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

Optimal. Leaf size=134 \[ \frac {\sqrt {d} (4 c d-3 b e) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b^3}-\frac {\sqrt {c d-b e} (4 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b^3 \sqrt {c}}-\frac {\sqrt {d+e x} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {738, 826, 1166, 208} \[ -\frac {\sqrt {d+e x} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}+\frac {\sqrt {d} (4 c d-3 b e) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b^3}-\frac {\sqrt {c d-b e} (4 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b^3 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), 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] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

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

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Mathematica [A]  time = 0.20, size = 127, normalized size = 0.95 \[ \frac {\frac {b \sqrt {d+e x} (-b d+b e x-2 c d x)}{x (b+c x)}+\sqrt {d} (4 c d-3 b e) \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )-\frac {\sqrt {c d-b e} (4 c d-b e) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{\sqrt {c}}}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sqrt[d + e*x]*(-(b*d) - 2*c*d*x + b*e*x))/(x*(b + c*x)) + Sqrt[d]*(4*c*d - 3*b*e)*ArcTanh[Sqrt[d + e*x]/Sq
rt[d]] - (Sqrt[c*d - b*e]*(4*c*d - b*e)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/Sqrt[c])/b^3

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fricas [A]  time = 1.08, size = 770, normalized size = 5.75 \[ \left [-\frac {{\left ({\left (4 \, c^{2} d - b c e\right )} x^{2} + {\left (4 \, b c d - b^{2} e\right )} x\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 ) + {\left ({\left (4 \, c^{2} d - 3 \, b c e\right )} x^{2} + {\left (4 \, b c d - 3 \, b^{2} e\right )} x\right )} \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, {\left (b^{2} d + {\left (2 \, b c d - b^{2} e\right )} x\right )} \sqrt {e x + d}}{2 \, {\left (b^{3} c x^{2} + b^{4} x\right )}}, -\frac {2 \, {\left ({\left (4 \, c^{2} d - b c e\right )} x^{2} + {\left (4 \, b c d - b^{2} e\right )} x\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 ({\left (4 \, c^{2} d - 3 \, b c e\right )} x^{2} + {\left (4 \, b c d - 3 \, b^{2} e\right )} x\right )} \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, {\left (b^{2} d + {\left (2 \, b c d - b^{2} e\right )} x\right )} \sqrt {e x + d}}{2 \, {\left (b^{3} c x^{2} + b^{4} x\right )}}, -\frac {2 \, {\left ({\left (4 \, c^{2} d - 3 \, b c e\right )} x^{2} + {\left (4 \, b c d - 3 \, b^{2} e\right )} x\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left ({\left (4 \, c^{2} d - b c e\right )} x^{2} + {\left (4 \, b c d - b^{2} e\right )} x\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^{2} d + {\left (2 \, b c d - b^{2} e\right )} x\right )} \sqrt {e x + d}}{2 \, {\left (b^{3} c x^{2} + b^{4} x\right )}}, -\frac {{\left ({\left (4 \, c^{2} d - b c e\right )} x^{2} + {\left (4 \, b c d - b^{2} e\right )} x\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 ({\left (4 \, c^{2} d - 3 \, b c e\right )} x^{2} + {\left (4 \, b c d - 3 \, b^{2} e\right )} x\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (b^{2} d + {\left (2 \, b c d - b^{2} e\right )} x\right )} \sqrt {e x + d}}{b^{3} c x^{2} + b^{4} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 