Integrand size = 21, antiderivative size = 92 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {2 e \sqrt {d+e x}}{c}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 (c d-b e)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{3/2}} \]
-2*d^(3/2)*arctanh((e*x+d)^(1/2)/d^(1/2))/b+2*(-b*e+c*d)^(3/2)*arctanh(c^( 1/2)*(e*x+d)^(1/2)/(-b*e+c*d)^(1/2))/b/c^(3/2)+2*e*(e*x+d)^(1/2)/c
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {2 \left (b \sqrt {c} e \sqrt {d+e x}-(-c d+b e)^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )-c^{3/2} d^{3/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )}{b c^{3/2}} \]
(2*(b*Sqrt[c]*e*Sqrt[d + e*x] - (-(c*d) + b*e)^(3/2)*ArcTan[(Sqrt[c]*Sqrt[ d + e*x])/Sqrt[-(c*d) + b*e]] - c^(3/2)*d^(3/2)*ArcTanh[Sqrt[d + e*x]/Sqrt [d]]))/(b*c^(3/2))
Time = 0.34 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1146, 1197, 25, 27, 1480, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx\) |
\(\Big \downarrow \) 1146 |
\(\displaystyle \frac {\int \frac {c d^2+e (2 c d-b e) x}{\sqrt {d+e x} \left (c x^2+b x\right )}dx}{c}+\frac {2 e \sqrt {d+e x}}{c}\) |
\(\Big \downarrow \) 1197 |
\(\displaystyle \frac {2 \int -\frac {e (d (c d-b e)-(2 c d-b e) (d+e x))}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{c}+\frac {2 e \sqrt {d+e x}}{c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 e \sqrt {d+e x}}{c}-\frac {2 \int \frac {e (d (c d-b e)-(2 c d-b e) (d+e x))}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 e \sqrt {d+e x}}{c}-\frac {2 e \int \frac {d (c d-b e)-(2 c d-b e) (d+e x)}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{c}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {2 e \sqrt {d+e x}}{c}-\frac {2 e \left (\frac {(c d-b e)^2 \int \frac {1}{-c d+b e+c (d+e x)}d\sqrt {d+e x}}{b e}-\frac {c^2 d^2 \int \frac {1}{c (d+e x)-c d}d\sqrt {d+e x}}{b e}\right )}{c}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 e \sqrt {d+e x}}{c}-\frac {2 e \left (\frac {c d^{3/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b e}-\frac {(c d-b e)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b \sqrt {c} e}\right )}{c}\) |
(2*e*Sqrt[d + e*x])/c - (2*e*((c*d^(3/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/( b*e) - ((c*d - b*e)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]] )/(b*Sqrt[c]*e)))/c
3.4.63.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol ] :> Simp[e*((d + e*x)^(m - 1)/(c*(m - 1))), x] + Simp[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] && GtQ[m, 1]
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)), x_Symbol] :> Simp[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] /; Fr eeQ[{a, b, c, d, e, f, g}, x]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(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]
Time = 1.96 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07
method | result | size |
pseudoelliptic | \(\frac {-2 \left (b e -c d \right )^{2} \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (b e -c d \right ) c}}\right )+2 \sqrt {\left (b e -c d \right ) c}\, \left (b e \sqrt {e x +d}-\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right ) d^{\frac {3}{2}} c \right )}{b c \sqrt {\left (b e -c d \right ) c}}\) | \(98\) |
derivativedivides | \(2 e \left (\frac {\sqrt {e x +d}}{c}+\frac {\left (-b^{2} e^{2}+2 b c d e -c^{2} d^{2}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (b e -c d \right ) c}}\right )}{c b e \sqrt {\left (b e -c d \right ) c}}-\frac {d^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e}\right )\) | \(106\) |
