c*d*e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(16*c^2*d^2*e
) + (a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2)/(5*c*d) + (3*(c*d^2 - a*e^2)^5*ArcTanh[(c*d^2 + a*e^2 + 2*c*
d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(256*c^(7/2)*d^(7/2)*e^(5/2))

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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 640

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

Rubi steps

\begin {align*} \int (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2} \, dx &=\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 c d}+\frac {\left (d^2-\frac {a e^2}{c}\right ) \int \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2} \, dx}{2 d}\\ &=\frac {\left (c d^2-a e^2\right ) \left (c d^2+a e^2+2 c d e x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{16 c^2 d^2 e}+\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 c d}-\frac {\left (3 \left (c d^2-a e^2\right )^3\right ) \int \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{32 c^2 d^2 e}\\ &=-\frac {3 \left (c d^2-a e^2\right )^3 \left (c d^2+a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{128 c^3 d^3 e^2}+\frac {\left (c d^2-a e^2\right ) \left (c d^2+a e^2+2 c d e x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{16 c^2 d^2 e}+\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 c d}+\frac {\left (3 \left (c d^2-a e^2\right )^5\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{256 c^3 d^3 e^2}\\ &=-\frac {3 \left (c d^2-a e^2\right )^3 \left (c d^2+a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{128 c^3 d^3 e^2}+\frac {\left (c d^2-a e^2\right ) \left (c d^2+a e^2+2 c d e x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{16 c^2 d^2 e}+\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 c d}+\frac {\left (3 \left (c d^2-a e^2\right )^5\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d e-x^2} \, dx,x,\frac {c d^2+a e^2+2 c d e x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{128 c^3 d^3 e^2}\\ &=-\frac {3 \left (c d^2-a e^2\right )^3 \left (c d^2+a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{128 c^3 d^3 e^2}+\frac {\left (c d^2-a e^2\right ) \left (c d^2+a e^2+2 c d e x\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{16 c^2 d^2 e}+\frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{5 c d}+\frac {3 \left (c d^2-a e^2\right )^5 \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{256 c^{7/2} d^{7/2} e^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 2.22, size = 299, normalized size = 1.06 \begin {gather*} \frac {(a e+c d x)^2 \sqrt {(d+e x) (a e+c d x)} \left (\frac {15 c^{3/2} d^{3/2} \sqrt {c d} \left (c d^2-a e^2\right )^{9/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d} \sqrt {c d^2-a e^2}}\right )}{e^{5/2} (a e+c d x)^{5/2} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}}}+80 c^3 d^3 (d+e x) \left (c d^2-a e^2\right )+40 \left (c^2 d^3-a c d e^2\right )^2-\frac {15 c^2 d^2 \left (c d^2-a e^2\right )^4}{e^2 (a e+c d x)^2}+\frac {10 c^2 d^2 \left (c d^2-a e^2\right )^3}{e (a e+c d x)}+128 c^4 d^4 (d+e x)^2\right )}{640 c^5 d^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a*e + c*d*x)^2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(40*(c^2*d^3 - a*c*d*e^2)^2 - (15*c^2*d^2*(c*d^2 - a*e^2)^4)/(e
^2*(a*e + c*d*x)^2) + (10*c^2*d^2*(c*d^2 - a*e^2)^3)/(e*(a*e + c*d*x)) + 80*c^3*d^3*(c*d^2 - a*e^2)*(d + e*x)
+ 128*c^4*d^4*(d + e*x)^2 + (15*c^(3/2)*d^(3/2)*Sqrt[c*d]*(c*d^2 - a*e^2)^(9/2)*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[
e]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*Sqrt[c*d^2 - a*e^2])])/(e^(5/2)*(a*e + c*d*x)^(5/2)*Sqrt[(c*d*(d + e*x))/(c*d
^2 - a*e^2)])))/(640*c^5*d^5)

