3.4.22 \(\int \frac {b x+c x^2}{(d+e x)^{7/2}} \, dx\)

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

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Rubi [A]  time = 0.03, 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 = {698} \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 698

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 = 0.03, 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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IntegrateAlgebraic [A]  time = 0.05, size = 56, normalized size = 0.85 \begin {gather*} -\frac {2 \left (5 b e (d+e x)-3 b d e+3 c d^2-10 c d (d+e x)+15 c (d+e x)^2\right )}{15 e^3 (d+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 79, normalized size = 1.20 \begin {gather*} -\frac {2 \, {\left (15 \, c e^{2} x^{2} + 8 \, c d^{2} + 2 \, b d e + 5 \, {\left (4 \, c d e + b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{6} x^{3} + 3 \, d e^{5} x^{2} + 3 \, d^{2} e^{4} x + 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*(15*c*e^2*x^2 + 8*c*d^2 + 2*b*d*e + 5*(4*c*d*e + b*e^2)*x)*sqrt(e*x + d)/(e^6*x^3 + 3*d*e^5*x^2 + 3*d^2*
e^4*x + d^3*e^3)

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giac [A]  time = 0.17, 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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maple [A]  time = 0.06, size = 47, normalized size = 0.71 \begin {gather*} -\frac {2 \left (15 c \,e^{2} x^{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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.27, size = 50, normalized size = 0.76 \begin {gather*} -\frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} c + 3 \, c d^{2} - 3 \, b d e - 5 \, {\left (2 \, c d - b e\right )} {\left (e x + d\right )}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{3}} \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*(e*x + d)^2*c + 3*c*d^2 - 3*b*d*e - 5*(2*c*d - b*e)*(e*x + d))/((e*x + d)^(5/2)*e^3)

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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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sympy [A]  time = 2.99, 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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