3.4.40 \(\int \frac {(d+e x)^{5/2}}{b x+c x^2} \, dx\)

Optimal. Leaf size=118 \[ \frac {2 (c d-b e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{5/2}}+\frac {2 e \sqrt {d+e x} (2 c d-b e)}{c^2}-\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 e (d+e x)^{3/2}}{3 c} \]

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Rubi [A]  time = 0.23, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {703, 824, 826, 1166, 208} \begin {gather*} \frac {2 e \sqrt {d+e x} (2 c d-b e)}{c^2}+\frac {2 (c d-b e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{5/2}}-\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 e (d+e x)^{3/2}}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(5/2)/(b*x + c*x^2),x]

[Out]

(2*e*(2*c*d - b*e)*Sqrt[d + e*x])/c^2 + (2*e*(d + e*x)^(3/2))/(3*c) - (2*d^(5/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]
])/b + (2*(c*d - b*e)^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*c^(5/2))

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 703

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

Rule 824

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g
*(d + e*x)^m)/(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x])
/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*
e^2, 0] && FractionQ[m] && GtQ[m, 0]

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 107, normalized size = 0.91 \begin {gather*} \frac {2 (c d-b e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{5/2}}+\frac {2 e \sqrt {d+e x} (-3 b e+7 c d+c e x)}{3 c^2}-\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*Sqrt[d + e*x]*(7*c*d - 3*b*e + c*e*x))/(3*c^2) - (2*d^(5/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b + (2*(c*d -
b*e)^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*c^(5/2))

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IntegrateAlgebraic [A]  time = 0.19, size = 120, normalized size = 1.02 \begin {gather*} -\frac {2 (b e-c d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x} \sqrt {b e-c d}}{c d-b e}\right )}{b c^{5/2}}+\frac {2 e \sqrt {d+e x} (-3 b e+c (d+e x)+6 c d)}{3 c^2}-\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(5/2)/(b*x + c*x^2),x]

[Out]

(2*e*Sqrt[d + e*x]*(6*c*d - 3*b*e + c*(d + e*x)))/(3*c^2) - (2*(-(c*d) + b*e)^(5/2)*ArcTan[(Sqrt[c]*Sqrt[-(c*d
) + b*e]*Sqrt[d + e*x])/(c*d - b*e)])/(b*c^(5/2)) - (2*d^(5/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b

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fricas [A]  time = 0.60, size = 598, normalized size = 5.07 \begin {gather*} \left [\frac {3 \, c^{2} d^{\frac {5}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 3 \, {\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e + 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right ) + 2 \, {\left (b c e^{2} x + 7 \, b c d e - 3 \, b^{2} e^{2}\right )} \sqrt {e x + d}}{3 \, b c^{2}}, \frac {3 \, c^{2} d^{\frac {5}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 6 \, {\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + 2 \, {\left (b c e^{2} x + 7 \, b c d e - 3 \, b^{2} e^{2}\right )} \sqrt {e x + d}}{3 \, b c^{2}}, \frac {6 \, c^{2} \sqrt {-d} d^{2} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + 3 \, {\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e + 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right ) + 2 \, {\left (b c e^{2} x + 7 \, b c d e - 3 \, b^{2} e^{2}\right )} \sqrt {e x + d}}{3 \, b c^{2}}, \frac {2 \, {\left (3 \, c^{2} \sqrt {-d} d^{2} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + 3 \, {\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + {\left (b c e^{2} x + 7 \, b c d e - 3 \, b^{2} e^{2}\right )} \sqrt {e x + d}\right )}}{3 \, b c^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

[1/3*(3*c^2*d^(5/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 3*(c^2*d^2 - 2*b*c*d*e + b^2*e^2)*sqrt((c*d
- b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + 2*(b*c*e^2*x + 7*b*c
*d*e - 3*b^2*e^2)*sqrt(e*x + d))/(b*c^2), 1/3*(3*c^2*d^(5/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 6*
(c^2*d^2 - 2*b*c*d*e + b^2*e^2)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e))
+ 2*(b*c*e^2*x + 7*b*c*d*e - 3*b^2*e^2)*sqrt(e*x + d))/(b*c^2), 1/3*(6*c^2*sqrt(-d)*d^2*arctan(sqrt(e*x + d)*
sqrt(-d)/d) + 3*(c^2*d^2 - 2*b*c*d*e + b^2*e^2)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)
*c*sqrt((c*d - b*e)/c))/(c*x + b)) + 2*(b*c*e^2*x + 7*b*c*d*e - 3*b^2*e^2)*sqrt(e*x + d))/(b*c^2), 2/3*(3*c^2*
sqrt(-d)*d^2*arctan(sqrt(e*x + d)*sqrt(-d)/d) + 3*(c^2*d^2 - 2*b*c*d*e + b^2*e^2)*sqrt(-(c*d - b*e)/c)*arctan(
-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + (b*c*e^2*x + 7*b*c*d*e - 3*b^2*e^2)*sqrt(e*x + d))/(b*c^2
)]