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IntegrateAlgebraic [F]  time = 180.01, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [A]  time = 0.47, size = 844, normalized size = 2.98 \begin {gather*} \left [\frac {15 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\right )} \sqrt {c d e} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {c d e} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) + 4 \, {\left (128 \, c^{5} d^{5} e^{5} x^{4} - 15 \, c^{5} d^{9} e + 70 \, a c^{4} d^{7} e^{3} + 128 \, a^{2} c^{3} d^{5} e^{5} - 70 \, a^{3} c^{2} d^{3} e^{7} + 15 \, a^{4} c d e^{9} + 16 \, {\left (21 \, c^{5} d^{6} e^{4} + 11 \, a c^{4} d^{4} e^{6}\right )} x^{3} + 8 \, {\left (31 \, c^{5} d^{7} e^{3} + 64 \, a c^{4} d^{5} e^{5} + a^{2} c^{3} d^{3} e^{7}\right )} x^{2} + 2 \, {\left (5 \, c^{5} d^{8} e^{2} + 233 \, a c^{4} d^{6} e^{4} + 23 \, a^{2} c^{3} d^{4} e^{6} - 5 \, a^{3} c^{2} d^{2} e^{8}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{2560 \, c^{4} d^{4} e^{3}}, -\frac {15 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\right )} \sqrt {-c d e} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-c d e}}{2 \, {\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} + {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )}}\right ) - 2 \, {\left (128 \, c^{5} d^{5} e^{5} x^{4} - 15 \, c^{5} d^{9} e + 70 \, a c^{4} d^{7} e^{3} + 128 \, a^{2} c^{3} d^{5} e^{5} - 70 \, a^{3} c^{2} d^{3} e^{7} + 15 \, a^{4} c d e^{9} + 16 \, {\left (21 \, c^{5} d^{6} e^{4} + 11 \, a c^{4} d^{4} e^{6}\right )} x^{3} + 8 \, {\left (31 \, c^{5} d^{7} e^{3} + 64 \, a c^{4} d^{5} e^{5} + a^{2} c^{3} d^{3} e^{7}\right )} x^{2} + 2 \, {\left (5 \, c^{5} d^{8} e^{2} + 233 \, a c^{4} d^{6} e^{4} + 23 \, a^{2} c^{3} d^{4} e^{6} - 5 \, a^{3} c^{2} d^{2} e^{8}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{1280 \, c^{4} d^{4} e^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2560*(15*(c^5*d^10 - 5*a*c^4*d^8*e^2 + 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8 - a^5*e^10
)*sqrt(c*d*e)*log(8*c^2*d^2*e^2*x^2 + c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4 + 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 +
a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(c*d*e) + 8*(c^2*d^3*e + a*c*d*e^3)*x) + 4*(128*c^5*d^5*e^5*x^4 - 15
*c^5*d^9*e + 70*a*c^4*d^7*e^3 + 128*a^2*c^3*d^5*e^5 - 70*a^3*c^2*d^3*e^7 + 15*a^4*c*d*e^9 + 16*(21*c^5*d^6*e^4
+ 11*a*c^4*d^4*e^6)*x^3 + 8*(31*c^5*d^7*e^3 + 64*a*c^4*d^5*e^5 + a^2*c^3*d^3*e^7)*x^2 + 2*(5*c^5*d^8*e^2 + 23
3*a*c^4*d^6*e^4 + 23*a^2*c^3*d^4*e^6 - 5*a^3*c^2*d^2*e^8)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^4
*d^4*e^3), -1/1280*(15*(c^5*d^10 - 5*a*c^4*d^8*e^2 + 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8
- a^5*e^10)*sqrt(-c*d*e)*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*s
qrt(-c*d*e)/(c^2*d^2*e^2*x^2 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)) - 2*(128*c^5*d^5*e^5*x^4 - 15*c^5*d^9
*e + 70*a*c^4*d^7*e^3 + 128*a^2*c^3*d^5*e^5 - 70*a^3*c^2*d^3*e^7 + 15*a^4*c*d*e^9 + 16*(21*c^5*d^6*e^4 + 11*a*
c^4*d^4*e^6)*x^3 + 8*(31*c^5*d^7*e^3 + 64*a*c^4*d^5*e^5 + a^2*c^3*d^3*e^7)*x^2 + 2*(5*c^5*d^8*e^2 + 233*a*c^4*
d^6*e^4 + 23*a^2*c^3*d^4*e^6 - 5*a^3*c^2*d^2*e^8)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^4*d^4*e^3
)]

