3.13.30 \(\int \frac {(A+B x) (d+e x)^{5/2}}{b x+c x^2} \, dx\) [1230]

Optimal. Leaf size=173 \[ \frac {2 \left (B (c d-b e)^2+A c e (2 c d-b e)\right ) \sqrt {d+e x}}{c^3}+\frac {2 (B c d-b B e+A c e) (d+e x)^{3/2}}{3 c^2}+\frac {2 B (d+e x)^{5/2}}{5 c}-\frac {2 A d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}-\frac {2 (b B-A c) (c d-b e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{7/2}} \]

[Out]

2/3*(A*c*e-B*b*e+B*c*d)*(e*x+d)^(3/2)/c^2+2/5*B*(e*x+d)^(5/2)/c-2*A*d^(5/2)*arctanh((e*x+d)^(1/2)/d^(1/2))/b-2
*(-A*c+B*b)*(-b*e+c*d)^(5/2)*arctanh(c^(1/2)*(e*x+d)^(1/2)/(-b*e+c*d)^(1/2))/b/c^(7/2)+2*(B*(-b*e+c*d)^2+A*c*e
*(-b*e+2*c*d))*(e*x+d)^(1/2)/c^3

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Rubi [A]
time = 0.25, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {838, 840, 1180, 214} \begin {gather*} -\frac {2 (b B-A c) (c d-b e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{7/2}}+\frac {2 \sqrt {d+e x} \left (A c e (2 c d-b e)+B (c d-b e)^2\right )}{c^3}+\frac {2 (d+e x)^{3/2} (A c e-b B e+B c d)}{3 c^2}-\frac {2 A d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}+\frac {2 B (d+e x)^{5/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 214

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]

Rule 838

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 840

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 1180

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)^{5/2}}{b x+c x^2} \, dx &=\frac {2 B (d+e x)^{5/2}}{5 c}+\frac {\int \frac {(d+e x)^{3/2} (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) (d+e x)^{3/2}}{3 c^2}+\frac {2 B (d+e x)^{5/2}}{5 c}+\frac {\int \frac {\sqrt {d+e x} \left (A c^2 d^2+\left (B (c d-b e)^2+A c e (2 c d-b e)\right ) x\right )}{b x+c x^2} \, dx}{c^2}\\ &=\frac {2 \left (B (c d-b e)^2+A c e (2 c d-b e)\right ) \sqrt {d+e x}}{c^3}+\frac {2 (B c d-b B e+A c e) (d+e x)^{3/2}}{3 c^2}+\frac {2 B (d+e x)^{5/2}}{5 c}+\frac {\int \frac {A c^3 d^3+\left (B (c d-b e)^3+A c e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )\right ) x}{\sqrt {d+e x} \left (b x+c x^2\right )} \, dx}{c^3}\\ &=\frac {2 \left (B (c d-b e)^2+A c e (2 c d-b e)\right ) \sqrt {d+e x}}{c^3}+\frac {2 (B c d-b B e+A c e) (d+e x)^{3/2}}{3 c^2}+\frac {2 B (d+e x)^{5/2}}{5 c}+\frac {2 \text {Subst}\left (\int \frac {A c^3 d^3 e-d \left (B (c d-b e)^3+A c e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )\right )+\left (B (c d-b e)^3+A c e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )\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^3}\\ &=\frac {2 \left (B (c d-b e)^2+A c e (2 c d-b e)\right ) \sqrt {d+e x}}{c^3}+\frac {2 (B c d-b B e+A c e) (d+e x)^{3/2}}{3 c^2}+\frac {2 B (d+e x)^{5/2}}{5 c}+\frac {\left (2 A c d^3\right ) \text {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)^3\right ) \text {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^3}\\ &=\frac {2 \left (B (c d-b e)^2+A c e (2 c d-b e)\right ) \sqrt {d+e x}}{c^3}+\frac {2 (B c d-b B e+A c e) (d+e x)^{3/2}}{3 c^2}+\frac {2 B (d+e x)^{5/2}}{5 c}-\frac {2 A d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}-\frac {2 (b B-A c) (c d-b e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 167, normalized size = 0.97 \begin {gather*} \frac {2 \sqrt {d+e x} \left (5 A c e (7 c d-3 b e+c e x)+B \left (15 b^2 e^2-5 b c e (7 d+e x)+c^2 \left (23 d^2+11 d e x+3 e^2 x^2\right )\right )\right )}{15 c^3}+\frac {2 (-b B+A c) (-c d+b e)^{5/2} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{b c^{7/2}}-\frac {2 A d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(5*A*c*e*(7*c*d - 3*b*e + c*e*x) + B*(15*b^2*e^2 - 5*b*c*e*(7*d + e*x) + c^2*(23*d^2 + 11*d*e
*x + 3*e^2*x^2))))/(15*c^3) + (2*(-(b*B) + A*c)*(-(c*d) + b*e)^(5/2)*ArcTan[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[-(c*d
) + b*e]])/(b*c^(7/2)) - (2*A*d^(5/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b

