3.4.44 \(\int \frac {b x+c x^2}{(d+e x)^{7/2}} \, dx\) [344]

Optimal. Leaf size=66 \[ -\frac {2 d (c d-b e)}{5 e^3 (d+e x)^{5/2}}+\frac {2 (2 c d-b e)}{3 e^3 (d+e x)^{3/2}}-\frac {2 c}{e^3 \sqrt {d+e x}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {712} \begin {gather*} \frac {2 (2 c d-b e)}{3 e^3 (d+e x)^{3/2}}-\frac {2 d (c d-b e)}{5 e^3 (d+e x)^{5/2}}-\frac {2 c}{e^3 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)/(d + e*x)^(7/2),x]

[Out]

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

Rule 712

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
 e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 49, normalized size = 0.74 \begin {gather*} -\frac {2 \left (b e (2 d+5 e x)+c \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )}{15 e^3 (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)/(d + e*x)^(7/2),x]

[Out]

(-2*(b*e*(2*d + 5*e*x) + c*(8*d^2 + 20*d*e*x + 15*e^2*x^2)))/(15*e^3*(d + e*x)^(5/2))

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Maple [A]
time = 0.41, size = 52, normalized size = 0.79

method result size
gosper \(-\frac {2 \left (15 c \,x^{2} e^{2}+5 b \,e^{2} x +20 c d e x +2 b d e +8 c \,d^{2}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\) \(47\)
trager \(-\frac {2 \left (15 c \,x^{2} e^{2}+5 b \,e^{2} x +20 c d e x +2 b d e +8 c \,d^{2}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{3}}\) \(47\)
derivativedivides \(\frac {-\frac {2 c}{\sqrt {e x +d}}-\frac {2 \left (b e -2 c d \right )}{3 \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 d \left (b e -c d \right )}{5 \left (e x +d \right )^{\frac {5}{2}}}}{e^{3}}\) \(52\)
default \(\frac {-\frac {2 c}{\sqrt {e x +d}}-\frac {2 \left (b e -2 c d \right )}{3 \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 d \left (b e -c d \right )}{5 \left (e x +d \right )^{\frac {5}{2}}}}{e^{3}}\) \(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)/(e*x+d)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 54, normalized size = 0.82 \begin {gather*} -\frac {2 \, {\left (15 \, {\left (x e + d\right )}^{2} c + 3 \, c d^{2} - 3 \, b d e - 5 \, {\left (2 \, c d - b e\right )} {\left (x e + d\right )}\right )} e^{\left (-3\right )}}{15 \, {\left (x e + d\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

-2/15*(15*(x*e + d)^2*c + 3*c*d^2 - 3*b*d*e - 5*(2*c*d - b*e)*(x*e + d))*e^(-3)/(x*e + d)^(5/2)

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Fricas [A]
time = 3.00, size = 75, normalized size = 1.14 \begin {gather*} -\frac {2 \, {\left (8 \, c d^{2} + 5 \, {\left (3 \, c x^{2} + b x\right )} e^{2} + 2 \, {\left (10 \, c d x + b d\right )} e\right )} \sqrt {x e + d}}{15 \, {\left (x^{3} e^{6} + 3 \, d x^{2} e^{5} + 3 \, d^{2} x e^{4} + d^{3} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

-2/15*(8*c*d^2 + 5*(3*c*x^2 + b*x)*e^2 + 2*(10*c*d*x + b*d)*e)*sqrt(x*e + d)/(x^3*e^6 + 3*d*x^2*e^5 + 3*d^2*x*
e^4 + d^3*e^3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (63) = 126\).
time = 0.61, size = 314, normalized size = 4.76 \begin {gather*} \begin {cases} - \frac {4 b d e}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {10 b e^{2} x}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {16 c d^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {40 c d e x}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} - \frac {30 c e^{2} x^{2}}{15 d^{2} e^{3} \sqrt {d + e x} + 30 d e^{4} x \sqrt {d + e x} + 15 e^{5} x^{2} \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {\frac {b x^{2}}{2} + \frac {c x^{3}}{3}}{d^{\frac {7}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)/(e*x+d)**(7/2),x)

[Out]

Piecewise((-4*b*d*e/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 10
*b*e**2*x/(15*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 16*c*d**2/(1
5*d**2*e**3*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 40*c*d*e*x/(15*d**2*e**3
*sqrt(d + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)) - 30*c*e**2*x**2/(15*d**2*e**3*sqrt(d
 + e*x) + 30*d*e**4*x*sqrt(d + e*x) + 15*e**5*x**2*sqrt(d + e*x)), Ne(e, 0)), ((b*x**2/2 + c*x**3/3)/d**(7/2),
 True))

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Giac [A]
time = 1.31, size = 57, normalized size = 0.86 \begin {gather*} -\frac {2 \, {\left (15 \, {\left (x e + d\right )}^{2} c - 10 \, {\left (x e + d\right )} c d + 3 \, c d^{2} + 5 \, {\left (x e + d\right )} b e - 3 \, b d e\right )} e^{\left (-3\right )}}{15 \, {\left (x e + d\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

-2/15*(15*(x*e + d)^2*c - 10*(x*e + d)*c*d + 3*c*d^2 + 5*(x*e + d)*b*e - 3*b*d*e)*e^(-3)/(x*e + d)^(5/2)

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Mupad [B]
time = 0.07, size = 49, normalized size = 0.74 \begin {gather*} -\frac {\left (\frac {2\,b\,e}{3}-\frac {4\,c\,d}{3}\right )\,\left (d+e\,x\right )+2\,c\,{\left (d+e\,x\right )}^2+\frac {2\,c\,d^2}{5}-\frac {2\,b\,d\,e}{5}}{e^3\,{\left (d+e\,x\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)/(d + e*x)^(7/2),x)

[Out]

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

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