3.5.80 \(\int (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}})^{3/2} \, dx\) [480]

Optimal. Leaf size=302 \[ \frac {f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{4 e^3}+\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}}{5 e}-\frac {f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{8 e^3 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}-\frac {3 f^2 \sqrt {2 d e-b f^2} \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\sqrt {2 d e-b f^2}}\right )}{8 \sqrt {2} e^{7/2}} \]

[Out]

-3/16*f^2*(-b^2*f^2+4*a*e^2)*arctanh(2^(1/2)*e^(1/2)*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2)/(-b*f^2+2*d*e)^
(1/2))*(-b*f^2+2*d*e)^(1/2)/e^(7/2)*2^(1/2)+1/5*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(5/2)/e+1/4*f^2*(-b^2*f^2+
4*a*e^2)*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2)/e^3-1/8*f^2*(-b*f^2+2*d*e)*(-b^2*f^2+4*a*e^2)*(d+e*x+f*(a+b
*x+e^2*x^2/f^2)^(1/2))^(1/2)/e^3/(b*f^2+2*e*(e*x+f*(a+x*(b*f^2+e^2*x)/f^2)^(1/2)))

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Rubi [A]
time = 0.30, antiderivative size = 302, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2141, 911, 1271, 1824, 214} \begin {gather*} -\frac {3 f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {2 d e-b f^2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {2 d e-b f^2}}\right )}{8 \sqrt {2} e^{7/2}}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{4 e^3}-\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{8 e^3 \left (2 e \left (f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac {\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{5/2}}{5 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 214

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

Rule 911

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - (2*c*d - b*e)*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1271

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Dist[1/(2*e^(2*p +
m/2)*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*e^(2*p + m/2)*(q + 1)*x^m*(a +
b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]