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giac [A]  time = 0.24, size = 161, normalized size = 1.36 \begin {gather*} \frac {2 \, d^{3} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} - \frac {2 \, {\left (c^{3} d^{3} - 3 \, b c^{2} d^{2} e + 3 \, b^{2} c d e^{2} - b^{3} e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b c^{2}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} c^{2} e + 6 \, \sqrt {x e + d} c^{2} d e - 3 \, \sqrt {x e + d} b c e^{2}\right )}}{3 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(c*x^2+b*x),x, algorithm="giac")

[Out]

2*d^3*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)) - 2*(c^3*d^3 - 3*b*c^2*d^2*e + 3*b^2*c*d*e^2 - b^3*e^3)*arct
an(sqrt(x*e + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b*c^2) + 2/3*((x*e + d)^(3/2)*c^2*e + 6*sqrt(x*
e + d)*c^2*d*e - 3*sqrt(x*e + d)*b*c*e^2)/c^3

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maple [B]  time = 0.06, size = 237, normalized size = 2.01 \begin {gather*} \frac {2 b^{2} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, c^{2}}-\frac {6 b d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, c}-\frac {2 c \,d^{3} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b}+\frac {6 d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}}-\frac {2 d^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b}-\frac {2 \sqrt {e x +d}\, b \,e^{2}}{c^{2}}+\frac {4 \sqrt {e x +d}\, d e}{c}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} e}{3 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(c*x^2+b*x),x)

[Out]

2/3*e*(e*x+d)^(3/2)/c-2/c^2*(e*x+d)^(1/2)*b*e^2+4*d*e*(e*x+d)^(1/2)/c+2/c^2*b^2*e^3/((b*e-c*d)*c)^(1/2)*arctan
((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)-6/c*b*e^2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c
)*d+6*e/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)*d^2-2*c/b/((b*e-c*d)*c)^(1/2)*arctan((
e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)*d^3-2*d^(5/2)*arctanh((e*x+d)^(1/2)/d^(1/2))/b

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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)^(5/2)/(c*x^2+b*x),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(b*e-c*d>0)', see `assume?` for
more details)Is b*e-c*d positive or negative?

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mupad [B]  time = 0.39, size = 2048, normalized size = 17.36

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(5/2)/(b*x + c*x^2),x)

[Out]

