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

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

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Rubi [A]  time = 0.30, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {828, 826, 1166, 208} \begin {gather*} \frac {2 (B d-A e)}{d \sqrt {d+e x} (c d-b e)}-\frac {2 \sqrt {c} (b B-A c) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b (c d-b e)^{3/2}}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 828

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

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

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Mathematica [C]  time = 0.03, size = 89, normalized size = 0.75 \begin {gather*} \frac {2 \left (d (b B-A c) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {c (d+e x)}{c d-b e}\right )+A (c d-b e) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {e x}{d}+1\right )\right )}{b d \sqrt {d+e x} (c d-b e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((b*B - A*c)*d*Hypergeometric2F1[-1/2, 1, 1/2, (c*(d + e*x))/(c*d - b*e)] + A*(c*d - b*e)*Hypergeometric2F1
[-1/2, 1, 1/2, 1 + (e*x)/d]))/(b*d*(c*d - b*e)*Sqrt[d + e*x])

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IntegrateAlgebraic [A]  time = 0.33, size = 132, normalized size = 1.12 \begin {gather*} -\frac {2 \left (A c^{3/2}-b B \sqrt {c}\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x} \sqrt {b e-c d}}{c d-b e}\right )}{b (b e-c d)^{3/2}}+\frac {2 (B d-A e)}{d \sqrt {d+e x} (c d-b e)}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.73, size = 835, normalized size = 7.08 \begin {gather*} \left [\frac {{\left ({\left (B b - A c\right )} d^{2} e x + {\left (B b - A c\right )} d^{3}\right )} \sqrt {\frac {c}{c d - b e}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, {\left (c d - b e\right )} \sqrt {e x + d} \sqrt {\frac {c}{c d - b e}}}{c x + b}\right ) + {\left (A c d^{2} - A b d e + {\left (A c d e - A b e^{2}\right )} x\right )} \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, {\left (B b d^{2} - A b d e\right )} \sqrt {e x + d}}{b c d^{4} - b^{2} d^{3} e + {\left (b c d^{3} e - b^{2} d^{2} e^{2}\right )} x}, -\frac {2 \, {\left ({\left (B b - A c\right )} d^{2} e x + {\left (B b - A c\right )} d^{3}\right )} \sqrt {-\frac {c}{c d - b e}} \arctan \left (-\frac {{\left (c d - b e\right )} \sqrt {e x + d} \sqrt {-\frac {c}{c d - b e}}}{c e x + c d}\right ) - {\left (A c d^{2} - A b d e + {\left (A c d e - A b e^{2}\right )} x\right )} \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - 2 \, {\left (B b d^{2} - A b d e\right )} \sqrt {e x + d}}{b c d^{4} - b^{2} d^{3} e + {\left (b c d^{3} e - b^{2} d^{2} e^{2}\right )} x}, \frac {2 \, {\left (A c d^{2} - A b d e + {\left (A c d e - A b e^{2}\right )} x\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left ({\left (B b - A c\right )} d^{2} e x + {\left (B b - A c\right )} d^{3}\right )} \sqrt {\frac {c}{c d - b e}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, {\left (c d - b e\right )} \sqrt {e x + d} \sqrt {\frac {c}{c d - b e}}}{c x + b}\right ) + 2 \, {\left (B b d^{2} - A b d e\right )} \sqrt {e x + d}}{b c d^{4} - b^{2} d^{3} e + {\left (b c d^{3} e - b^{2} d^{2} e^{2}\right )} x}, -\frac {2 \, {\left ({\left ({\left (B b - A c\right )} d^{2} e x + {\left (B b - A c\right )} d^{3}\right )} \sqrt {-\frac {c}{c d - b e}} \arctan \left (-\frac {{\left (c d - b e\right )} \sqrt {e x + d} \sqrt {-\frac {c}{c d - b e}}}{c e x + c d}\right ) - {\left (A c d^{2} - A b d e + {\left (A c d e - A b e^{2}\right )} x\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) - {\left (B b d^{2} - A b d e\right )} \sqrt {e x + d}\right )}}{b c d^{4} - b^{2} d^{3} e + {\left (b c d^{3} e - b^{2} d^{2} e^{2}\right )} x}\right ] \end {gather*}

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]

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

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giac [A]  time = 0.22, size = 129, normalized size = 1.09 \begin {gather*} \frac {2 \, {\left (B b c - A c^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{{\left (b c d - b^{2} e\right )} \sqrt {-c^{2} d + b c e}} + \frac {2 \, {\left (B d - A e\right )}}{{\left (c d^{2} - b d e\right )} \sqrt {x e + d}} + \frac {2 \, A \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right )}{b \sqrt {-d} d} \end {gather*}

