3.2034 \(\int \frac{(a+b x) (d+e x)^4}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=201 \[ -\frac{4 e^2 (b d-a e)^2}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{4 e^3 (a+b x) (b d-a e) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 e (b d-a e)^3}{3 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{4 e^4 x (a+b x)}{3 b^4 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

-(d + e*x)^4/(3*b*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - (4*e^2*(b*d - a*e)^2)/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
- (2*e*(b*d - a*e)^3)/(3*b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (4*e^4*x*(a + b*x))/(3*b^4*Sqrt[a^2 +
2*a*b*x + b^2*x^2]) + (4*e^3*(b*d - a*e)*(a + b*x)*Log[a + b*x])/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.144315, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {768, 646, 43} \[ -\frac{4 e^2 (b d-a e)^2}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{4 e^3 (a+b x) (b d-a e) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 e (b d-a e)^3}{3 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{4 e^4 x (a+b x)}{3 b^4 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(d + e*x)^4)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

-(d + e*x)^4/(3*b*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - (4*e^2*(b*d - a*e)^2)/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
- (2*e*(b*d - a*e)^3)/(3*b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (4*e^4*x*(a + b*x))/(3*b^4*Sqrt[a^2 +
2*a*b*x + b^2*x^2]) + (4*e^3*(b*d - a*e)*(a + b*x)*Log[a + b*x])/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=-\frac{(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{(4 e) \int \frac{(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx}{3 b}\\ &=-\frac{(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{\left (4 b e \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^3}{\left (a b+b^2 x\right )^3} \, dx}{3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{\left (4 b e \left (a b+b^2 x\right )\right ) \int \left (\frac{e^3}{b^6}+\frac{(b d-a e)^3}{b^6 (a+b x)^3}+\frac{3 e (b d-a e)^2}{b^6 (a+b x)^2}+\frac{3 e^2 (b d-a e)}{b^6 (a+b x)}\right ) \, dx}{3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{4 e^2 (b d-a e)^2}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 e (b d-a e)^3}{3 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{4 e^4 x (a+b x)}{3 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{4 e^3 (b d-a e) (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.102086, size = 170, normalized size = 0.85 \[ \frac{-3 a^2 b^2 e^2 \left (2 d^2-18 d e x+3 e^2 x^2\right )+a^3 b e^3 (22 d-27 e x)-13 a^4 e^4+a b^3 e \left (-18 d^2 e x-2 d^3+36 d e^2 x^2+9 e^3 x^3\right )-12 e^3 (a+b x)^3 (a e-b d) \log (a+b x)+b^4 \left (-\left (18 d^2 e^2 x^2+6 d^3 e x+d^4-3 e^4 x^4\right )\right )}{3 b^5 \left ((a+b x)^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(d + e*x)^4)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-13*a^4*e^4 + a^3*b*e^3*(22*d - 27*e*x) - 3*a^2*b^2*e^2*(2*d^2 - 18*d*e*x + 3*e^2*x^2) + a*b^3*e*(-2*d^3 - 18
*d^2*e*x + 36*d*e^2*x^2 + 9*e^3*x^3) - b^4*(d^4 + 6*d^3*e*x + 18*d^2*e^2*x^2 - 3*e^4*x^4) - 12*e^3*(-(b*d) + a
*e)*(a + b*x)^3*Log[a + b*x])/(3*b^5*((a + b*x)^2)^(3/2))

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Maple [B]  time = 0.013, size = 322, normalized size = 1.6 \begin{align*} -{\frac{ \left ( 12\,\ln \left ( bx+a \right ){x}^{3}a{b}^{3}{e}^{4}-12\,\ln \left ( bx+a \right ){x}^{3}{b}^{4}d{e}^{3}-3\,{x}^{4}{b}^{4}{e}^{4}+36\,\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}{e}^{4}-36\,\ln \left ( bx+a \right ){x}^{2}a{b}^{3}d{e}^{3}-9\,{x}^{3}a{b}^{3}{e}^{4}+36\,\ln \left ( bx+a \right ) x{a}^{3}b{e}^{4}-36\,\ln \left ( bx+a \right ) x{a}^{2}{b}^{2}d{e}^{3}+9\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}-36\,{x}^{2}a{b}^{3}d{e}^{3}+18\,{x}^{2}{b}^{4}{d}^{2}{e}^{2}+12\,\ln \left ( bx+a \right ){a}^{4}{e}^{4}-12\,\ln \left ( bx+a \right ){a}^{3}bd{e}^{3}+27\,x{a}^{3}b{e}^{4}-54\,x{a}^{2}{b}^{2}d{e}^{3}+18\,xa{b}^{3}{d}^{2}{e}^{2}+6\,x{b}^{4}{d}^{3}e+13\,{a}^{4}{e}^{4}-22\,d{e}^{3}{a}^{3}b+6\,{a}^{2}{b}^{2}{d}^{2}{e}^{2}+2\,a{b}^{3}{d}^{3}e+{b}^{4}{d}^{4} \right ) \left ( bx+a \right ) ^{2}}{3\,{b}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