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Maple [A]
time = 0.69, size = 285, normalized size = 1.65

method	result	size
derivativedivides	\(-\frac {2 \left (-\frac {B \,c^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {A \,c^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {B b c e \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {B \,c^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+A b c \,e^{2} \sqrt {e x +d}-2 A \,c^{2} d e \sqrt {e x +d}-B \,b^{2} e^{2} \sqrt {e x +d}+2 B b c d e \sqrt {e x +d}-B \,c^{2} d^{2} \sqrt {e x +d}\right )}{c^{3}}-\frac {2 A \,d^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b}+\frac {2 \left (A \,b^{3} c \,e^{3}-3 A \,b^{2} c^{2} d \,e^{2}+3 A b \,c^{3} d^{2} e -A \,c^{4} d^{3}-b^{4} B \,e^{3}+3 b^{3} B c d \,e^{2}-3 B \,b^{2} c^{2} d^{2} e +B b \,c^{3} d^{3}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (b e -c d \right ) c}}\right )}{c^{3} b \sqrt {\left (b e -c d \right ) c}}\)	\(285\)
default	\(-\frac {2 \left (-\frac {B \,c^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {A \,c^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {B b c e \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {B \,c^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+A b c \,e^{2} \sqrt {e x +d}-2 A \,c^{2} d e \sqrt {e x +d}-B \,b^{2} e^{2} \sqrt {e x +d}+2 B b c d e \sqrt {e x +d}-B \,c^{2} d^{2} \sqrt {e x +d}\right )}{c^{3}}-\frac {2 A \,d^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b}+\frac {2 \left (A \,b^{3} c \,e^{3}-3 A \,b^{2} c^{2} d \,e^{2}+3 A b \,c^{3} d^{2} e -A \,c^{4} d^{3}-b^{4} B \,e^{3}+3 b^{3} B c d \,e^{2}-3 B \,b^{2} c^{2} d^{2} e +B b \,c^{3} d^{3}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (b e -c d \right ) c}}\right )}{c^{3} b \sqrt {\left (b e -c d \right ) c}}\)	\(285\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(5/2)/(c*x^2+b*x),x,method=_RETURNVERBOSE)

[Out]

