3.4.30 \(\int \frac {(d+e x)^3}{(b x+c x^2)^{5/2}} \, dx\) [330]

Optimal. Leaf size=87 \[ -\frac {2 (d+e x)^2 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {16 d (c d-b e) (b d+(2 c d-b e) x)}{3 b^4 \sqrt {b x+c x^2}} \]

[Out]

-2/3*(e*x+d)^2*(b*d+(-b*e+2*c*d)*x)/b^2/(c*x^2+b*x)^(3/2)+16/3*d*(-b*e+c*d)*(b*d+(-b*e+2*c*d)*x)/b^4/(c*x^2+b*
x)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {736, 650} \begin {gather*} \frac {16 d (c d-b e) (x (2 c d-b e)+b d)}{3 b^4 \sqrt {b x+c x^2}}-\frac {2 (d+e x)^2 (x (2 c d-b e)+b d)}{3 b^2 \left (b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(b*x + c*x^2)^(5/2),x]

[Out]

(-2*(d + e*x)^2*(b*d + (2*c*d - b*e)*x))/(3*b^2*(b*x + c*x^2)^(3/2)) + (16*d*(c*d - b*e)*(b*d + (2*c*d - b*e)*
x))/(3*b^4*Sqrt[b*x + c*x^2])

Rule 650

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-2*((b*d - 2*a*e + (2*c*
d - b*e)*x)/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
NeQ[b^2 - 4*a*c, 0]

Rule 736

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[2*(2*p + 3)*((c*d
^2 - b*d*e + a*e^2)/((p + 1)*(b^2 - 4*a*c))), Int[(d + e*x)^(m - 2)*(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] && EqQ[m +
2*p + 2, 0] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\left (b x+c x^2\right )^{5/2}} \, dx &=-\frac {2 (d+e x)^2 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}-\frac {(8 d (c d-b e)) \int \frac {d+e x}{\left (b x+c x^2\right )^{3/2}} \, dx}{3 b^2}\\ &=-\frac {2 (d+e x)^2 (b d+(2 c d-b e) x)}{3 b^2 \left (b x+c x^2\right )^{3/2}}+\frac {16 d (c d-b e) (b d+(2 c d-b e) x)}{3 b^4 \sqrt {b x+c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 105, normalized size = 1.21 \begin {gather*} \frac {2 \left (16 c^3 d^3 x^3+24 b c^2 d^2 x^2 (d-e x)+6 b^2 c d x \left (d^2-6 d e x+e^2 x^2\right )+b^3 \left (-d^3-9 d^2 e x+9 d e^2 x^2+e^3 x^3\right )\right )}{3 b^4 (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/(b*x + c*x^2)^(5/2),x]

[Out]

(2*(16*c^3*d^3*x^3 + 24*b*c^2*d^2*x^2*(d - e*x) + 6*b^2*c*d*x*(d^2 - 6*d*e*x + e^2*x^2) + b^3*(-d^3 - 9*d^2*e*
x + 9*d*e^2*x^2 + e^3*x^3)))/(3*b^4*(x*(b + c*x))^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(348\) vs. \(2(79)=158\).
time = 0.47, size = 349, normalized size = 4.01

