3.1231 \(\int \frac {(A+B x) (d+e x)^{3/2}}{b x+c x^2} \, dx\)

Optimal. Leaf size=131 \[ -\frac {2 (b B-A c) (c d-b e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{5/2}}+\frac {2 \sqrt {d+e x} (A c e-b B e+B c d)}{c^2}-\frac {2 A d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 B (d+e x)^{3/2}}{3 c} \]

[Out]

2/3*B*(e*x+d)^(3/2)/c-2*A*d^(3/2)*arctanh((e*x+d)^(1/2)/d^(1/2))/b-2*(-A*c+B*b)*(-b*e+c*d)^(3/2)*arctanh(c^(1/
2)*(e*x+d)^(1/2)/(-b*e+c*d)^(1/2))/b/c^(5/2)+2*(A*c*e-B*b*e+B*c*d)*(e*x+d)^(1/2)/c^2

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Rubi [A]  time = 0.30, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {824, 826, 1166, 208} \[ \frac {2 \sqrt {d+e x} (A c e-b B e+B c d)}{c^2}-\frac {2 (b B-A c) (c d-b e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{5/2}}-\frac {2 A d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 B (d+e x)^{3/2}}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(B*c*d - b*B*e + A*c*e)*Sqrt[d + e*x])/c^2 + (2*B*(d + e*x)^(3/2))/(3*c) - (2*A*d^(3/2)*ArcTanh[Sqrt[d + e*
x]/Sqrt[d]])/b - (2*(b*B - A*c)*(c*d - b*e)^(3/2)*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 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 {(A+B x) (d+e x)^{3/2}}{b x+c x^2} \, dx &=\frac {2 B (d+e x)^{3/2}}{3 c}+\frac {\int \frac {\sqrt {d+e x} (A c d+(B c d-b B e+A c e) x)}{b x+c x^2} \, dx}{c}\\ &=\frac {2 (B c d-b B e+A c e) \sqrt {d+e x}}{c^2}+\frac {2 B (d+e x)^{3/2}}{3 c}+\frac {\int \frac {A c^2 d^2+\left (B (c d-b e)^2+A c e (2 c d-b e)\right ) x}{\sqrt {d+e x} \left (b x+c x^2\right )} \, dx}{c^2}\\ &=\frac {2 (B c d-b B e+A c e) \sqrt {d+e x}}{c^2}+\frac {2 B (d+e x)^{3/2}}{3 c}+\frac {2 \operatorname {Subst}\left (\int \frac {A c^2 d^2 e-d \left (B (c d-b e)^2+A c e (2 c d-b e)\right )+\left (B (c d-b e)^2+A c e (2 c d-b e)\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 (B c d-b B e+A c e) \sqrt {d+e x}}{c^2}+\frac {2 B (d+e x)^{3/2}}{3 c}+\frac {\left (2 A c d^2\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 (b B-A c) (c d-b e)^2\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 (B c d-b B e+A c e) \sqrt {d+e x}}{c^2}+\frac {2 B (d+e x)^{3/2}}{3 c}-\frac {2 A d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}-\frac {2 (b B-A c) (c d-b e)^{3/2} \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 = 0.15, size = 123, normalized size = 0.94 \[ \frac {2 (A c-b B) (c d-b e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{5/2}}+\frac {2 \sqrt {d+e x} (3 A c e+B (-3 b e+4 c d+c e x))}{3 c^2}-\frac {2 A d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(3*A*c*e + B*(4*c*d - 3*b*e + c*e*x)))/(3*c^2) - (2*A*d^(3/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])
/b + (2*(-(b*B) + A*c)*(c*d - b*e)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*c^(5/2))