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

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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)^(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(c*d-%e*b>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 12.22, size = 1010, normalized size = 5.84 \begin {gather*} \left [\frac {15 \, A c^{3} d^{\frac {5}{2}} \log \left (\frac {x e - 2 \, \sqrt {x e + d} \sqrt {d} + 2 \, d}{x}\right ) + 15 \, {\left ({\left (B b c^{2} - A c^{3}\right )} d^{2} - 2 \, {\left (B b^{2} c - A b c^{2}\right )} d e + {\left (B b^{3} - A b^{2} c\right )} e^{2}\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {2 \, c d - 2 \, \sqrt {x e + d} c \sqrt {\frac {c d - b e}{c}} + {\left (c x - b\right )} e}{c x + b}\right ) + 2 \, {\left (23 \, B b c^{2} d^{2} + {\left (3 \, B b c^{2} x^{2} + 15 \, B b^{3} - 15 \, A b^{2} c - 5 \, {\left (B b^{2} c - A b c^{2}\right )} x\right )} e^{2} + {\left (11 \, B b c^{2} d x - 35 \, {\left (B b^{2} c - A b c^{2}\right )} d\right )} e\right )} \sqrt {x e + d}}{15 \, b c^{3}}, \frac {15 \, A c^{3} d^{\frac {5}{2}} \log \left (\frac {x e - 2 \, \sqrt {x e + d} \sqrt {d} + 2 \, d}{x}\right ) - 30 \, {\left ({\left (B b c^{2} - A c^{3}\right )} d^{2} - 2 \, {\left (B b^{2} c - A b c^{2}\right )} d e + {\left (B b^{3} - A b^{2} c\right )} e^{2}\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {x e + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + 2 \, {\left (23 \, B b c^{2} d^{2} + {\left (3 \, B b c^{2} x^{2} + 15 \, B b^{3} - 15 \, A b^{2} c - 5 \, {\left (B b^{2} c - A b c^{2}\right )} x\right )} e^{2} + {\left (11 \, B b c^{2} d x - 35 \, {\left (B b^{2} c - A b c^{2}\right )} d\right )} e\right )} \sqrt {x e + d}}{15 \, b c^{3}}, \frac {30 \, A c^{3} \sqrt {-d} d^{2} \arctan \left (\frac {\sqrt {x e + d} \sqrt {-d}}{d}\right ) + 15 \, {\left ({\left (B b c^{2} - A c^{3}\right )} d^{2} - 2 \, {\left (B b^{2} c - A b c^{2}\right )} d e + {\left (B b^{3} - A b^{2} c\right )} e^{2}\right )} \sqrt {\frac {c d - b e}{c}} \log \left (\frac {2 \, c d - 2 \, \sqrt {x e + d} c \sqrt {\frac {c d - b e}{c}} + {\left (c x - b\right )} e}{c x + b}\right ) + 2 \, {\left (23 \, B b c^{2} d^{2} + {\left (3 \, B b c^{2} x^{2} + 15 \, B b^{3} - 15 \, A b^{2} c - 5 \, {\left (B b^{2} c - A b c^{2}\right )} x\right )} e^{2} + {\left (11 \, B b c^{2} d x - 35 \, {\left (B b^{2} c - A b c^{2}\right )} d\right )} e\right )} \sqrt {x e + d}}{15 \, b c^{3}}, \frac {2 \, {\left (15 \, A c^{3} \sqrt {-d} d^{2} \arctan \left (\frac {\sqrt {x e + d} \sqrt {-d}}{d}\right ) - 15 \, {\left ({\left (B b c^{2} - A c^{3}\right )} d^{2} - 2 \, {\left (B b^{2} c - A b c^{2}\right )} d e + {\left (B b^{3} - A b^{2} c\right )} e^{2}\right )} \sqrt {-\frac {c d - b e}{c}} \arctan \left (-\frac {\sqrt {x e + d} c \sqrt {-\frac {c d - b e}{c}}}{c d - b e}\right ) + {\left (23 \, B b c^{2} d^{2} + {\left (3 \, B b c^{2} x^{2} + 15 \, B b^{3} - 15 \, A b^{2} c - 5 \, {\left (B b^{2} c - A b c^{2}\right )} x\right )} e^{2} + {\left (11 \, B b c^{2} d x - 35 \, {\left (B b^{2} c - A b c^{2}\right )} d\right )} e\right )} \sqrt {x e + d}\right )}}{15 \, b c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*A*c^3*d^(5/2)*log((x*e - 2*sqrt(x*e + d)*sqrt(d) + 2*d)/x) + 15*((B*b*c^2 - A*c^3)*d^2 - 2*(B*b^2*c
- A*b*c^2)*d*e + (B*b^3 - A*b^2*c)*e^2)*sqrt((c*d - b*e)/c)*log((2*c*d - 2*sqrt(x*e + d)*c*sqrt((c*d - b*e)/c)
+ (c*x - b)*e)/(c*x + b)) + 2*(23*B*b*c^2*d^2 + (3*B*b*c^2*x^2 + 15*B*b^3 - 15*A*b^2*c - 5*(B*b^2*c - A*b*c^2
)*x)*e^2 + (11*B*b*c^2*d*x - 35*(B*b^2*c - A*b*c^2)*d)*e)*sqrt(x*e + d))/(b*c^3), 1/15*(15*A*c^3*d^(5/2)*log((
x*e - 2*sqrt(x*e + d)*sqrt(d) + 2*d)/x) - 30*((B*b*c^2 - A*c^3)*d^2 - 2*(B*b^2*c - A*b*c^2)*d*e + (B*b^3 - A*b
^2*c)*e^2)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(x*e + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + 2*(23*B*b*c^2*d^2
+ (3*B*b*c^2*x^2 + 15*B*b^3 - 15*A*b^2*c - 5*(B*b^2*c - A*b*c^2)*x)*e^2 + (11*B*b*c^2*d*x - 35*(B*b^2*c - A*b*
c^2)*d)*e)*sqrt(x*e + d))/(b*c^3), 1/15*(30*A*c^3*sqrt(-d)*d^2*arctan(sqrt(x*e + d)*sqrt(-d)/d) + 15*((B*b*c^2
- A*c^3)*d^2 - 2*(B*b^2*c - A*b*c^2)*d*e + (B*b^3 - A*b^2*c)*e^2)*sqrt((c*d - b*e)/c)*log((2*c*d - 2*sqrt(x*e
+ d)*c*sqrt((c*d - b*e)/c) + (c*x - b)*e)/(c*x + b)) + 2*(23*B*b*c^2*d^2 + (3*B*b*c^2*x^2 + 15*B*b^3 - 15*A*b
^2*c - 5*(B*b^2*c - A*b*c^2)*x)*e^2 + (11*B*b*c^2*d*x - 35*(B*b^2*c - A*b*c^2)*d)*e)*sqrt(x*e + d))/(b*c^3), 2
/15*(15*A*c^3*sqrt(-d)*d^2*arctan(sqrt(x*e + d)*sqrt(-d)/d) - 15*((B*b*c^2 - A*c^3)*d^2 - 2*(B*b^2*c - A*b*c^2
)*d*e + (B*b^3 - A*b^2*c)*e^2)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(x*e + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e))
+ (23*B*b*c^2*d^2 + (3*B*b*c^2*x^2 + 15*B*b^3 - 15*A*b^2*c - 5*(B*b^2*c - A*b*c^2)*x)*e^2 + (11*B*b*c^2*d*x -
35*(B*b^2*c - A*b*c^2)*d)*e)*sqrt(x*e + d))/(b*c^3)]

