3.479 \(\int (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}})^{5/2} \, dx\)

Optimal. Leaf size=370 \[ -\frac {5 f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {2 d e-b f^2}}\right )}{16 \sqrt {2} e^{9/2}}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{4 e^4}-\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^2 \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{16 e^4 \left (2 e \left (f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2}}{12 e^3}+\frac {\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{7/2}}{7 e} \]

[Out]

-5/32*f^2*(-b*f^2+2*d*e)^(3/2)*(-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))/e^(9/2)*2^(1/2)+1/12*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))^(3
/2)/e^3+1/7*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(7/2)/e+1/4*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^4-1/16*f^2*(-b*f^2+2*d*e)^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^4/(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 = 0.60, antiderivative size = 370, 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 = {2116, 897, 1257, 1810, 208} \[ \frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2}}{12 e^3}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{4 e^4}-\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^2 \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{16 e^4 \left (2 e \left (f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}+e x\right )+b f^2\right )}-\frac {5 f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {2 d e-b f^2}}\right )}{16 \sqrt {2} e^{9/2}}+\frac {\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{7/2}}{7 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(5/2),x]

[Out]

(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]])/(4*e^4) + (f^2*(4*a*
e^2 - b^2*f^2)*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(3/2))/(12*e^3) + (d + e*x + f*Sqrt[a + b*x + (e^2*
x^2)/f^2])^(7/2)/(7*e) - (f^2*(2*d*e - b*f^2)^2*(4*a*e^2 - b^2*f^2)*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/
f^2]])/(16*e^4*(b*f^2 + 2*e*(e*x + f*Sqrt[a + (x*(b*f^2 + e^2*x))/f^2]))) - (5*f^2*(2*d*e - b*f^2)^(3/2)*(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]])
/(16*Sqrt[2]*e^(9/2))

Rule 208

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

Rule 897

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 1257

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*(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)))/(d + e*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 1810

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 2116

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 )^{5/2} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^{5/2} \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 \operatorname {Subst}\left (\int \frac {x^6 \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 )^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}}}}{16 e^4 \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 {\operatorname {Subst}\left (\int \frac {-e f^2 \left (2 d e-b f^2\right )^2 \left (4 a e^2-b^2 f^2\right )-4 e^2 f^2 \left (2 d e-b f^2\right ) \left (4 a e^2-b^2 f^2\right ) x^2-8 e^3 f^2 \left (4 a e^2-b^2 f^2\right ) x^4+16 e^4 \left (2 d e-b f^2\right ) x^6-32 e^5 x^8}{-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 )}{16 e^5}\\ &=-\frac {f^2 \left (2 d e-b f^2\right )^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}}}}{16 e^4 \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 {\operatorname {Subst}\left (\int \left (-4 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^2-16 e^4 x^6-\frac {5 \left (16 a d^2 e^5 f^2-4 b^2 d^2 e^3 f^4-16 a b d e^4 f^4+4 b^3 d e^2 f^6+4 a b^2 e^3 f^6-b^4 e f^8\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 )}{16 e^5}\\ &=\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}}}}{4 e^4}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}}{12 e^3}+\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{7/2}}{7 e}-\frac {f^2 \left (2 d e-b f^2\right )^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}}}}{16 e^4 \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 (5 f^2 \left (2 d e-b f^2\right )^2 \left (4 a e^2-b^2 f^2\right )\right ) \operatorname {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 )}{16 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}}}}{4 e^4}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}}{12 e^3}+\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{7/2}}{7 e}-\frac {f^2 \left (2 d e-b f^2\right )^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}}}}{16 e^4 \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 {5 f^2 \left (2 d e-b f^2\right )^{3/2} \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 )}{16 \sqrt {2} e^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 1.06, size = 357, normalized size = 0.96 \[ \frac {\frac {4}{3} e^2 f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+d+e x\right )^{3/2}+4 e f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right ) \sqrt {f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+d+e x}-\frac {5 \sqrt {e} f^2 \left (4 a e^2-b^2 f^2\right ) \left (2 d e-b f^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+d+e x}}{\sqrt {2 d e-b f^2}}\right )}{\sqrt {2}}-\frac {\left (4 a e^3 f^2-b^2 e f^4\right ) \left (b f^2-2 d e\right )^2 \sqrt {f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+d+e x}}{2 e \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+e x\right )+b f^2}+\frac {16}{7} e^4 \left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+d+e x\right )^{7/2}}{16 e^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(5/2),x]