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

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maple [A]  time = 0.07, size = 168, normalized size = 1.42 \begin {gather*} \frac {2 A \,c^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\left (b e -c d \right ) \sqrt {\left (b e -c d \right ) c}\, b}-\frac {2 B c \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\left (b e -c d \right ) \sqrt {\left (b e -c d \right ) c}}+\frac {2 A e}{\left (b e -c d \right ) \sqrt {e x +d}\, d}-\frac {2 B}{\left (b e -c d \right ) \sqrt {e x +d}}-\frac {2 A \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b \,d^{\frac {3}{2}}} \end {gather*}

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/(b*e-c*d)*c^2/b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)*A-2/(b*e-c*d)*c/((b*e-c*d)*c
)^(1/2)*arctan((e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2)*c)*B-2*A*arctanh((e*x+d)^(1/2)/d^(1/2))/b/d^(3/2)+2/(b*e-c*d)
/d/(e*x+d)^(1/2)*A*e-2/(b*e-c*d)/(e*x+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((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 = 3.12, size = 3674, normalized size = 31.14

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((((A*c - B*b)*(-c*(b*e - c*d)^3)^(1/2)*((d + e*x)^(1/2)*(16*A^2*c^8*d^8*e^2 + 104*A^2*b^2*c^6*d^6*e^4 -
88*A^2*b^3*c^5*d^5*e^5 + 40*A^2*b^4*c^4*d^4*e^6 - 8*A^2*b^5*c^3*d^3*e^7 + 8*B^2*b^2*c^6*d^8*e^2 - 24*B^2*b^3*c
^5*d^7*e^3 + 24*B^2*b^4*c^4*d^6*e^4 - 8*B^2*b^5*c^3*d^5*e^5 - 64*A^2*b*c^7*d^7*e^3 - 16*A*B*b*c^7*d^8*e^2 + 48
*A*B*b^2*c^6*d^7*e^3 - 48*A*B*b^3*c^5*d^6*e^4 + 16*A*B*b^4*c^4*d^5*e^5) - ((A*c - B*b)*(-c*(b*e - c*d)^3)^(1/2
)*(((A*c - B*b)*(-c*(b*e - c*d)^3)^(1/2)*(d + e*x)^(1/2)*(16*b^2*c^8*d^11*e^2 - 88*b^3*c^7*d^10*e^3 + 200*b^4*
c^6*d^9*e^4 - 240*b^5*c^5*d^8*e^5 + 160*b^6*c^4*d^7*e^6 - 56*b^7*c^3*d^6*e^7 + 8*b^8*c^2*d^5*e^8))/(b^4*e^3 -
b*c^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3*c*d*e^2) - 16*A*b^2*c^7*d^9*e^3 + 72*A*b^3*c^6*d^8*e^4 - 128*A*b^4*c^5*d^7
*e^5 + 112*A*b^5*c^4*d^6*e^6 - 48*A*b^6*c^3*d^5*e^7 + 8*A*b^7*c^2*d^4*e^8 + 8*B*b^2*c^7*d^10*e^2 - 32*B*b^3*c^
6*d^9*e^3 + 48*B*b^4*c^5*d^8*e^4 - 32*B*b^5*c^4*d^7*e^5 + 8*B*b^6*c^3*d^6*e^6))/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c
^2*d^2*e - 3*b^3*c*d*e^2))*1i)/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3*c*d*e^2) + ((A*c - B*b)*(-c*(b*e
- c*d)^3)^(1/2)*((d + e*x)^(1/2)*(16*A^2*c^8*d^8*e^2 + 104*A^2*b^2*c^6*d^6*e^4 - 88*A^2*b^3*c^5*d^5*e^5 + 40*
A^2*b^4*c^4*d^4*e^6 - 8*A^2*b^5*c^3*d^3*e^7 + 8*B^2*b^2*c^6*d^8*e^2 - 24*B^2*b^3*c^5*d^7*e^3 + 24*B^2*b^4*c^4*