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Sympy [A]
time = 54.56, size = 199, normalized size = 1.15 \begin {gather*} \frac {2 A d^{3} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}} + \frac {2 B \left (d + e x\right )^{\frac {5}{2}}}{5 c} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (2 A c e - 2 B b e + 2 B c d\right )}{3 c^{2}} + \frac {\sqrt {d + e x} \left (- 2 A b c e^{2} + 4 A c^{2} d e + 2 B b^{2} e^{2} - 4 B b c d e + 2 B c^{2} d^{2}\right )}{c^{3}} - \frac {2 \left (- A c + B b\right ) \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^{4} \sqrt {\frac {b e - c d}{c}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*d**3*atan(sqrt(d + e*x)/sqrt(-d))/(b*sqrt(-d)) + 2*B*(d + e*x)**(5/2)/(5*c) + (d + e*x)**(3/2)*(2*A*c*e -
2*B*b*e + 2*B*c*d)/(3*c**2) + sqrt(d + e*x)*(-2*A*b*c*e**2 + 4*A*c**2*d*e + 2*B*b**2*e**2 - 4*B*b*c*d*e + 2*B*
c**2*d**2)/c**3 - 2*(-A*c + B*b)*(b*e - c*d)**3*atan(sqrt(d + e*x)/sqrt((b*e - c*d)/c))/(b*c**4*sqrt((b*e - c*
d)/c))