Rule 1824

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2141

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol]
:> Dist[2, Subst[Int[(g + h*x^n)^p*((d^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2)/(-2*d*e + b*f^2 + 2
*e*x)^2), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e, f, g, h, n}, x] && EqQ[e^2 -
c*f^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx &=2 \text {Subst}\left (\int \frac {x^{3/2} \left (d^2 e-(b d-a e) f^2-\left (2 d e-b f^2\right ) x+e x^2\right )}{\left (-2 d e+b f^2+2 e x\right )^2} \, dx,x,d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )\\ &=4 \text {Subst}\left (\int \frac {x^4 \left (d^2 e-(b d-a e) f^2+\left (-2 d e+b f^2\right ) x^2+e x^4\right )}{\left (-2 d e+b f^2+2 e x^2\right )^2} \, dx,x,\sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}\right )\\ &=-\frac {f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{8 e^3 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}-\frac {\text {Subst}\left (\int \frac {-e f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right )-4 e^2 f^2 \left (4 a e^2-b^2 f^2\right ) x^2+8 e^3 \left (2 d e-b f^2\right ) x^4-16 e^4 x^6}{-2 d e+b f^2+2 e x^2} \, dx,x,\sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}\right )}{8 e^4}\\ &=-\frac {f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{8 e^3 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}-\frac {\text {Subst}\left (\int \left (-2 e f^2 \left (4 a e^2-b^2 f^2\right )-8 e^3 x^4-\frac {3 \left (8 a d e^4 f^2-2 b^2 d e^2 f^4-4 a b e^3 f^4+b^3 e f^6\right )}{-2 d e+b f^2+2 e x^2}\right ) \, dx,x,\sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}\right )}{8 e^4}\\ &=\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{4 e^3}+\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}}{5 e}-\frac {f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{8 e^3 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac {\left (3 f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right )\right ) \text {Subst}\left (\int \frac {1}{-2 d e+b f^2+2 e x^2} \, dx,x,\sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}\right )}{8 e^3}\\ &=\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{4 e^3}+\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}}{5 e}-\frac {f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{8 e^3 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}-\frac {3 f^2 \sqrt {2 d e-b f^2} \left (4 a e^2-b^2 f^2\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\sqrt {2 d e-b f^2}}\right )}{8 \sqrt {2} e^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 2.84, size = 443, normalized size = 1.47 \begin {gather*} \frac {\sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}} \left (-15 b^3 f^6-2 b^2 e f^4 \left (-5 d+6 e x+10 f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+4 b e^2 f^2 \left (2 d^2+17 a f^2+8 e x \left (2 e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+4 d \left (3 e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )+8 e^3 \left (2 (d+2 e x)^2 \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+a f^2 \left (-d+16 e x+12 f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )\right )}{40 e^3 \left (b f^2+2 e \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )}-\frac {3 a f^2 \sqrt {-d e+\frac {b f^2}{2}} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{2 e^{3/2}}+\frac {3 b^2 f^4 \sqrt {-d e+\frac {b f^2}{2}} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{8 e^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]]*(-15*b^3*f^6 - 2*b^2*e*f^4*(-5*d + 6*e*x + 10*f*Sqrt[a + x*(b
+ (e^2*x)/f^2)]) + 4*b*e^2*f^2*(2*d^2 + 17*a*f^2 + 8*e*x*(2*e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + 4*d*(3*e
*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])) + 8*e^3*(2*(d + 2*e*x)^2*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + a*f^
2*(-d + 16*e*x + 12*f*Sqrt[a + x*(b + (e^2*x)/f^2)]))))/(40*e^3*(b*f^2 + 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/
f^2)]))) - (3*a*f^2*Sqrt[-(d*e) + (b*f^2)/2]*ArcTan[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/
f^2)]])/Sqrt[-2*d*e + b*f^2]])/(2*e^(3/2)) + (3*b^2*f^4*Sqrt[-(d*e) + (b*f^2)/2]*ArcTan[(Sqrt[2]*Sqrt[e]*Sqrt[
d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]])/Sqrt[-2*d*e + b*f^2]])/(8*e^(7/2))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (d +e x +f \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+e*x+f*(a+b*x+1/f^2*e^2*x^2)^(1/2))^(3/2),x)

[Out]

int((d+e*x+f*(a+b*x+1/f^2*e^2*x^2)^(1/2))^(3/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(3/2),x, algorithm="maxima")

[Out]

integrate((x*e + sqrt(b*x + a + x^2*e^2/f^2)*f + d)^(3/2), x)