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fricas [A]  time = 2.21, size = 646, normalized size = 4.93 \[ \left [\frac {3 \, A c^{2} d^{\frac {3}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - 3 \, {\left ({\left (B b c - A c^{2}\right )} d - {\left (B b^{2} - A b c\right )} e\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e + 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right ) + 2 \, {\left (B b c e x + 4 \, B b c d - 3 \, {\left (B b^{2} - A b c\right )} e\right )} \sqrt {e x + d}}{3 \, b c^{2}}, \frac {3 \, A c^{2} d^{\frac {3}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - 6 \, {\left ({\left (B b c - A c^{2}\right )} d - {\left (B b^{2} - A b c\right )} e\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + 2 \, {\left (B b c e x + 4 \, B b c d - 3 \, {\left (B b^{2} - A b c\right )} e\right )} \sqrt {e x + d}}{3 \, b c^{2}}, \frac {6 \, A c^{2} \sqrt {-d} d \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) - 3 \, {\left ({\left (B b c - A c^{2}\right )} d - {\left (B b^{2} - A b c\right )} e\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e + 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right ) + 2 \, {\left (B b c e x + 4 \, B b c d - 3 \, {\left (B b^{2} - A b c\right )} e\right )} \sqrt {e x + d}}{3 \, b c^{2}}, \frac {2 \, {\left (3 \, A c^{2} \sqrt {-d} d \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) - 3 \, {\left ({\left (B b c - A c^{2}\right )} d - {\left (B b^{2} - A b c\right )} e\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + {\left (B b c e x + 4 \, B b c d - 3 \, {\left (B b^{2} - A b c\right )} e\right )} \sqrt {e x + d}\right )}}{3 \, b c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*A*c^2*d^(3/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 3*((B*b*c - A*c^2)*d - (B*b^2 - A*b*c)*e)
*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*b*c*e
*x + 4*B*b*c*d - 3*(B*b^2 - A*b*c)*e)*sqrt(e*x + d))/(b*c^2), 1/3*(3*A*c^2*d^(3/2)*log((e*x - 2*sqrt(e*x + d)*
sqrt(d) + 2*d)/x) - 6*((B*b*c - A*c^2)*d - (B*b^2 - A*b*c)*e)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqr
t(-(c*d - b*e)/c)/(c*d - b*e)) + 2*(B*b*c*e*x + 4*B*b*c*d - 3*(B*b^2 - A*b*c)*e)*sqrt(e*x + d))/(b*c^2), 1/3*(
6*A*c^2*sqrt(-d)*d*arctan(sqrt(e*x + d)*sqrt(-d)/d) - 3*((B*b*c - A*c^2)*d - (B*b^2 - A*b*c)*e)*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*b*c*e*x + 4*B*b*c*d
- 3*(B*b^2 - A*b*c)*e)*sqrt(e*x + d))/(b*c^2), 2/3*(3*A*c^2*sqrt(-d)*d*arctan(sqrt(e*x + d)*sqrt(-d)/d) - 3*((
B*b*c - A*c^2)*d - (B*b^2 - A*b*c)*e)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d -
b*e)) + (B*b*c*e*x + 4*B*b*c*d - 3*(B*b^2 - A*b*c)*e)*sqrt(e*x + d))/(b*c^2)]

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giac [A]  time = 0.19, size = 197, normalized size = 1.50 \[ \frac {2 \, A d^{2} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} + \frac {2 \, {\left (B b c^{2} d^{2} - A c^{3} d^{2} - 2 \, B b^{2} c d e + 2 \, A b c^{2} d e + B b^{3} e^{2} - A b^{2} c e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b c^{2}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} B c^{2} + 3 \, \sqrt {x e + d} B c^{2} d - 3 \, \sqrt {x e + d} B b c e + 3 \, \sqrt {x e + d} A c^{2} e\right )}}{3 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*d^2*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)) + 2*(B*b*c^2*d^2 - A*c^3*d^2 - 2*B*b^2*c*d*e + 2*A*b*c^2*d
*e + B*b^3*e^2 - A*b^2*c*e^2)*arctan(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)*B*c^2 + 3*sqrt(x*e + d)*B*c^2*d - 3*sqrt(x*e + d)*B*b*c*e + 3*sqrt(x*e + d)*A*c^2*e)/c^3

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maple [B]  time = 0.09, size = 335, normalized size = 2.56 \[ -\frac {2 A b \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, c}-\frac {2 A c \,d^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b}+\frac {4 A d e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}}+\frac {2 B \,b^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, c^{2}}-\frac {4 B b d e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, c}+\frac {2 B \,d^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}}-\frac {2 A \,d^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b}+\frac {2 \sqrt {e x +d}\, A e}{c}-\frac {2 \sqrt {e x +d}\, B b e}{c^{2}}+\frac {2 \sqrt {e x +d}\, B d}{c}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} B}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(3/2)/(c*x^2+b*x),x)

[Out]

2/3*B*(e*x+d)^(3/2)/c+2/c*A*e*(e*x+d)^(1/2)-2/c^2*B*b*e*(e*x+d)^(1/2)+2/c*B*d*(e*x+d)^(1/2)-2*b/c/((b*e-c*d)*c
)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)*A*e^2+4/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d
)*c)^(1/2)*c)*A*d*e-2/b*c/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)*A*d^2+2*b^2/c^2/((b*
e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)*B*e^2-4*b/c/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/
2)/((b*e-c*d)*c)^(1/2)*c)*B*d*e+2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)*B*d^2-2*A*d^
(3/2)*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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(3/2)/(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 = 0.59, size = 3810, normalized size = 29.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^(3/2))/(b*x + c*x^2),x)