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Giac [A]
time = 1.41, size = 316, normalized size = 1.83 \begin {gather*} \frac {2 \, A d^{3} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} + \frac {2 \, {\left (B b c^{3} d^{3} - A c^{4} d^{3} - 3 \, B b^{2} c^{2} d^{2} e + 3 \, A b c^{3} d^{2} e + 3 \, B b^{3} c d e^{2} - 3 \, A b^{2} c^{2} d e^{2} - B b^{4} e^{3} + A b^{3} c 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^{3}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} B c^{4} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} B c^{4} d + 15 \, \sqrt {x e + d} B c^{4} d^{2} - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} B b c^{3} e + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A c^{4} e - 30 \, \sqrt {x e + d} B b c^{3} d e + 30 \, \sqrt {x e + d} A c^{4} d e + 15 \, \sqrt {x e + d} B b^{2} c^{2} e^{2} - 15 \, \sqrt {x e + d} A b c^{3} e^{2}\right )}}{15 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*d^3*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)) + 2*(B*b*c^3*d^3 - A*c^4*d^3 - 3*B*b^2*c^2*d^2*e + 3*A*b*c
^3*d^2*e + 3*B*b^3*c*d*e^2 - 3*A*b^2*c^2*d*e^2 - B*b^4*e^3 + A*b^3*c*e^3)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d +
b*c*e))/(sqrt(-c^2*d + b*c*e)*b*c^3) + 2/15*(3*(x*e + d)^(5/2)*B*c^4 + 5*(x*e + d)^(3/2)*B*c^4*d + 15*sqrt(x*
e + d)*B*c^4*d^2 - 5*(x*e + d)^(3/2)*B*b*c^3*e + 5*(x*e + d)^(3/2)*A*c^4*e - 30*sqrt(x*e + d)*B*b*c^3*d*e + 30
*sqrt(x*e + d)*A*c^4*d*e + 15*sqrt(x*e + d)*B*b^2*c^2*e^2 - 15*sqrt(x*e + d)*A*b*c^3*e^2)/c^5