(atan(((((8*(d + e*x)^(1/2)*(b^6*e^8 + 2*c^6*d^6*e^2 - 6*b*c^5*d^5*e^3 + 15*b^2*c^4*d^4*e^4 - 20*b^3*c^3*d^3*e
^5 + 15*b^4*c^2*d^2*e^6 - 6*b^5*c*d*e^7))/c^3 + (((8*(b^4*c^3*d*e^5 + 2*b^2*c^5*d^3*e^3 - 3*b^3*c^4*d^2*e^4))/
c^3 + (8*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(d^5)^(1/2)*(d + e*x)^(1/2))/(b*c^3))*(d^5)^(1/2))/b)*(d^5)^(1/2)*1i)
/b + (((8*(d + e*x)^(1/2)*(b^6*e^8 + 2*c^6*d^6*e^2 - 6*b*c^5*d^5*e^3 + 15*b^2*c^4*d^4*e^4 - 20*b^3*c^3*d^3*e^5
+ 15*b^4*c^2*d^2*e^6 - 6*b^5*c*d*e^7))/c^3 - (((8*(b^4*c^3*d*e^5 + 2*b^2*c^5*d^3*e^3 - 3*b^3*c^4*d^2*e^4))/c^
3 - (8*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(d^5)^(1/2)*(d + e*x)^(1/2))/(b*c^3))*(d^5)^(1/2))/b)*(d^5)^(1/2)*1i)/b
)/((16*(b^5*d^3*e^8 - 3*c^5*d^8*e^3 + 12*b*c^4*d^7*e^4 - 6*b^4*c*d^4*e^7 - 19*b^2*c^3*d^6*e^5 + 15*b^3*c^2*d^5
*e^6))/c^3 - (((8*(d + e*x)^(1/2)*(b^6*e^8 + 2*c^6*d^6*e^2 - 6*b*c^5*d^5*e^3 + 15*b^2*c^4*d^4*e^4 - 20*b^3*c^3
*d^3*e^5 + 15*b^4*c^2*d^2*e^6 - 6*b^5*c*d*e^7))/c^3 + (((8*(b^4*c^3*d*e^5 + 2*b^2*c^5*d^3*e^3 - 3*b^3*c^4*d^2*
e^4))/c^3 + (8*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(d^5)^(1/2)*(d + e*x)^(1/2))/(b*c^3))*(d^5)^(1/2))/b)*(d^5)^(1/
2))/b + (((8*(d + e*x)^(1/2)*(b^6*e^8 + 2*c^6*d^6*e^2 - 6*b*c^5*d^5*e^3 + 15*b^2*c^4*d^4*e^4 - 20*b^3*c^3*d^3*
e^5 + 15*b^4*c^2*d^2*e^6 - 6*b^5*c*d*e^7))/c^3 - (((8*(b^4*c^3*d*e^5 + 2*b^2*c^5*d^3*e^3 - 3*b^3*c^4*d^2*e^4))
/c^3 - (8*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(d^5)^(1/2)*(d + e*x)^(1/2))/(b*c^3))*(d^5)^(1/2))/b)*(d^5)^(1/2))/b
))*(d^5)^(1/2)*2i)/b + (2*e*(d + e*x)^(3/2))/(3*c) + (atan((((-c^5*(b*e - c*d)^5)^(1/2)*((8*(d + e*x)^(1/2)*(b
^6*e^8 + 2*c^6*d^6*e^2 - 6*b*c^5*d^5*e^3 + 15*b^2*c^4*d^4*e^4 - 20*b^3*c^3*d^3*e^5 + 15*b^4*c^2*d^2*e^6 - 6*b^
5*c*d*e^7))/c^3 + ((-c^5*(b*e - c*d)^5)^(1/2)*((8*(b^4*c^3*d*e^5 + 2*b^2*c^5*d^3*e^3 - 3*b^3*c^4*d^2*e^4))/c^3
+ (8*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(-c^5*(b*e - c*d)^5)^(1/2)*(d + e*x)^(1/2))/(b*c^8)))/(b*c^5))*1i)/(b*c^
5) + ((-c^5*(b*e - c*d)^5)^(1/2)*((8*(d + e*x)^(1/2)*(b^6*e^8 + 2*c^6*d^6*e^2 - 6*b*c^5*d^5*e^3 + 15*b^2*c^4*d
^4*e^4 - 20*b^3*c^3*d^3*e^5 + 15*b^4*c^2*d^2*e^6 - 6*b^5*c*d*e^7))/c^3 - ((-c^5*(b*e - c*d)^5)^(1/2)*((8*(b^4*
c^3*d*e^5 + 2*b^2*c^5*d^3*e^3 - 3*b^3*c^4*d^2*e^4))/c^3 - (8*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(-c^5*(b*e - c*d)
^5)^(1/2)*(d + e*x)^(1/2))/(b*c^8)))/(b*c^5))*1i)/(b*c^5))/((16*(b^5*d^3*e^8 - 3*c^5*d^8*e^3 + 12*b*c^4*d^7*e^
4 - 6*b^4*c*d^4*e^7 - 19*b^2*c^3*d^6*e^5 + 15*b^3*c^2*d^5*e^6))/c^3 - ((-c^5*(b*e - c*d)^5)^(1/2)*((8*(d + e*x
)^(1/2)*(b^6*e^8 + 2*c^6*d^6*e^2 - 6*b*c^5*d^5*e^3 + 15*b^2*c^4*d^4*e^4 - 20*b^3*c^3*d^3*e^5 + 15*b^4*c^2*d^2*
e^6 - 6*b^5*c*d*e^7))/c^3 + ((-c^5*(b*e - c*d)^5)^(1/2)*((8*(b^4*c^3*d*e^5 + 2*b^2*c^5*d^3*e^3 - 3*b^3*c^4*d^2
*e^4))/c^3 + (8*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(-c^5*(b*e - c*d)^5)^(1/2)*(d + e*x)^(1/2))/(b*c^8)))/(b*c^5))
)/(b*c^5) + ((-c^5*(b*e - c*d)^5)^(1/2)*((8*(d + e*x)^(1/2)*(b^6*e^8 + 2*c^6*d^6*e^2 - 6*b*c^5*d^5*e^3 + 15*b^
2*c^4*d^4*e^4 - 20*b^3*c^3*d^3*e^5 + 15*b^4*c^2*d^2*e^6 - 6*b^5*c*d*e^7))/c^3 - ((-c^5*(b*e - c*d)^5)^(1/2)*((
8*(b^4*c^3*d*e^5 + 2*b^2*c^5*d^3*e^3 - 3*b^3*c^4*d^2*e^4))/c^3 - (8*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(-c^5*(b*e
- c*d)^5)^(1/2)*(d + e*x)^(1/2))/(b*c^8)))/(b*c^5)))/(b*c^5)))*(-c^5*(b*e - c*d)^5)^(1/2)*2i)/(b*c^5) - (2*e*
(b*e - 2*c*d)*(d + e*x)^(1/2))/c^2

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sympy [A]  time = 58.86, size = 119, normalized size = 1.01 \begin {gather*} \frac {2 e \left (d + e x\right )^{\frac {3}{2}}}{3 c} + \frac {\sqrt {d + e x} \left (- 2 b e^{2} + 4 c d e\right )}{c^{2}} + \frac {2 d^{3} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}} + \frac {2 \left (b e - c d\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b c^{3} \sqrt {\frac {b e - c d}{c}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(5/2)/(c*x**2+b*x),x)

[Out]

2*e*(d + e*x)**(3/2)/(3*c) + sqrt(d + e*x)*(-2*b*e**2 + 4*c*d*e)/c**2 + 2*d**3*atan(sqrt(d + e*x)/sqrt(-d))/(b
*sqrt(-d)) + 2*(b*e - c*d)**3*atan(sqrt(d + e*x)/sqrt((b*e - c*d)/c))/(b*c**3*sqrt((b*e - c*d)/c))

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