method	result	size
risch	\(-\frac {2 d^{2} \left (c x +b \right ) \left (9 b e x -8 c d x +b d \right )}{3 b^{4} x \sqrt {x \left (c x +b \right )}}+\frac {2 x \left (b e x +8 c d x +9 b d \right ) \left (b^{2} e^{2}-2 b c d e +d^{2} c^{2}\right )}{3 \sqrt {x \left (c x +b \right )}\, \left (c x +b \right ) b^{4}}\)	\(98\)
gosper	\(-\frac {2 x \left (c x +b \right ) \left (-b^{3} e^{3} x^{3}-6 b^{2} c d \,e^{2} x^{3}+24 b \,c^{2} d^{2} e \,x^{3}-16 c^{3} d^{3} x^{3}-9 b^{3} d \,e^{2} x^{2}+36 b^{2} c \,d^{2} e \,x^{2}-24 b \,c^{2} d^{3} x^{2}+9 b^{3} d^{2} e x -6 b^{2} c \,d^{3} x +b^{3} d^{3}\right )}{3 b^{4} \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}\)	\(136\)
trager	\(-\frac {2 \left (-b^{3} e^{3} x^{3}-6 b^{2} c d \,e^{2} x^{3}+24 b \,c^{2} d^{2} e \,x^{3}-16 c^{3} d^{3} x^{3}-9 b^{3} d \,e^{2} x^{2}+36 b^{2} c \,d^{2} e \,x^{2}-24 b \,c^{2} d^{3} x^{2}+9 b^{3} d^{2} e x -6 b^{2} c \,d^{3} x +b^{3} d^{3}\right ) \sqrt {c \,x^{2}+b x}}{3 b^{4} x^{2} \left (c x +b \right )^{2}}\)	\(140\)
default	\(e^{3} \left (-\frac {x^{2}}{c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {b \left (-\frac {x}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )}{4 c}\right )}{2 c}\right )+3 d \,e^{2} \left (-\frac {x}{2 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )}{4 c}\right )+3 d^{2} e \left (-\frac {1}{3 c \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}-\frac {b \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )}{2 c}\right )+d^{3} \left (-\frac {2 \left (2 c x +b \right )}{3 b^{2} \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}+\frac {16 c \left (2 c x +b \right )}{3 b^{4} \sqrt {c \,x^{2}+b x}}\right )\)	\(349\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

e^3*(-x^2/c/(c*x^2+b*x)^(3/2)+1/2*b/c*(-1/2*x/c/(c*x^2+b*x)^(3/2)-1/4*b/c*(-1/3/c/(c*x^2+b*x)^(3/2)-1/2*b/c*(-
2/3*(2*c*x+b)/b^2/(c*x^2+b*x)^(3/2)+16/3*c*(2*c*x+b)/b^4/(c*x^2+b*x)^(1/2)))))+3*d*e^2*(-1/2*x/c/(c*x^2+b*x)^(
3/2)-1/4*b/c*(-1/3/c/(c*x^2+b*x)^(3/2)-1/2*b/c*(-2/3*(2*c*x+b)/b^2/(c*x^2+b*x)^(3/2)+16/3*c*(2*c*x+b)/b^4/(c*x
^2+b*x)^(1/2))))+3*d^2*e*(-1/3/c/(c*x^2+b*x)^(3/2)-1/2*b/c*(-2/3*(2*c*x+b)/b^2/(c*x^2+b*x)^(3/2)+16/3*c*(2*c*x
+b)/b^4/(c*x^2+b*x)^(1/2)))+d^3*(-2/3*(2*c*x+b)/b^2/(c*x^2+b*x)^(3/2)+16/3*c*(2*c*x+b)/b^4/(c*x^2+b*x)^(1/2))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (83) = 166\).
time = 0.27, size = 293, normalized size = 3.37 \begin {gather*} -\frac {4 \, c d^{3} x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} + \frac {32 \, c^{2} d^{3} x}{3 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {2 \, d^{2} x e}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} - \frac {16 \, c d^{2} x e}{\sqrt {c x^{2} + b x} b^{3}} - \frac {2 \, d^{3}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} + \frac {16 \, c d^{3}}{3 \, \sqrt {c x^{2} + b x} b^{3}} - \frac {x^{2} e^{3}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} + \frac {4 \, d x e^{2}}{\sqrt {c x^{2} + b x} b^{2}} - \frac {2 \, d x e^{2}}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} c} - \frac {8 \, d^{2} e}{\sqrt {c x^{2} + b x} b^{2}} - \frac {b x e^{3}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} c^{2}} + \frac {2 \, x e^{3}}{3 \, \sqrt {c x^{2} + b x} b c} + \frac {2 \, d e^{2}}{\sqrt {c x^{2} + b x} b c} + \frac {e^{3}}{3 \, \sqrt {c x^{2} + b x} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

-4/3*c*d^3*x/((c*x^2 + b*x)^(3/2)*b^2) + 32/3*c^2*d^3*x/(sqrt(c*x^2 + b*x)*b^4) + 2*d^2*x*e/((c*x^2 + b*x)^(3/
2)*b) - 16*c*d^2*x*e/(sqrt(c*x^2 + b*x)*b^3) - 2/3*d^3/((c*x^2 + b*x)^(3/2)*b) + 16/3*c*d^3/(sqrt(c*x^2 + b*x)
*b^3) - x^2*e^3/((c*x^2 + b*x)^(3/2)*c) + 4*d*x*e^2/(sqrt(c*x^2 + b*x)*b^2) - 2*d*x*e^2/((c*x^2 + b*x)^(3/2)*c
) - 8*d^2*e/(sqrt(c*x^2 + b*x)*b^2) - 1/3*b*x*e^3/((c*x^2 + b*x)^(3/2)*c^2) + 2/3*x*e^3/(sqrt(c*x^2 + b*x)*b*c
) + 2*d*e^2/(sqrt(c*x^2 + b*x)*b*c) + 1/3*e^3/(sqrt(c*x^2 + b*x)*c^2)