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Mupad [B]
time = 2.19, size = 2500, normalized size = 14.45 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*A*e - 2*B*d)/(3*c) - (2*B*(b*e - 2*c*d))/(3*c^2))*(d + e*x)^(3/2) - (((b*e - 2*c*d)*((2*A*e - 2*B*d)/c - (
2*B*(b*e - 2*c*d))/c^2))/c + (2*B*(c*d^2 - b*d*e))/c^2)*(d + e*x)^(1/2) + (2*B*(d + e*x)^(5/2))/(5*c) - (A*ata
n(((A*((8*(d + e*x)^(1/2)*(B^2*b^8*e^8 + A^2*b^6*c^2*e^8 + 2*A^2*c^8*d^6*e^2 + 15*A^2*b^2*c^6*d^4*e^4 - 20*A^2
*b^3*c^5*d^3*e^5 + 15*A^2*b^4*c^4*d^2*e^6 + B^2*b^2*c^6*d^6*e^2 - 6*B^2*b^3*c^5*d^5*e^3 + 15*B^2*b^4*c^4*d^4*e
^4 - 20*B^2*b^5*c^3*d^3*e^5 + 15*B^2*b^6*c^2*d^2*e^6 - 6*B^2*b^7*c*d*e^7 - 6*A^2*b*c^7*d^5*e^3 - 6*A^2*b^5*c^3
*d*e^7 - 2*A*B*b^7*c*e^8 - 2*A*B*b*c^7*d^6*e^2 + 12*A*B*b^6*c^2*d*e^7 + 12*A*B*b^2*c^6*d^5*e^3 - 30*A*B*b^3*c^
5*d^4*e^4 + 40*A*B*b^4*c^4*d^3*e^5 - 30*A*B*b^5*c^3*d^2*e^6))/c^5 + (A*((8*(A*b^4*c^5*d*e^5 - B*b^5*c^4*d*e^5
+ 2*A*b^2*c^7*d^3*e^3 - 3*A*b^3*c^6*d^2*e^4 + B*b^2*c^7*d^4*e^2 - 3*B*b^3*c^6*d^3*e^3 + 3*B*b^4*c^5*d^2*e^4))/
c^5 + (8*A*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(d^5)^(1/2)*(d + e*x)^(1/2))/(b*c^5))*(d^5)^(1/2))/b)*(d^5)^(1/2)*1
i)/b + (A*((8*(d + e*x)^(1/2)*(B^2*b^8*e^8 + A^2*b^6*c^2*e^8 + 2*A^2*c^8*d^6*e^2 + 15*A^2*b^2*c^6*d^4*e^4 - 20
*A^2*b^3*c^5*d^3*e^5 + 15*A^2*b^4*c^4*d^2*e^6 + B^2*b^2*c^6*d^6*e^2 - 6*B^2*b^3*c^5*d^5*e^3 + 15*B^2*b^4*c^4*d
^4*e^4 - 20*B^2*b^5*c^3*d^3*e^5 + 15*B^2*b^6*c^2*d^2*e^6 - 6*B^2*b^7*c*d*e^7 - 6*A^2*b*c^7*d^5*e^3 - 6*A^2*b^5
*c^3*d*e^7 - 2*A*B*b^7*c*e^8 - 2*A*B*b*c^7*d^6*e^2 + 12*A*B*b^6*c^2*d*e^7 + 12*A*B*b^2*c^6*d^5*e^3 - 30*A*B*b^
3*c^5*d^4*e^4 + 40*A*B*b^4*c^4*d^3*e^5 - 30*A*B*b^5*c^3*d^2*e^6))/c^5 - (A*((8*(A*b^4*c^5*d*e^5 - B*b^5*c^4*d*
e^5 + 2*A*b^2*c^7*d^3*e^3 - 3*A*b^3*c^6*d^2*e^4 + B*b^2*c^7*d^4*e^2 - 3*B*b^3*c^6*d^3*e^3 + 3*B*b^4*c^5*d^2*e^
4))/c^5 - (8*A*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(d^5)^(1/2)*(d + e*x)^(1/2))/(b*c^5))*(d^5)^(1/2))/b)*(d^5)^(1/
2)*1i)/b)/((16*(3*A^3*c^7*d^8*e^3 + 19*A^3*b^2*c^5*d^6*e^5 - 15*A^3*b^3*c^4*d^5*e^6 + 6*A^3*b^4*c^3*d^4*e^7 -
A^3*b^5*c^2*d^3*e^8 - A*B^2*b^7*d^3*e^8 + A^2*B*c^7*d^9*e^2 - 12*A^3*b*c^6*d^7*e^4 + 6*A*B^2*b^2*c^5*d^8*e^3 -
15*A*B^2*b^3*c^4*d^7*e^4 + 20*A*B^2*b^4*c^3*d^6*e^5 - 15*A*B^2*b^5*c^2*d^5*e^6 + 27*A^2*B*b^2*c^5*d^7*e^4 - 3
9*A^2*B*b^3*c^4*d^6*e^5 + 30*A^2*B*b^4*c^3*d^5*e^6 - 12*A^2*B*b^5*c^2*d^4*e^7 - A*B^2*b*c^6*d^9*e^2 + 6*A*B^2*
b^6*c*d^4*e^7 - 9*A^2*B*b*c^6*d^8*e^3 + 2*A^2*B*b^6*c*d^3*e^8))/c^5 + (A*((8*(d + e*x)^(1/2)*(B^2*b^8*e^8 + A^
2*b^6*c^2*e^8 + 2*A^2*c^8*d^6*e^2 + 15*A^2*b^2*c^6*d^4*e^4 - 20*A^2*b^3*c^5*d^3*e^5 + 15*A^2*b^4*c^4*d^2*e^6 +