default | \(2 e \left (\frac {\sqrt {e x +d}}{c}+\frac {\left (-b^{2} e^{2}+2 b c d e -c^{2} d^{2}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (b e -c d \right ) c}}\right )}{c b e \sqrt {\left (b e -c d \right ) c}}-\frac {d^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e}\right )\) | \(106\) |
2*(-(b*e-c*d)^2*arctan(c*(e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2))+((b*e-c*d)*c)^ (1/2)*(b*e*(e*x+d)^(1/2)-arctanh((e*x+d)^(1/2)/d^(1/2))*d^(3/2)*c))/((b*e- c*d)*c)^(1/2)/c/b
Time = 0.32 (sec) , antiderivative size = 447, normalized size of antiderivative = 4.86 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\left [\frac {c d^{\frac {3}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, \sqrt {e x + d} b e - {\left (c d - b e\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right )}{b c}, \frac {c d^{\frac {3}{2}} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + 2 \, \sqrt {e x + d} b e + 2 \, {\left (c d - b e\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right )}{b c}, \frac {2 \, c \sqrt {-d} d \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + 2 \, \sqrt {e x + d} b e - {\left (c d - b e\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {c e x + 2 \, c d - b e - 2 \, \sqrt {e x + d} c \sqrt {\frac {c d - b e}{c}}}{c x + b}\right )}{b c}, \frac {2 \, {\left (c \sqrt {-d} d \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + \sqrt {e x + d} b e + {\left (c d - b e\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {e x + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right )\right )}}{b c}\right ] \]
[(c*d^(3/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*sqrt(e*x + d) *b*e - (c*d - b*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)))/(b*c), (c*d^(3/2)*log((e*x - 2* sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*sqrt(e*x + d)*b*e + 2*(c*d - b*e)*sqrt (-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) )/(b*c), (2*c*sqrt(-d)*d*arctan(sqrt(e*x + d)*sqrt(-d)/d) + 2*sqrt(e*x + d )*b*e - (c*d - b*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)))/(b*c), 2*(c*sqrt(-d)*d*arctan( sqrt(e*x + d)*sqrt(-d)/d) + sqrt(e*x + d)*b*e + (c*d - b*e)*sqrt(-(c*d - b *e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)))/(b*c)]
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (80) = 160\).
Time = 2.39 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.83 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\begin {cases} \frac {2 \left (\frac {e^{2} \sqrt {d + e x}}{c} + \frac {d^{2} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}} - \frac {e \left (b e - c d\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b c^{2} \sqrt {\frac {b e - c d}{c}}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {3}{2}} \left (- \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\- \frac {\log {\left (b - 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b} - \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\\frac {\log {\left (b + 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b}\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(e**2*sqrt(d + e*x)/c + d**2*e*atan(sqrt(d + e*x)/sqrt(-d))/( b*sqrt(-d)) - e*(b*e - c*d)**2*atan(sqrt(d + e*x)/sqrt((b*e - c*d)/c))/(b* c**2*sqrt((b*e - c*d)/c)))/e, Ne(e, 0)), (d**(3/2)*(-2*c*Piecewise(((b/(2* c) + x)/b, Eq(c, 0)), (-log(b - 2*c*(b/(2*c) + x))/(2*c), True))/b - 2*c*P iecewise(((b/(2*c) + x)/b, Eq(c, 0)), (log(b + 2*c*(b/(2*c) + x))/(2*c), T rue))/b), True))