d^6*e^4 - 8*B^2*b^5*c^3*d^5*e^5 - 64*A^2*b*c^7*d^7*e^3 - 16*A*B*b*c^7*d^8*e^2 + 48*A*B*b^2*c^6*d^7*e^3 - 48*A*
B*b^3*c^5*d^6*e^4 + 16*A*B*b^4*c^4*d^5*e^5) - ((A*c - B*b)*(-c*(b*e - c*d)^3)^(1/2)*(((A*c - B*b)*(-c*(b*e - c
*d)^3)^(1/2)*(d + e*x)^(1/2)*(16*b^2*c^8*d^11*e^2 - 88*b^3*c^7*d^10*e^3 + 200*b^4*c^6*d^9*e^4 - 240*b^5*c^5*d^
8*e^5 + 160*b^6*c^4*d^7*e^6 - 56*b^7*c^3*d^6*e^7 + 8*b^8*c^2*d^5*e^8))/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e
- 3*b^3*c*d*e^2) + 16*A*b^2*c^7*d^9*e^3 - 72*A*b^3*c^6*d^8*e^4 + 128*A*b^4*c^5*d^7*e^5 - 112*A*b^5*c^4*d^6*e^6
+ 48*A*b^6*c^3*d^5*e^7 - 8*A*b^7*c^2*d^4*e^8 - 8*B*b^2*c^7*d^10*e^2 + 32*B*b^3*c^6*d^9*e^3 - 48*B*b^4*c^5*d^8
*e^4 + 32*B*b^5*c^4*d^7*e^5 - 8*B*b^6*c^3*d^6*e^6))/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3*c*d*e^2))*1
i)/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3*c*d*e^2))/(16*A^3*c^7*d^6*e^3 + ((A*c - B*b)*(-c*(b*e - c*d)
^3)^(1/2)*((d + e*x)^(1/2)*(16*A^2*c^8*d^8*e^2 + 104*A^2*b^2*c^6*d^6*e^4 - 88*A^2*b^3*c^5*d^5*e^5 + 40*A^2*b^4
*c^4*d^4*e^6 - 8*A^2*b^5*c^3*d^3*e^7 + 8*B^2*b^2*c^6*d^8*e^2 - 24*B^2*b^3*c^5*d^7*e^3 + 24*B^2*b^4*c^4*d^6*e^4
- 8*B^2*b^5*c^3*d^5*e^5 - 64*A^2*b*c^7*d^7*e^3 - 16*A*B*b*c^7*d^8*e^2 + 48*A*B*b^2*c^6*d^7*e^3 - 48*A*B*b^3*c
^5*d^6*e^4 + 16*A*B*b^4*c^4*d^5*e^5) - ((A*c - B*b)*(-c*(b*e - c*d)^3)^(1/2)*(((A*c - B*b)*(-c*(b*e - c*d)^3)^
(1/2)*(d + e*x)^(1/2)*(16*b^2*c^8*d^11*e^2 - 88*b^3*c^7*d^10*e^3 + 200*b^4*c^6*d^9*e^4 - 240*b^5*c^5*d^8*e^5 +
160*b^6*c^4*d^7*e^6 - 56*b^7*c^3*d^6*e^7 + 8*b^8*c^2*d^5*e^8))/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3
*c*d*e^2) - 16*A*b^2*c^7*d^9*e^3 + 72*A*b^3*c^6*d^8*e^4 - 128*A*b^4*c^5*d^7*e^5 + 112*A*b^5*c^4*d^6*e^6 - 48*A
*b^6*c^3*d^5*e^7 + 8*A*b^7*c^2*d^4*e^8 + 8*B*b^2*c^7*d^10*e^2 - 32*B*b^3*c^6*d^9*e^3 + 48*B*b^4*c^5*d^8*e^4 -
32*B*b^5*c^4*d^7*e^5 + 8*B*b^6*c^3*d^6*e^6))/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3*c*d*e^2)))/(b^4*e^
3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3*c*d*e^2) - ((A*c - B*b)*(-c*(b*e - c*d)^3)^(1/2)*((d + e*x)^(1/2)*(16*
A^2*c^8*d^8*e^2 + 104*A^2*b^2*c^6*d^6*e^4 - 88*A^2*b^3*c^5*d^5*e^5 + 40*A^2*b^4*c^4*d^4*e^6 - 8*A^2*b^5*c^3*d^
3*e^7 + 8*B^2*b^2*c^6*d^8*e^2 - 24*B^2*b^3*c^5*d^7*e^3 + 24*B^2*b^4*c^4*d^6*e^4 - 8*B^2*b^5*c^3*d^5*e^5 - 64*A
^2*b*c^7*d^7*e^3 - 16*A*B*b*c^7*d^8*e^2 + 48*A*B*b^2*c^6*d^7*e^3 - 48*A*B*b^3*c^5*d^6*e^4 + 16*A*B*b^4*c^4*d^5
*e^5) - ((A*c - B*b)*(-c*(b*e - c*d)^3)^(1/2)*(((A*c - B*b)*(-c*(b*e - c*d)^3)^(1/2)*(d + e*x)^(1/2)*(16*b^2*c