[Out]

((2*A*e - 2*B*d)/c - (2*B*(b*e - 2*c*d))/c^2)*(d + e*x)^(1/2) + (2*B*(d + e*x)^(3/2))/(3*c) - (A*atan(((A*((8*
(d + e*x)^(1/2)*(B^2*b^6*e^6 + A^2*b^4*c^2*e^6 + 2*A^2*c^6*d^4*e^2 + 6*A^2*b^2*c^4*d^2*e^4 + B^2*b^2*c^4*d^4*e
^2 - 4*B^2*b^3*c^3*d^3*e^3 + 6*B^2*b^4*c^2*d^2*e^4 - 4*B^2*b^5*c*d*e^5 - 4*A^2*b*c^5*d^3*e^3 - 4*A^2*b^3*c^3*d
*e^5 - 2*A*B*b^5*c*e^6 - 2*A*B*b*c^5*d^4*e^2 + 8*A*B*b^4*c^2*d*e^5 + 8*A*B*b^2*c^4*d^3*e^3 - 12*A*B*b^3*c^3*d^
2*e^4))/c^3 + (A*((8*(B*b^4*c^3*d*e^4 - A*b^3*c^4*d*e^4 + A*b^2*c^5*d^2*e^3 + B*b^2*c^5*d^3*e^2 - 2*B*b^3*c^4*
d^2*e^3))/c^3 + (8*A*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(d^3)^(1/2)*(d + e*x)^(1/2))/(b*c^3))*(d^3)^(1/2))/b)*(d^
3)^(1/2)*1i)/b + (A*((8*(d + e*x)^(1/2)*(B^2*b^6*e^6 + A^2*b^4*c^2*e^6 + 2*A^2*c^6*d^4*e^2 + 6*A^2*b^2*c^4*d^2
*e^4 + B^2*b^2*c^4*d^4*e^2 - 4*B^2*b^3*c^3*d^3*e^3 + 6*B^2*b^4*c^2*d^2*e^4 - 4*B^2*b^5*c*d*e^5 - 4*A^2*b*c^5*d
^3*e^3 - 4*A^2*b^3*c^3*d*e^5 - 2*A*B*b^5*c*e^6 - 2*A*B*b*c^5*d^4*e^2 + 8*A*B*b^4*c^2*d*e^5 + 8*A*B*b^2*c^4*d^3
*e^3 - 12*A*B*b^3*c^3*d^2*e^4))/c^3 - (A*((8*(B*b^4*c^3*d*e^4 - A*b^3*c^4*d*e^4 + A*b^2*c^5*d^2*e^3 + B*b^2*c^
5*d^3*e^2 - 2*B*b^3*c^4*d^2*e^3))/c^3 - (8*A*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(d^3)^(1/2)*(d + e*x)^(1/2))/(b*c
^3))*(d^3)^(1/2))/b)*(d^3)^(1/2)*1i)/b)/((16*(2*A^3*c^5*d^5*e^3 + 4*A^3*b^2*c^3*d^3*e^5 - A^3*b^3*c^2*d^2*e^6
- A*B^2*b^5*d^2*e^6 + A^2*B*c^5*d^6*e^2 - 5*A^3*b*c^4*d^4*e^4 + 4*A*B^2*b^2*c^3*d^5*e^3 - 6*A*B^2*b^3*c^2*d^4*
e^4 + 11*A^2*B*b^2*c^3*d^4*e^4 - 8*A^2*B*b^3*c^2*d^3*e^5 - A*B^2*b*c^4*d^6*e^2 + 4*A*B^2*b^4*c*d^3*e^5 - 6*A^2
*B*b*c^4*d^5*e^3 + 2*A^2*B*b^4*c*d^2*e^6))/c^3 + (A*((8*(d + e*x)^(1/2)*(B^2*b^6*e^6 + A^2*b^4*c^2*e^6 + 2*A^2
*c^6*d^4*e^2 + 6*A^2*b^2*c^4*d^2*e^4 + B^2*b^2*c^4*d^4*e^2 - 4*B^2*b^3*c^3*d^3*e^3 + 6*B^2*b^4*c^2*d^2*e^4 - 4
*B^2*b^5*c*d*e^5 - 4*A^2*b*c^5*d^3*e^3 - 4*A^2*b^3*c^3*d*e^5 - 2*A*B*b^5*c*e^6 - 2*A*B*b*c^5*d^4*e^2 + 8*A*B*b
^4*c^2*d*e^5 + 8*A*B*b^2*c^4*d^3*e^3 - 12*A*B*b^3*c^3*d^2*e^4))/c^3 + (A*((8*(B*b^4*c^3*d*e^4 - A*b^3*c^4*d*e^
4 + A*b^2*c^5*d^2*e^3 + B*b^2*c^5*d^3*e^2 - 2*B*b^3*c^4*d^2*e^3))/c^3 + (8*A*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(