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giac [A]  time = 0.54, size = 388, normalized size = 1.37 \begin {gather*} \frac {1}{640} \, \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c d x e^{2} + \frac {{\left (21 \, c^{5} d^{6} e^{5} + 11 \, a c^{4} d^{4} e^{7}\right )} e^{\left (-4\right )}}{c^{4} d^{4}}\right )} x + \frac {{\left (31 \, c^{5} d^{7} e^{4} + 64 \, a c^{4} d^{5} e^{6} + a^{2} c^{3} d^{3} e^{8}\right )} e^{\left (-4\right )}}{c^{4} d^{4}}\right )} x + \frac {{\left (5 \, c^{5} d^{8} e^{3} + 233 \, a c^{4} d^{6} e^{5} + 23 \, a^{2} c^{3} d^{4} e^{7} - 5 \, a^{3} c^{2} d^{2} e^{9}\right )} e^{\left (-4\right )}}{c^{4} d^{4}}\right )} x - \frac {{\left (15 \, c^{5} d^{9} e^{2} - 70 \, a c^{4} d^{7} e^{4} - 128 \, a^{2} c^{3} d^{5} e^{6} + 70 \, a^{3} c^{2} d^{3} e^{8} - 15 \, a^{4} c d e^{10}\right )} e^{\left (-4\right )}}{c^{4} d^{4}}\right )} - \frac {3 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\right )} e^{\left (-\frac {5}{2}\right )} \log \left ({\left | -c d^{2} - 2 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )} \sqrt {c d} e^{\frac {1}{2}} - a e^{2} \right |}\right )}{256 \, \sqrt {c d} c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/640*sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*(2*(4*(2*(8*c*d*x*e^2 + (21*c^5*d^6*e^5 + 11*a*c^4*d^4*e^7)*
e^(-4)/(c^4*d^4))*x + (31*c^5*d^7*e^4 + 64*a*c^4*d^5*e^6 + a^2*c^3*d^3*e^8)*e^(-4)/(c^4*d^4))*x + (5*c^5*d^8*e
^3 + 233*a*c^4*d^6*e^5 + 23*a^2*c^3*d^4*e^7 - 5*a^3*c^2*d^2*e^9)*e^(-4)/(c^4*d^4))*x - (15*c^5*d^9*e^2 - 70*a*
c^4*d^7*e^4 - 128*a^2*c^3*d^5*e^6 + 70*a^3*c^2*d^3*e^8 - 15*a^4*c*d*e^10)*e^(-4)/(c^4*d^4)) - 3/256*(c^5*d^10
- 5*a*c^4*d^8*e^2 + 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8 - a^5*e^10)*e^(-5/2)*log(abs(-c*
d^2 - 2*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e))*sqrt(c*d)*e^(1/2) - a*e^2))/(sqrt(
c*d)*c^3*d^3)

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maple [B]  time = 0.05, size = 917, normalized size = 3.24 \begin {gather*} -\frac {3 a^{5} e^{8} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{256 \sqrt {c d e}\, c^{3} d^{3}}+\frac {15 a^{4} e^{6} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{256 \sqrt {c d e}\, c^{2} d}-\frac {15 a^{3} d \,e^{4} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{128 \sqrt {c d e}\, c}+\frac {15 a^{2} d^{3} e^{2} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{128 \sqrt {c d e}}-\frac {15 a c \,d^{5} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{256 \sqrt {c d e}}+\frac {3 c^{2} d^{7} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{256 \sqrt {c d e}\, e^{2}}+\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a^{3} e^{5} x}{64 c^{2} d^{2}}-\frac {9 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a^{2} e^{3} x}{64 c}+\frac {9 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a \,d^{2} e x}{64}-\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c \,d^{4} x}{64 e}+\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a^{4} e^{6}}{128 c^{3} d^{3}}-\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a^{3} e^{4}}{64 c^{2} d}+\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a \,d^{3}}{64}-\frac {3 \sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, c \,d^{5}}{128 e^{2}}-\frac {\left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{\frac {3}{2}} a \,e^{2} x}{8 c d}+\frac {\left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{\frac {3}{2}} d x}{8}-\frac {\left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{\frac {3}{2}} a^{2} e^{3}}{16 c^{2} d^{2}}+\frac {\left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{\frac {3}{2}} d^{2}}{16 e}+\frac {\left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{\frac {5}{2}}}{5 c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/c/d*e^2*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*x*a-3/256/c^3/d^3*e^8*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c
*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*a^5+15/256/c^2/d*e^6*ln((c*d*e*x+1/2*a*e^2+
1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*a^4-15/128/c*d*e^4*ln((c*d*e*x
+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*a^3+3/64/c^2/d^2*e^
5*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a^3+3/64*d^3*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a-9/64/c*e^3*
(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a^2+9/64*e*d^2*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a+15/128*d^
3*e^2*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*a^
2-1/16/c^2/d^2*e^3*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*a^2-3/64*c*d^4/e*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^
(1/2)*x-3/128*c*d^5/e^2*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)+1/16*d^2/e*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(
3/2)+1/5*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(5/2)/c/d+1/8*d*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)*x+3/256*c^2
*d^7/e^2*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)
+3/128/c^3/d^3*e^6*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^4-15/256*c*d^5*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(
c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*a-3/64/c^2/d*e^4*(c*d*e*x^2+a*d*e+(a*e^2+c
*d^2)*x)^(1/2)*a^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e^2-c*d^2>0)', see `assume?`
for more details)Is a*e^2-c*d^2 zero or nonzero?