^8*d^11*e^2 - 88*b^3*c^7*d^10*e^3 + 200*b^4*c^6*d^9*e^4 - 240*b^5*c^5*d^8*e^5 + 160*b^6*c^4*d^7*e^6 - 56*b^7*c
^3*d^6*e^7 + 8*b^8*c^2*d^5*e^8))/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3*c*d*e^2) + 16*A*b^2*c^7*d^9*e^
3 - 72*A*b^3*c^6*d^8*e^4 + 128*A*b^4*c^5*d^7*e^5 - 112*A*b^5*c^4*d^6*e^6 + 48*A*b^6*c^3*d^5*e^7 - 8*A*b^7*c^2*
d^4*e^8 - 8*B*b^2*c^7*d^10*e^2 + 32*B*b^3*c^6*d^9*e^3 - 48*B*b^4*c^5*d^8*e^4 + 32*B*b^5*c^4*d^7*e^5 - 8*B*b^6*
c^3*d^6*e^6))/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3*c*d*e^2)))/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e
- 3*b^3*c*d*e^2) + 48*A^3*b^2*c^5*d^4*e^5 - 16*A^3*b^3*c^4*d^3*e^6 - 16*A^2*B*c^7*d^7*e^2 - 48*A^3*b*c^6*d^5*
e^4 - 48*A*B^2*b^2*c^5*d^6*e^3 + 48*A*B^2*b^3*c^4*d^5*e^4 - 16*A*B^2*b^4*c^3*d^4*e^5 - 32*A^2*B*b^3*c^4*d^4*e^
5 + 16*A^2*B*b^4*c^3*d^3*e^6 + 16*A*B^2*b*c^6*d^7*e^2 + 32*A^2*B*b*c^6*d^6*e^3))*(A*c - B*b)*(-c*(b*e - c*d)^3
)^(1/2)*2i)/(b^4*e^3 - b*c^3*d^3 + 3*b^2*c^2*d^2*e - 3*b^3*c*d*e^2) - (2*(A*e - B*d))/((c*d^2 - b*d*e)*(d + e*
x)^(1/2)) + (A*atan((B^2*b^2*c^4*d^11*(d + e*x)^(1/2)*1i - A^2*b^6*d^5*e^6*(d + e*x)^(1/2)*1i + A^2*b^5*c*d^6*
e^5*(d + e*x)^(1/2)*6i - B^2*b^3*c^3*d^10*e*(d + e*x)^(1/2)*3i - B^2*b^5*c*d^8*e^3*(d + e*x)^(1/2)*1i - A*B*b*
c^5*d^11*(d + e*x)^(1/2)*2i - A^2*b^2*c^4*d^9*e^2*(d + e*x)^(1/2)*12i + A^2*b^3*c^3*d^8*e^3*(d + e*x)^(1/2)*19
i - A^2*b^4*c^2*d^7*e^4*(d + e*x)^(1/2)*15i + B^2*b^4*c^2*d^9*e^2*(d + e*x)^(1/2)*3i + A^2*b*c^5*d^10*e*(d + e
*x)^(1/2)*3i - A*B*b^3*c^3*d^9*e^2*(d + e*x)^(1/2)*6i + A*B*b^4*c^2*d^8*e^3*(d + e*x)^(1/2)*2i + A*B*b^2*c^4*d
^10*e*(d + e*x)^(1/2)*6i)/(d^3*(d^3)^(1/2)*(d^3*(d^3*(3*A^2*b*c^5*e + B^2*b^2*c^4*d - 3*B^2*b^3*c^3*e - 2*A*B*
b*c^5*d + 6*A*B*b^2*c^4*e) - 15*A^2*b^4*c^2*e^4 - 12*A^2*b^2*c^4*d^2*e^2 + 3*B^2*b^4*c^2*d^2*e^2 - B^2*b^5*c*d
*e^3 + 19*A^2*b^3*c^3*d*e^3 + 2*A*B*b^4*c^2*d*e^3 - 6*A*B*b^3*c^3*d^2*e^2) - A^2*b^6*d*e^6 + 6*A^2*b^5*c*d^2*e
^5)))*2i)/(b*(d^3)^(1/2))

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sympy [A]  time = 61.49, size = 107, normalized size = 0.91 \begin {gather*} \frac {2 A \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b d \sqrt {- d}} - \frac {2 \left (- A e + B d\right )}{d \sqrt {d + e x} \left (b e - c d\right )} - \frac {2 \left (- A c + B b\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b \sqrt {\frac {b e - c d}{c}} \left (b e - c d\right )} \end {gather*}

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

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