211, normalized size = 1.57 \[ \frac {{\left (4 \, c^{2} d^{2} - 5 \, b c d e + b^{2} e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b^{3}} - \frac {{\left (4 \, c d^{2} - 3 \, b d e\right )} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right )}{b^{3} \sqrt {-d}} - \frac {2 \, {\left (x e + d\right )}^{\frac {3}{2}} c d e - 2 \, \sqrt {x e + d} c d^{2} e - {\left (x e + d\right )}^{\frac {3}{2}} b e^{2} + 2 \, \sqrt {x e + d} b d e^{2}}{{\left ({\left (x e + d\right )}^{2} c - 2 \, {\left (x e + d\right )} c d + c d^{2} + {\left (x e + d\right )} b e - b d e\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 237, normalized size = 1.77 \[ \frac {e^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b}-\frac {5 c d e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b^{2}}+\frac {4 c^{2} d^{2} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (b e -c d \right ) c}}\right )}{\sqrt {\left (b e -c d \right ) c}\, b^{3}}+\frac {\sqrt {e x +d}\, e^{2}}{\left (c e x +b e \right ) b}-\frac {\sqrt {e x +d}\, c d e}{\left (c e x +b e \right ) b^{2}}-\frac {3 \sqrt {d}\, e \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b^{2}}+\frac {4 c \,d^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b^{3}}-\frac {\sqrt {e x +d}\, d}{b^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((e*x+d)^(3/2)/(c*x^2+b*x)^2,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 = 0.42, size = 429, normalized size = 3.20 \[ -\frac {\frac {2\,\left (b\,d\,e^2-c\,d^2\,e\right )\,\sqrt {d+e\,x}}{b^2}-\frac {e\,\left (b\,e-2\,c\,d\right )\,{\left (d+e\,x\right )}^{3/2}}{b^2}}{\left (b\,e-2\,c\,d\right )\,\left (d+e\,x\right )+c\,{\left (d+e\,x\right )}^2+c\,d^2-b\,d\,e}-\frac {\sqrt {d}\,\mathrm {atanh}\left (\frac {6\,c\,\sqrt {d}\,e^7\,\sqrt {d+e\,x}}{6\,c\,d\,e^7-\frac {14\,c^2\,d^2\,e^6}{b}+\frac {8\,c^3\,d^3\,e^5}{b^2}}-\frac {14\,c^2\,d^{3/2}\,e^6\,\sqrt {d+e\,x}}{6\,b\,c\,d\,e^7-14\,c^2\,d^2\,e^6+\frac {8\,c^3\,d^3\,e^5}{b}}+\frac {8\,c^3\,d^{5/2}\,e^5\,\sqrt {d+e\,x}}{6\,b^2\,c\,d\,e^7-14\,b\,c^2\,d^2\,e^6+8\,c^3\,d^3\,e^5}\right )\,\left (3\,b\,e-4\,c\,d\right )}{b^3}-\frac {\mathrm {atanh}\left (\frac {2\,c\,d\,e^6\,\sqrt {c^2\,d-b\,c\,e}\,\sqrt {d+e\,x}}{2\,b\,c\,d\,e^7-10\,c^2\,d^2\,e^6+\frac {8\,c^3\,d^3\,e^5}{b}}-\frac {8\,c^2\,d^2\,e^5\,\sqrt {c^2\,d-b\,c\,e}\,\sqrt {d+e\,x}}{2\,b^2\,c\,d\,e^7-10\,b\,c^2\,d^2\,e^6+8\,c^3\,d^3\,e^5}\right )\,\sqrt {-c\,\left (b\,e-c\,d\right )}\,\left (b\,e-4\,c\,d\right )}{b^3\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*(b*d*e^2 - c*d^2*e)*(d + e*x)^(1/2))/b^2 - (e*(b*e - 2*c*d)*(d + e*x)^(3/2))/b^2)/((b*e - 2*c*d)*(d + e*
x) + c*(d + e*x)^2 + c*d^2 - b*d*e) - (d^(1/2)*atanh((6*c*d^(1/2)*e^7*(d + e*x)^(1/2))/(6*c*d*e^7 - (14*c^2*d^
2*e^6)/b + (8*c^3*d^3*e^5)/b^2) - (14*c^2*d^(3/2)*e^6*(d + e*x)^(1/2))/(6*b*c*d*e^7 - 14*c^2*d^2*e^6 + (8*c^3*
d^3*e^5)/b) + (8*c^3*d^(5/2)*e^5*(d + e*x)^(1/2))/(8*c^3*d^3*e^5 - 14*b*c^2*d^2*e^6 + 6*b^2*c*d*e^7))*(3*b*e -
4*c*d))/b^3 - (atanh((2*c*d*e^6*(c^2*d - b*c*e)^(1/2)*(d + e*x)^(1/2))/(2*b*c*d*e^7 - 10*c^2*d^2*e^6 + (8*c^3
*d^3*e^5)/b) - (8*c^2*d^2*e^5*(c^2*d - b*c*e)^(1/2)*(d + e*x)^(1/2))/(8*c^3*d^3*e^5 - 10*b*c^2*d^2*e^6 + 2*b^2
*c*d*e^7))*(-c*(b*e - c*d))^(1/2)*(b*e - 4*c*d))/(b^3*c)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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