B^2*b^2*c^6*d^6*e^2 - 6*B^2*b^3*c^5*d^5*e^3 + 15*B^2*b^4*c^4*d^4*e^4 - 20*B^2*b^5*c^3*d^3*e^5 + 15*B^2*b^6*c^
2*d^2*e^6 - 6*B^2*b^7*c*d*e^7 - 6*A^2*b*c^7*d^5*e^3 - 6*A^2*b^5*c^3*d*e^7 - 2*A*B*b^7*c*e^8 - 2*A*B*b*c^7*d^6*
e^2 + 12*A*B*b^6*c^2*d*e^7 + 12*A*B*b^2*c^6*d^5*e^3 - 30*A*B*b^3*c^5*d^4*e^4 + 40*A*B*b^4*c^4*d^3*e^5 - 30*A*B
*b^5*c^3*d^2*e^6))/c^5 + (A*((8*(A*b^4*c^5*d*e^5 - B*b^5*c^4*d*e^5 + 2*A*b^2*c^7*d^3*e^3 - 3*A*b^3*c^6*d^2*e^4
+ B*b^2*c^7*d^4*e^2 - 3*B*b^3*c^6*d^3*e^3 + 3*B*b^4*c^5*d^2*e^4))/c^5 + (8*A*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*
(d^5)^(1/2)*(d + e*x)^(1/2))/(b*c^5))*(d^5)^(1/2))/b)*(d^5)^(1/2))/b - (A*((8*(d + e*x)^(1/2)*(B^2*b^8*e^8 + A
^2*b^6*c^2*e^8 + 2*A^2*c^8*d^6*e^2 + 15*A^2*b^2*c^6*d^4*e^4 - 20*A^2*b^3*c^5*d^3*e^5 + 15*A^2*b^4*c^4*d^2*e^6
+ B^2*b^2*c^6*d^6*e^2 - 6*B^2*b^3*c^5*d^5*e^3 + 15*B^2*b^4*c^4*d^4*e^4 - 20*B^2*b^5*c^3*d^3*e^5 + 15*B^2*b^6*c
^2*d^2*e^6 - 6*B^2*b^7*c*d*e^7 - 6*A^2*b*c^7*d^5*e^3 - 6*A^2*b^5*c^3*d*e^7 - 2*A*B*b^7*c*e^8 - 2*A*B*b*c^7*d^6
*e^2 + 12*A*B*b^6*c^2*d*e^7 + 12*A*B*b^2*c^6*d^5*e^3 - 30*A*B*b^3*c^5*d^4*e^4 + 40*A*B*b^4*c^4*d^3*e^5 - 30*A*
B*b^5*c^3*d^2*e^6))/c^5 - (A*((8*(A*b^4*c^5*d*e^5 - B*b^5*c^4*d*e^5 + 2*A*b^2*c^7*d^3*e^3 - 3*A*b^3*c^6*d^2*e^
4 + B*b^2*c^7*d^4*e^2 - 3*B*b^3*c^6*d^3*e^3 + 3*B*b^4*c^5*d^2*e^4))/c^5 - (8*A*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)
*(d^5)^(1/2)*(d + e*x)^(1/2))/(b*c^5))*(d^5)^(1/2))/b)*(d^5)^(1/2))/b))*(d^5)^(1/2)*2i)/b - (atan((((-c^7*(b*e
- c*d)^5)^(1/2)*(A*c - B*b)*((8*(d + e*x)^(1/2)*(B^2*b^8*e^8 + A^2*b^6*c^2*e^8 + 2*A^2*c^8*d^6*e^2 + 15*A^2*b
^2*c^6*d^4*e^4 - 20*A^2*b^3*c^5*d^3*e^5 + 15*A^2*b^4*c^4*d^2*e^6 + B^2*b^2*c^6*d^6*e^2 - 6*B^2*b^3*c^5*d^5*e^3
+ 15*B^2*b^4*c^4*d^4*e^4 - 20*B^2*b^5*c^3*d^3*e^5 + 15*B^2*b^6*c^2*d^2*e^6 - 6*B^2*b^7*c*d*e^7 - 6*A^2*b*c^7*
d^5*e^3 - 6*A^2*b^5*c^3*d*e^7 - 2*A*B*b^7*c*e^8 - 2*A*B*b*c^7*d^6*e^2 + 12*A*B*b^6*c^2*d*e^7 + 12*A*B*b^2*c^6*
d^5*e^3 - 30*A*B*b^3*c^5*d^4*e^4 + 40*A*B*b^4*c^4*d^3*e^5 - 30*A*B*b^5*c^3*d^2*e^6))/c^5 + (((8*(A*b^4*c^5*d*e
^5 - B*b^5*c^4*d*e^5 + 2*A*b^2*c^7*d^3*e^3 - 3*A*b^3*c^6*d^2*e^4 + B*b^2*c^7*d^4*e^2 - 3*B*b^3*c^6*d^3*e^3 + 3
*B*b^4*c^5*d^2*e^4))/c^5 + (8*(b^3*c^7*e^3 - 2*b^2*c^8*d*e^2)*(-c^7*(b*e - c*d)^5)^(1/2)*(A*c - B*b)*(d + e*x)
^(1/2))/(b*c^12))*(-c^7*(b*e - c*d)^5)^(1/2)*(A*c - B*b))/(b*c^7))*1i)/(b*c^7) + ((-c^7*(b*e - c*d)^5)^(1/2)*(
A*c - B*b)*((8*(d + e*x)^(1/2)*(B^2*b^8*e^8 + A^2*b^6*c^2*e^8 + 2*A^2*c^8*d^6*e^2 + 15*A^2*b^2*c^6*d^4*e^4 - 2
0*A^2*b^3*c^5*d^3*e^5 + 15*A^2*b^4*c^4*d^2*e^6 + B^2*b^2*c^6*d^6*e^2 - 6*B^2*b^3*c^5*d^5*e^3 + 15*B^2*b^4*c^4*
d^4*e^4 - 20*B^2*b^5*c^3*d^3*e^5 + 15*B^2*b^6*c...

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