d^3)^(1/2)*(d + e*x)^(1/2))/(b*c^3))*(d^3)^(1/2))/b)*(d^3)^(1/2))/b - (A*((8*(d + e*x)^(1/2)*(B^2*b^6*e^6 + A^
2*b^4*c^2*e^6 + 2*A^2*c^6*d^4*e^2 + 6*A^2*b^2*c^4*d^2*e^4 + B^2*b^2*c^4*d^4*e^2 - 4*B^2*b^3*c^3*d^3*e^3 + 6*B^
2*b^4*c^2*d^2*e^4 - 4*B^2*b^5*c*d*e^5 - 4*A^2*b*c^5*d^3*e^3 - 4*A^2*b^3*c^3*d*e^5 - 2*A*B*b^5*c*e^6 - 2*A*B*b*
c^5*d^4*e^2 + 8*A*B*b^4*c^2*d*e^5 + 8*A*B*b^2*c^4*d^3*e^3 - 12*A*B*b^3*c^3*d^2*e^4))/c^3 - (A*((8*(B*b^4*c^3*d
*e^4 - A*b^3*c^4*d*e^4 + A*b^2*c^5*d^2*e^3 + B*b^2*c^5*d^3*e^2 - 2*B*b^3*c^4*d^2*e^3))/c^3 - (8*A*(b^3*c^5*e^3
- 2*b^2*c^6*d*e^2)*(d^3)^(1/2)*(d + e*x)^(1/2))/(b*c^3))*(d^3)^(1/2))/b)*(d^3)^(1/2))/b))*(d^3)^(1/2)*2i)/b -
(atan((((-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B*b)*((8*(d + e*x)^(1/2)*(B^2*b^6*e^6 + A^2*b^4*c^2*e^6 + 2*A^2*c^6
*d^4*e^2 + 6*A^2*b^2*c^4*d^2*e^4 + B^2*b^2*c^4*d^4*e^2 - 4*B^2*b^3*c^3*d^3*e^3 + 6*B^2*b^4*c^2*d^2*e^4 - 4*B^2
*b^5*c*d*e^5 - 4*A^2*b*c^5*d^3*e^3 - 4*A^2*b^3*c^3*d*e^5 - 2*A*B*b^5*c*e^6 - 2*A*B*b*c^5*d^4*e^2 + 8*A*B*b^4*c
^2*d*e^5 + 8*A*B*b^2*c^4*d^3*e^3 - 12*A*B*b^3*c^3*d^2*e^4))/c^3 + ((-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B*b)*((8*
(B*b^4*c^3*d*e^4 - A*b^3*c^4*d*e^4 + A*b^2*c^5*d^2*e^3 + B*b^2*c^5*d^3*e^2 - 2*B*b^3*c^4*d^2*e^3))/c^3 + (8*(b
^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B*b)*(d + e*x)^(1/2))/(b*c^8)))/(b*c^5))*1i)/(
b*c^5) + ((-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B*b)*((8*(d + e*x)^(1/2)*(B^2*b^6*e^6 + A^2*b^4*c^2*e^6 + 2*A^2*c^
6*d^4*e^2 + 6*A^2*b^2*c^4*d^2*e^4 + B^2*b^2*c^4*d^4*e^2 - 4*B^2*b^3*c^3*d^3*e^3 + 6*B^2*b^4*c^2*d^2*e^4 - 4*B^
2*b^5*c*d*e^5 - 4*A^2*b*c^5*d^3*e^3 - 4*A^2*b^3*c^3*d*e^5 - 2*A*B*b^5*c*e^6 - 2*A*B*b*c^5*d^4*e^2 + 8*A*B*b^4*
c^2*d*e^5 + 8*A*B*b^2*c^4*d^3*e^3 - 12*A*B*b^3*c^3*d^2*e^4))/c^3 - ((-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B*b)*((8
*(B*b^4*c^3*d*e^4 - A*b^3*c^4*d*e^4 + A*b^2*c^5*d^2*e^3 + B*b^2*c^5*d^3*e^2 - 2*B*b^3*c^4*d^2*e^3))/c^3 - (8*(
b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B*b)*(d + e*x)^(1/2))/(b*c^8)))/(b*c^5))*1i)/
(b*c^5))/((16*(2*A^3*c^5*d^5*e^3 + 4*A^3*b^2*c^3*d^3*e^5 - A^3*b^3*c^2*d^2*e^6 - A*B^2*b^5*d^2*e^6 + A^2*B*c^5
*d^6*e^2 - 5*A^3*b*c^4*d^4*e^4 + 4*A*B^2*b^2*c^3*d^5*e^3 - 6*A*B^2*b^3*c^2*d^4*e^4 + 11*A^2*B*b^2*c^3*d^4*e^4