[Out]

(4*e*f^2*(2*d*e - b*f^2)*(4*a*e^2 - b^2*f^2)*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]] + (4*e^2*f^2*(4*a
*e^2 - b^2*f^2)*(d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])^(3/2))/3 + (16*e^4*(d + e*x + f*Sqrt[a + x*(b + (e
^2*x)/f^2)])^(7/2))/7 - ((-2*d*e + b*f^2)^2*(4*a*e^3*f^2 - b^2*e*f^4)*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)
/f^2)]])/(b*f^2 + 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])) - (5*Sqrt[e]*f^2*(2*d*e - b*f^2)^(3/2)*(4*a*e^2
- b^2*f^2)*ArcTanh[(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]])/Sq
rt[2])/(16*e^5)

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fricas [A]  time = 0.70, size = 923, normalized size = 2.49 \[ \left [\frac {105 \, \sqrt {\frac {1}{2}} {\left (b^{3} f^{6} + 8 \, a d e^{3} f^{2} - 2 \, {\left (b^{2} d e + 2 \, a b e^{2}\right )} f^{4}\right )} \sqrt {-\frac {b f^{2} - 2 \, d e}{e}} \log \left (-b^{2} f^{4} + 4 \, {\left (b d e - a e^{2}\right )} f^{2} - 4 \, {\left (b e^{2} f^{2} - 2 \, d e^{3}\right )} x + 4 \, {\left (2 \, \sqrt {\frac {1}{2}} e^{2} f \sqrt {-\frac {b f^{2} - 2 \, d e}{e}} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} - \sqrt {\frac {1}{2}} {\left (b e f^{2} + 2 \, e^{3} x\right )} \sqrt {-\frac {b f^{2} - 2 \, d e}{e}}\right )} \sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d} + 4 \, {\left (b e f^{3} - 2 \, d e^{2} f\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right ) + 2 \, {\left (105 \, b^{3} f^{6} + 192 \, e^{6} x^{3} + 48 \, d^{3} e^{3} - 56 \, {\left (5 \, b^{2} d e + 6 \, a b e^{2}\right )} f^{4} + 4 \, {\left (21 \, b d^{2} e^{2} + 232 \, a d e^{3}\right )} f^{2} + 144 \, {\left (b e^{4} f^{2} + 2 \, d e^{5}\right )} x^{2} + 2 \, {\left (7 \, b^{2} e^{2} f^{4} + 156 \, d^{2} e^{4} - 4 \, {\left (3 \, b d e^{3} - 32 \, a e^{4}\right )} f^{2}\right )} x - 2 \, {\left (35 \, b^{2} e f^{5} - 96 \, e^{5} f x^{2} + 12 \, d^{2} e^{3} f - 4 \, {\left (21 \, b d e^{2} + 20 \, a e^{3}\right )} f^{3} - 24 \, {\left (b e^{3} f^{3} + 6 \, d e^{4} f\right )} x\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{672 \, e^{4}}, -\frac {105 \, \sqrt {\frac {1}{2}} {\left (b^{3} f^{6} + 8 \, a d e^{3} f^{2} - 2 \, {\left (b^{2} d e + 2 \, a b e^{2}\right )} f^{4}\right )} \sqrt {\frac {b f^{2} - 2 \, d e}{e}} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d} e \sqrt {\frac {b f^{2} - 2 \, d e}{e}}}{b f^{2} - 2 \, d e}\right ) - {\left (105 \, b^{3} f^{6} + 192 \, e^{6} x^{3} + 48 \, d^{3} e^{3} - 56 \, {\left (5 \, b^{2} d e + 6 \, a b e^{2}\right )} f^{4} + 4 \, {\left (21 \, b d^{2} e^{2} + 232 \, a d e^{3}\right )} f^{2} + 144 \, {\left (b e^{4} f^{2} + 2 \, d e^{5}\right )} x^{2} + 2 \, {\left (7 \, b^{2} e^{2} f^{4} + 156 \, d^{2} e^{4} - 4 \, {\left (3 \, b d e^{3} - 32 \, a e^{4}\right )} f^{2}\right )} x - 2 \, {\left (35 \, b^{2} e f^{5} - 96 \, e^{5} f x^{2} + 12 \, d^{2} e^{3} f - 4 \, {\left (21 \, b d e^{2} + 20 \, a e^{3}\right )} f^{3} - 24 \, {\left (b e^{3} f^{3} + 6 \, d e^{4} f\right )} x\right )} \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {b f^{2} x + e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{336 \, e^{4}}\right ] \]