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Fricas [A]
time = 1.44, size = 155, normalized size = 1.78 \begin {gather*} \frac {2 \, {\left (16 \, c^{3} d^{3} x^{3} + 24 \, b c^{2} d^{3} x^{2} + 6 \, b^{2} c d^{3} x + b^{3} x^{3} e^{3} - b^{3} d^{3} + 3 \, {\left (2 \, b^{2} c d x^{3} + 3 \, b^{3} d x^{2}\right )} e^{2} - 3 \, {\left (8 \, b c^{2} d^{2} x^{3} + 12 \, b^{2} c d^{2} x^{2} + 3 \, b^{3} d^{2} x\right )} e\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

2/3*(16*c^3*d^3*x^3 + 24*b*c^2*d^3*x^2 + 6*b^2*c*d^3*x + b^3*x^3*e^3 - b^3*d^3 + 3*(2*b^2*c*d*x^3 + 3*b^3*d*x^
2)*e^2 - 3*(8*b*c^2*d^2*x^3 + 12*b^2*c*d^2*x^2 + 3*b^3*d^2*x)*e)*sqrt(c*x^2 + b*x)/(b^4*c^2*x^4 + 2*b^5*c*x^3
+ b^6*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{3}}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*x**2+b*x)**(5/2),x)

[Out]

Integral((d + e*x)**3/(x*(b + c*x))**(5/2), x)

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Giac [A]
time = 1.26, size = 127, normalized size = 1.46 \begin {gather*} -\frac {2 \, {\left (\frac {d^{3}}{b} - {\left (x {\left (\frac {{\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 6 \, b^{2} c d e^{2} + b^{3} e^{3}\right )} x}{b^{4}} + \frac {3 \, {\left (8 \, b c^{2} d^{3} - 12 \, b^{2} c d^{2} e + 3 \, b^{3} d e^{2}\right )}}{b^{4}}\right )} + \frac {3 \, {\left (2 \, b^{2} c d^{3} - 3 \, b^{3} d^{2} e\right )}}{b^{4}}\right )} x\right )}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

-2/3*(d^3/b - (x*((16*c^3*d^3 - 24*b*c^2*d^2*e + 6*b^2*c*d*e^2 + b^3*e^3)*x/b^4 + 3*(8*b*c^2*d^3 - 12*b^2*c*d^
2*e + 3*b^3*d*e^2)/b^4) + 3*(2*b^2*c*d^3 - 3*b^3*d^2*e)/b^4)*x)/(c*x^2 + b*x)^(3/2)

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Mupad [B]
time = 0.48, size = 129, normalized size = 1.48 \begin {gather*} \frac {2\,\left (-b^3\,d^3-9\,b^3\,d^2\,e\,x+9\,b^3\,d\,e^2\,x^2+b^3\,e^3\,x^3+6\,b^2\,c\,d^3\,x-36\,b^2\,c\,d^2\,e\,x^2+6\,b^2\,c\,d\,e^2\,x^3+24\,b\,c^2\,d^3\,x^2-24\,b\,c^2\,d^2\,e\,x^3+16\,c^3\,d^3\,x^3\right )}{3\,b^4\,{\left (c\,x^2+b\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(b*x + c*x^2)^(5/2),x)

[Out]

(2*(b^3*e^3*x^3 - b^3*d^3 + 16*c^3*d^3*x^3 + 24*b*c^2*d^3*x^2 + 9*b^3*d*e^2*x^2 + 6*b^2*c*d^3*x - 9*b^3*d^2*e*
x - 36*b^2*c*d^2*e*x^2 - 24*b*c^2*d^2*e*x^3 + 6*b^2*c*d*e^2*x^3))/(3*b^4*(b*x + c*x^2)^(3/2))

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