- 8*A^2*B*b^3*c^2*d^3*e^5 - A*B^2*b*c^4*d^6*e^2 + 4*A*B^2*b^4*c*d^3*e^5 - 6*A^2*B*b*c^4*d^5*e^3 + 2*A^2*B*b^4*
c*d^2*e^6))/c^3 + ((-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B*b)*((8*(d + e*x)^(1/2)*(B^2*b^6*e^6 + A^2*b^4*c^2*e^6 +
2*A^2*c^6*d^4*e^2 + 6*A^2*b^2*c^4*d^2*e^4 + B^2*b^2*c^4*d^4*e^2 - 4*B^2*b^3*c^3*d^3*e^3 + 6*B^2*b^4*c^2*d^2*e
^4 - 4*B^2*b^5*c*d*e^5 - 4*A^2*b*c^5*d^3*e^3 - 4*A^2*b^3*c^3*d*e^5 - 2*A*B*b^5*c*e^6 - 2*A*B*b*c^5*d^4*e^2 + 8
*A*B*b^4*c^2*d*e^5 + 8*A*B*b^2*c^4*d^3*e^3 - 12*A*B*b^3*c^3*d^2*e^4))/c^3 + ((-c^5*(b*e - c*d)^3)^(1/2)*(A*c -
B*b)*((8*(B*b^4*c^3*d*e^4 - A*b^3*c^4*d*e^4 + A*b^2*c^5*d^2*e^3 + B*b^2*c^5*d^3*e^2 - 2*B*b^3*c^4*d^2*e^3))/c
^3 + (8*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B*b)*(d + e*x)^(1/2))/(b*c^8)))/(b*c
^5)))/(b*c^5) - ((-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B*b)*((8*(d + e*x)^(1/2)*(B^2*b^6*e^6 + A^2*b^4*c^2*e^6 + 2
*A^2*c^6*d^4*e^2 + 6*A^2*b^2*c^4*d^2*e^4 + B^2*b^2*c^4*d^4*e^2 - 4*B^2*b^3*c^3*d^3*e^3 + 6*B^2*b^4*c^2*d^2*e^4
- 4*B^2*b^5*c*d*e^5 - 4*A^2*b*c^5*d^3*e^3 - 4*A^2*b^3*c^3*d*e^5 - 2*A*B*b^5*c*e^6 - 2*A*B*b*c^5*d^4*e^2 + 8*A
*B*b^4*c^2*d*e^5 + 8*A*B*b^2*c^4*d^3*e^3 - 12*A*B*b^3*c^3*d^2*e^4))/c^3 - ((-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B
*b)*((8*(B*b^4*c^3*d*e^4 - A*b^3*c^4*d*e^4 + A*b^2*c^5*d^2*e^3 + B*b^2*c^5*d^3*e^2 - 2*B*b^3*c^4*d^2*e^3))/c^3
- (8*(b^3*c^5*e^3 - 2*b^2*c^6*d*e^2)*(-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B*b)*(d + e*x)^(1/2))/(b*c^8)))/(b*c^5
)))/(b*c^5)))*(-c^5*(b*e - c*d)^3)^(1/2)*(A*c - B*b)*2i)/(b*c^5)

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sympy [A]  time = 82.05, size = 134, normalized size = 1.02 \[ \frac {2 A d^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}} + \frac {2 B \left (d + e x\right )^{\frac {3}{2}}}{3 c} + \frac {\sqrt {d + e x} \left (2 A c e - 2 B b e + 2 B c d\right )}{c^{2}} + \frac {2 \left (- A c + B b\right ) \left (b e - c d\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b c^{3} \sqrt {\frac {b e - c d}{c}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(3/2)/(c*x**2+b*x),x)

[Out]

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

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