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mupad [B]  time = 1.43, size = 646, normalized size = 2.28 \begin {gather*} \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}}{5\,c\,d}+\frac {\left (\frac {c\,d^2}{2}+c\,x\,d\,e+\frac {a\,e^2}{2}\right )\,{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}}{4\,c\,e}-\frac {\left (\frac {3\,{\left (c\,d^2+a\,e^2\right )}^2}{4}-3\,a\,c\,d^2\,e^2\right )\,\left (\left (\frac {x}{2}+\frac {c\,d^2+a\,e^2}{4\,c\,d\,e}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}-\frac {\ln \left (2\,\sqrt {\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )}\,\sqrt {c\,d\,e}+a\,e^2+c\,d^2+2\,c\,d\,e\,x\right )\,\left (\frac {{\left (c\,d^2+a\,e^2\right )}^2}{4}-a\,c\,d^2\,e^2\right )}{2\,{\left (c\,d\,e\right )}^{3/2}}\right )}{4\,c\,e}-\frac {\left (c\,d^2+a\,e^2\right )\,\left (\frac {x\,{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}}{4}-\frac {3\,a\,d\,e\,\left (\ln \left (2\,\sqrt {\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )}\,\sqrt {c\,d\,e}+a\,e^2+c\,d^2+2\,c\,d\,e\,x\right )\,\left (\frac {{\left (c\,d^2+a\,e^2\right )}^2}{8\,{\left (c\,d\,e\right )}^{3/2}}-\frac {a\,d\,e}{2\,\sqrt {c\,d\,e}}\right )-\frac {\left (c\,d^2+2\,c\,x\,d\,e+a\,e^2\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{4\,c\,d\,e}\right )}{4}+\frac {\left (c\,d^2+a\,e^2\right )\,{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}}{8\,c\,d\,e}+\frac {3\,{\left (c\,d^2+a\,e^2\right )}^2\,\left (\ln \left (2\,\sqrt {\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )}\,\sqrt {c\,d\,e}+a\,e^2+c\,d^2+2\,c\,d\,e\,x\right )\,\left (\frac {{\left (c\,d^2+a\,e^2\right )}^2}{8\,{\left (c\,d\,e\right )}^{3/2}}-\frac {a\,d\,e}{2\,\sqrt {c\,d\,e}}\right )-\frac {\left (c\,d^2+2\,c\,x\,d\,e+a\,e^2\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{4\,c\,d\,e}\right )}{16\,c\,d\,e}\right )}{2\,c\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2)/(5*c*d) + (((a*e^2)/2 + (c*d^2)/2 + c*d*e*x)*(x*(a*e^2 + c*d^2)
+ a*d*e + c*d*e*x^2)^(3/2))/(4*c*e) - (((3*(a*e^2 + c*d^2)^2)/4 - 3*a*c*d^2*e^2)*((x/2 + (a*e^2 + c*d^2)/(4*c*
d*e))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2) - (log(2*((a*e + c*d*x)*(d + e*x))^(1/2)*(c*d*e)^(1/2) + a
*e^2 + c*d^2 + 2*c*d*e*x)*((a*e^2 + c*d^2)^2/4 - a*c*d^2*e^2))/(2*(c*d*e)^(3/2))))/(4*c*e) - ((a*e^2 + c*d^2)*
((x*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(3/2))/4 - (3*a*d*e*(log(2*((a*e + c*d*x)*(d + e*x))^(1/2)*(c*d*e)
^(1/2) + a*e^2 + c*d^2 + 2*c*d*e*x)*((a*e^2 + c*d^2)^2/(8*(c*d*e)^(3/2)) - (a*d*e)/(2*(c*d*e)^(1/2))) - ((a*e^
2 + c*d^2 + 2*c*d*e*x)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(4*c*d*e)))/4 + ((a*e^2 + c*d^2)*(x*(a*e
^2 + c*d^2) + a*d*e + c*d*e*x^2)^(3/2))/(8*c*d*e) + (3*(a*e^2 + c*d^2)^2*(log(2*((a*e + c*d*x)*(d + e*x))^(1/2
)*(c*d*e)^(1/2) + a*e^2 + c*d^2 + 2*c*d*e*x)*((a*e^2 + c*d^2)^2/(8*(c*d*e)^(3/2)) - (a*d*e)/(2*(c*d*e)^(1/2)))
- ((a*e^2 + c*d^2 + 2*c*d*e*x)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(4*c*d*e)))/(16*c*d*e)))/(2*c*d
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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