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))^(5/2),x, algorithm="fricas")

[Out]

[1/672*(105*sqrt(1/2)*(b^3*f^6 + 8*a*d*e^3*f^2 - 2*(b^2*d*e + 2*a*b*e^2)*f^4)*sqrt(-(b*f^2 - 2*d*e)/e)*log(-b^
2*f^4 + 4*(b*d*e - a*e^2)*f^2 - 4*(b*e^2*f^2 - 2*d*e^3)*x + 4*(2*sqrt(1/2)*e^2*f*sqrt(-(b*f^2 - 2*d*e)/e)*sqrt
((b*f^2*x + e^2*x^2 + a*f^2)/f^2) - sqrt(1/2)*(b*e*f^2 + 2*e^3*x)*sqrt(-(b*f^2 - 2*d*e)/e))*sqrt(e*x + f*sqrt(
(b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d) + 4*(b*e*f^3 - 2*d*e^2*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2)) + 2*(10
5*b^3*f^6 + 192*e^6*x^3 + 48*d^3*e^3 - 56*(5*b^2*d*e + 6*a*b*e^2)*f^4 + 4*(21*b*d^2*e^2 + 232*a*d*e^3)*f^2 + 1
44*(b*e^4*f^2 + 2*d*e^5)*x^2 + 2*(7*b^2*e^2*f^4 + 156*d^2*e^4 - 4*(3*b*d*e^3 - 32*a*e^4)*f^2)*x - 2*(35*b^2*e*
f^5 - 96*e^5*f*x^2 + 12*d^2*e^3*f - 4*(21*b*d*e^2 + 20*a*e^3)*f^3 - 24*(b*e^3*f^3 + 6*d*e^4*f)*x)*sqrt((b*f^2*
x + e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/e^4, -1/336*(105*sqrt(1/2)
*(b^3*f^6 + 8*a*d*e^3*f^2 - 2*(b^2*d*e + 2*a*b*e^2)*f^4)*sqrt((b*f^2 - 2*d*e)/e)*arctan(2*sqrt(1/2)*sqrt(e*x +
f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d)*e*sqrt((b*f^2 - 2*d*e)/e)/(b*f^2 - 2*d*e)) - (105*b^3*f^6 + 192*
e^6*x^3 + 48*d^3*e^3 - 56*(5*b^2*d*e + 6*a*b*e^2)*f^4 + 4*(21*b*d^2*e^2 + 232*a*d*e^3)*f^2 + 144*(b*e^4*f^2 +
2*d*e^5)*x^2 + 2*(7*b^2*e^2*f^4 + 156*d^2*e^4 - 4*(3*b*d*e^3 - 32*a*e^4)*f^2)*x - 2*(35*b^2*e*f^5 - 96*e^5*f*x
^2 + 12*d^2*e^3*f - 4*(21*b*d*e^2 + 20*a*e^3)*f^3 - 24*(b*e^3*f^3 + 6*d*e^4*f)*x)*sqrt((b*f^2*x + e^2*x^2 + a*
f^2)/f^2))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/e^4]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {5}{2}}\,{d x} \]

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))^(5/2),x, algorithm="giac")

[Out]

integrate((e*x + sqrt(b*x + e^2*x^2/f^2 + a)*f + d)^(5/2), x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \left (e x +d +\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\, f \right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d+(b*x+e^2/f^2*x^2+a)^(1/2)*f)^(5/2),x)

[Out]

int((e*x+d+(b*x+e^2/f^2*x^2+a)^(1/2)*f)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {5}{2}}\,{d x} \]

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))^(5/2),x, algorithm="maxima")

[Out]

integrate((e*x + sqrt(b*x + e^2*x^2/f^2 + a)*f + d)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (d+e\,x+f\,\sqrt {a+b\,x+\frac {e^2\,x^2}{f^2}}\right )}^{5/2} \,d x \]

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))^(5/2),x)

[Out]

int((d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d + e x + f \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {5}{2}}\, dx \]

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))**(5/2),x)

[Out]

Integral((d + e*x + f*sqrt(a + b*x + e**2*x**2/f**2))**(5/2), x)

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