Exception generated. \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for m ore detail
Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {2 \, d^{2} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} + \frac {2 \, \sqrt {e x + d} e}{c} - \frac {2 \, {\left (c^{2} d^{2} - 2 \, b c d e + b^{2} e^{2}\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b c} \]
2*d^2*arctan(sqrt(e*x + d)/sqrt(-d))/(b*sqrt(-d)) + 2*sqrt(e*x + d)*e/c - 2*(c^2*d^2 - 2*b*c*d*e + b^2*e^2)*arctan(sqrt(e*x + d)*c/sqrt(-c^2*d + b*c *e))/(sqrt(-c^2*d + b*c*e)*b*c)
Time = 0.23 (sec) , antiderivative size = 697, normalized size of antiderivative = 7.58 \[ \int \frac {(d+e x)^{3/2}}{b x+c x^2} \, dx=\frac {2\,e\,\sqrt {d+e\,x}}{c}-\frac {2\,\mathrm {atanh}\left (\frac {16\,b^3\,e^6\,\sqrt {d^3}\,\sqrt {d+e\,x}}{16\,b^3\,d^2\,e^6-64\,b^2\,c\,d^3\,e^5+96\,b\,c^2\,d^4\,e^4-48\,c^3\,d^5\,e^3}+\frac {48\,c^2\,d^3\,e^3\,\sqrt {d^3}\,\sqrt {d+e\,x}}{64\,b^2\,d^3\,e^5+48\,c^2\,d^5\,e^3-\frac {16\,b^3\,d^2\,e^6}{c}-96\,b\,c\,d^4\,e^4}+\frac {64\,b^2\,d\,e^5\,\sqrt {d^3}\,\sqrt {d+e\,x}}{64\,b^2\,d^3\,e^5+48\,c^2\,d^5\,e^3-\frac {16\,b^3\,d^2\,e^6}{c}-96\,b\,c\,d^4\,e^4}-\frac {96\,b\,c\,d^2\,e^4\,\sqrt {d^3}\,\sqrt {d+e\,x}}{64\,b^2\,d^3\,e^5+48\,c^2\,d^5\,e^3-\frac {16\,b^3\,d^2\,e^6}{c}-96\,b\,c\,d^4\,e^4}\right )\,\sqrt {d^3}}{b}+\frac {2\,\mathrm {atanh}\left (\frac {48\,d^3\,e^3\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{48\,c^3\,d^5\,e^3-80\,b^3\,d^2\,e^6-144\,b\,c^2\,d^4\,e^4+160\,b^2\,c\,d^3\,e^5+\frac {16\,b^4\,d\,e^7}{c}}+\frac {16\,b^2\,d\,e^5\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{16\,b^4\,c\,d\,e^7-80\,b^3\,c^2\,d^2\,e^6+160\,b^2\,c^3\,d^3\,e^5-144\,b\,c^4\,d^4\,e^4+48\,c^5\,d^5\,e^3}-\frac {48\,b\,d^2\,e^4\,\sqrt {d+e\,x}\,\sqrt {-b^3\,c^3\,e^3+3\,b^2\,c^4\,d\,e^2-3\,b\,c^5\,d^2\,e+c^6\,d^3}}{16\,b^4\,d\,e^7-80\,b^3\,c\,d^2\,e^6+160\,b^2\,c^2\,d^3\,e^5-144\,b\,c^3\,d^4\,e^4+48\,c^4\,d^5\,e^3}\right )\,\sqrt {-c^3\,{\left (b\,e-c\,d\right )}^3}}{b\,c^3} \]
(2*e*(d + e*x)^(1/2))/c - (2*atanh((16*b^3*e^6*(d^3)^(1/2)*(d + e*x)^(1/2) )/(16*b^3*d^2*e^6 - 48*c^3*d^5*e^3 + 96*b*c^2*d^4*e^4 - 64*b^2*c*d^3*e^5) + (48*c^2*d^3*e^3*(d^3)^(1/2)*(d + e*x)^(1/2))/(64*b^2*d^3*e^5 + 48*c^2*d^ 5*e^3 - (16*b^3*d^2*e^6)/c - 96*b*c*d^4*e^4) + (64*b^2*d*e^5*(d^3)^(1/2)*( d + e*x)^(1/2))/(64*b^2*d^3*e^5 + 48*c^2*d^5*e^3 - (16*b^3*d^2*e^6)/c - 96 *b*c*d^4*e^4) - (96*b*c*d^2*e^4*(d^3)^(1/2)*(d + e*x)^(1/2))/(64*b^2*d^3*e ^5 + 48*c^2*d^5*e^3 - (16*b^3*d^2*e^6)/c - 96*b*c*d^4*e^4))*(d^3)^(1/2))/b + (2*atanh((48*d^3*e^3*(d + e*x)^(1/2)*(c^6*d^3 - b^3*c^3*e^3 + 3*b^2*c^4 *d*e^2 - 3*b*c^5*d^2*e)^(1/2))/(48*c^3*d^5*e^3 - 80*b^3*d^2*e^6 - 144*b*c^ 2*d^4*e^4 + 160*b^2*c*d^3*e^5 + (16*b^4*d*e^7)/c) + (16*b^2*d*e^5*(d + e*x )^(1/2)*(c^6*d^3 - b^3*c^3*e^3 + 3*b^2*c^4*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/( 48*c^5*d^5*e^3 - 144*b*c^4*d^4*e^4 + 160*b^2*c^3*d^3*e^5 - 80*b^3*c^2*d^2* e^6 + 16*b^4*c*d*e^7) - (48*b*d^2*e^4*(d + e*x)^(1/2)*(c^6*d^3 - b^3*c^3*e ^3 + 3*b^2*c^4*d*e^2 - 3*b*c^5*d^2*e)^(1/2))/(16*b^4*d*e^7 + 48*c^4*d^5*e^ 3 - 144*b*c^3*d^4*e^4 - 80*b^3*c*d^2*e^6 + 160*b^2*c^2*d^3*e^5))*(-c^3*(b* e - c*d)^3)^(1/2))/(b*c^3)