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Fricas [A]
time = 0.50, size = 612, normalized size = 2.03 \begin {gather*} \left [-\frac {1}{80} \, {\left (15 \, \sqrt {\frac {1}{2}} {\left (b^{2} f^{4} - 4 \, a f^{2} e^{2}\right )} \sqrt {-{\left (b f^{2} - 2 \, d e\right )} e^{\left (-1\right )}} \log \left (b^{2} f^{4} - 4 \, b d f^{2} e - 8 \, d x e^{3} + 4 \, {\left (b f^{2} x + a f^{2}\right )} e^{2} - 4 \, {\left (2 \, \sqrt {\frac {1}{2}} \sqrt {-{\left (b f^{2} - 2 \, d e\right )} e^{\left (-1\right )}} f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} e^{2} - \sqrt {\frac {1}{2}} {\left (b f^{2} e + 2 \, x e^{3}\right )} \sqrt {-{\left (b f^{2} - 2 \, d e\right )} e^{\left (-1\right )}}\right )} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d} - 4 \, {\left (b f^{3} e - 2 \, d f e^{2}\right )} \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}}\right ) + 2 \, {\left (15 \, b^{2} f^{4} - 10 \, b d f^{2} e - 16 \, x^{2} e^{4} - 36 \, d x e^{3} + 2 \, {\left (b f^{2} x - 24 \, a f^{2} - 4 \, d^{2}\right )} e^{2} - 2 \, {\left (5 \, b f^{3} e + 8 \, f x e^{3} - 2 \, d f e^{2}\right )} \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d}\right )} e^{\left (-3\right )}, -\frac {1}{40} \, {\left (15 \, \sqrt {\frac {1}{2}} {\left (b^{2} f^{4} - 4 \, a f^{2} e^{2}\right )} \sqrt {b f^{2} - 2 \, d e} \arctan \left (-\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d} e^{\frac {1}{2}}}{\sqrt {b f^{2} - 2 \, d e}}\right ) e^{\left (-\frac {1}{2}\right )} + {\left (15 \, b^{2} f^{4} - 10 \, b d f^{2} e - 16 \, x^{2} e^{4} - 36 \, d x e^{3} + 2 \, {\left (b f^{2} x - 24 \, a f^{2} - 4 \, d^{2}\right )} e^{2} - 2 \, {\left (5 \, b f^{3} e + 8 \, f x e^{3} - 2 \, d f e^{2}\right )} \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}}\right )} \sqrt {x e + f \sqrt {\frac {b f^{2} x + a f^{2} + x^{2} e^{2}}{f^{2}}} + d}\right )} e^{\left (-3\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/80*(15*sqrt(1/2)*(b^2*f^4 - 4*a*f^2*e^2)*sqrt(-(b*f^2 - 2*d*e)*e^(-1))*log(b^2*f^4 - 4*b*d*f^2*e - 8*d*x*e
^3 + 4*(b*f^2*x + a*f^2)*e^2 - 4*(2*sqrt(1/2)*sqrt(-(b*f^2 - 2*d*e)*e^(-1))*f*sqrt((b*f^2*x + a*f^2 + x^2*e^2)
/f^2)*e^2 - sqrt(1/2)*(b*f^2*e + 2*x*e^3)*sqrt(-(b*f^2 - 2*d*e)*e^(-1)))*sqrt(x*e + f*sqrt((b*f^2*x + a*f^2 +
x^2*e^2)/f^2) + d) - 4*(b*f^3*e - 2*d*f*e^2)*sqrt((b*f^2*x + a*f^2 + x^2*e^2)/f^2)) + 2*(15*b^2*f^4 - 10*b*d*f
^2*e - 16*x^2*e^4 - 36*d*x*e^3 + 2*(b*f^2*x - 24*a*f^2 - 4*d^2)*e^2 - 2*(5*b*f^3*e + 8*f*x*e^3 - 2*d*f*e^2)*sq
rt((b*f^2*x + a*f^2 + x^2*e^2)/f^2))*sqrt(x*e + f*sqrt((b*f^2*x + a*f^2 + x^2*e^2)/f^2) + d))*e^(-3), -1/40*(1
5*sqrt(1/2)*(b^2*f^4 - 4*a*f^2*e^2)*sqrt(b*f^2 - 2*d*e)*arctan(-2*sqrt(1/2)*sqrt(x*e + f*sqrt((b*f^2*x + a*f^2
+ x^2*e^2)/f^2) + d)*e^(1/2)/sqrt(b*f^2 - 2*d*e))*e^(-1/2) + (15*b^2*f^4 - 10*b*d*f^2*e - 16*x^2*e^4 - 36*d*x
*e^3 + 2*(b*f^2*x - 24*a*f^2 - 4*d^2)*e^2 - 2*(5*b*f^3*e + 8*f*x*e^3 - 2*d*f*e^2)*sqrt((b*f^2*x + a*f^2 + x^2*
e^2)/f^2))*sqrt(x*e + f*sqrt((b*f^2*x + a*f^2 + x^2*e^2)/f^2) + d))*e^(-3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x*e + sqrt(b*x + a + x^2*e^2/f^2)*f + d)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (d+e\,x+f\,\sqrt {a+b\,x+\frac {e^2\,x^2}{f^2}}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^(3/2),x)

[Out]

int((d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^(3/2), x)

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