-1/3*(12*ln(b*x+a)*x^3*a*b^3*e^4-12*ln(b*x+a)*x^3*b^4*d*e^3-3*x^4*b^4*e^4+36*ln(b*x+a)*x^2*a^2*b^2*e^4-36*ln(b
*x+a)*x^2*a*b^3*d*e^3-9*x^3*a*b^3*e^4+36*ln(b*x+a)*x*a^3*b*e^4-36*ln(b*x+a)*x*a^2*b^2*d*e^3+9*x^2*a^2*b^2*e^4-
36*x^2*a*b^3*d*e^3+18*x^2*b^4*d^2*e^2+12*ln(b*x+a)*a^4*e^4-12*ln(b*x+a)*a^3*b*d*e^3+27*x*a^3*b*e^4-54*x*a^2*b^
2*d*e^3+18*x*a*b^3*d^2*e^2+6*x*b^4*d^3*e+13*a^4*e^4-22*d*e^3*a^3*b+6*a^2*b^2*d^2*e^2+2*a*b^3*d^3*e+b^4*d^4)*(b
*x+a)^2/b^5/((b*x+a)^2)^(5/2)

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Maxima [B]  time = 1.2197, size = 1149, normalized size = 5.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/12*b*e^4*((12*b^5*x^5 + 48*a*b^4*x^4 - 48*a^2*b^3*x^3 - 252*a^3*b^2*x^2 - 248*a^4*b*x - 77*a^5)/(b^10*x^4 +
4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x + a^4*b^6) - 60*a*log(b*x + a)/b^6) + 1/3*b*d*e^3*((48*a*b^3*x^3 + 1
08*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5) + 12*log
(b*x + a)/b^5) + 1/12*a*e^4*((48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6
*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5) + 12*log(b*x + a)/b^5) - 1/2*b*d^2*e^2*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2
)^(3/2)*b^2) + 8*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 3*a^3*b/((b^2)^(9/2)*(x + a/b)^4) - 8*a^2/((b^2)^
(7/2)*(x + a/b)^3) + 6*a/((b^2)^(5/2)*b*(x + a/b)^2) - 6*a^3/((b^2)^(5/2)*b^3*(x + a/b)^4)) - 1/3*a*d*e^3*(12*
x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 3*a^3*b/((b^2)^(9/2)
*(x + a/b)^4) - 8*a^2/((b^2)^(7/2)*(x + a/b)^3) + 6*a/((b^2)^(5/2)*b*(x + a/b)^2) - 6*a^3/((b^2)^(5/2)*b^3*(x
+ a/b)^4)) - 1/12*b*d^4*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/((b^2)^(5/2)*b*(x + a/b)^4)) - 1/3*a*d^
3*e*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/((b^2)^(5/2)*b*(x + a/b)^4)) - 1/3*b*d^3*e*(3*a^2*b^2/((b^2
)^(9/2)*(x + a/b)^4) - 8*a*b/((b^2)^(7/2)*(x + a/b)^3) + 6/((b^2)^(5/2)*(x + a/b)^2)) - 1/2*a*d^2*e^2*(3*a^2*b
^2/((b^2)^(9/2)*(x + a/b)^4) - 8*a*b/((b^2)^(7/2)*(x + a/b)^3) + 6/((b^2)^(5/2)*(x + a/b)^2)) - 1/4*a*d^4/((b^
2)^(5/2)*(x + a/b)^4)

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Fricas [A]  time = 1.55416, size = 581, normalized size = 2.89 \begin{align*} \frac{3 \, b^{4} e^{4} x^{4} + 9 \, a b^{3} e^{4} x^{3} - b^{4} d^{4} - 2 \, a b^{3} d^{3} e - 6 \, a^{2} b^{2} d^{2} e^{2} + 22 \, a^{3} b d e^{3} - 13 \, a^{4} e^{4} - 9 \,{\left (2 \, b^{4} d^{2} e^{2} - 4 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} - 3 \,{\left (2 \, b^{4} d^{3} e + 6 \, a b^{3} d^{2} e^{2} - 18 \, a^{2} b^{2} d e^{3} + 9 \, a^{3} b e^{4}\right )} x + 12 \,{\left (a^{3} b d e^{3} - a^{4} e^{4} +{\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 3 \,{\left (a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 3 \,{\left (a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \log \left (b x + a\right )}{3 \,{\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/3*(3*b^4*e^4*x^4 + 9*a*b^3*e^4*x^3 - b^4*d^4 - 2*a*b^3*d^3*e - 6*a^2*b^2*d^2*e^2 + 22*a^3*b*d*e^3 - 13*a^4*e
^4 - 9*(2*b^4*d^2*e^2 - 4*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 - 3*(2*b^4*d^3*e + 6*a*b^3*d^2*e^2 - 18*a^2*b^2*d*e^3
+ 9*a^3*b*e^4)*x + 12*(a^3*b*d*e^3 - a^4*e^4 + (b^4*d*e^3 - a*b^3*e^4)*x^3 + 3*(a*b^3*d*e^3 - a^2*b^2*e^4)*x^
2 + 3*(a^2*b^2*d*e^3 - a^3*b*e^4)*x)*log(b*x + a))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right ) \left (d + e x\right )^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)**4/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((a + b*x)*(d + e*x)**4/((a + b*x)**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}{\left (e x + d\right )}^{4}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x + a)*(e*x + d)^4/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)