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

[Out]

-1/3*(e*x+d)^4/b/(b^2*x^2+2*a*b*x+a^2)^(3/2)-4*e^2*(-a*e+b*d)^2/b^5/((b*x+a)^2)^(1/2)-2/3*e*(-a*e+b*d)^3/b^5/(
b*x+a)/((b*x+a)^2)^(1/2)+4/3*e^4*x*(b*x+a)/b^4/((b*x+a)^2)^(1/2)+4*e^3*(-a*e+b*d)*(b*x+a)*ln(b*x+a)/b^5/((b*x+
a)^2)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 201, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.08, size = 170, normalized size = 0.85 \[ \frac {-13 a^4 e^4+a^3 b e^3 (22 d-27 e x)-3 a^2 b^2 e^2 \left (2 d^2-18 d e x+3 e^2 x^2\right )+a b^3 e \left (-2 d^3-18 d^2 e x+36 d e^2 x^2+9 e^3 x^3\right )-12 e^3 (a+b x)^3 (a e-b d) \log (a+b x)-\left (b^4 \left (d^4+6 d^3 e x+18 d^2 e^2 x^2-3 e^4 x^4\right )\right )}{3 b^5 \left ((a+b x)^2\right )^{3/2}} \]

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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fricas [A]  time = 1.43, size = 292, normalized size = 1.45 \[ \frac {3 \, b^{4} e^{4} x^{4} + 9 \, a b^{3} e^{4} x^{3} - b^{4} d^{4} - 2 \, a b^{3} d^{3} e - 6 \, a^{2} b^{2} d^{2} e^{2} + 22 \, a^{3} b d e^{3} - 13 \, a^{4} e^{4} - 9 \, {\left (2 \, b^{4} d^{2} e^{2} - 4 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} - 3 \, {\left (2 \, b^{4} d^{3} e + 6 \, a b^{3} d^{2} e^{2} - 18 \, a^{2} b^{2} d e^{3} + 9 \, a^{3} b e^{4}\right )} x + 12 \, {\left (a^{3} b d e^{3} - a^{4} e^{4} + {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \log \left (b x + a\right )}{3 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{4}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}\,{d x} \]

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)

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maple [B]  time = 0.07, size = 322, normalized size = 1.60 \[ -\frac {\left (12 a \,b^{3} e^{4} x^{3} \ln \left (b x +a \right )-12 b^{4} d \,e^{3} x^{3} \ln \left (b x +a \right )-3 b^{4} e^{4} x^{4}+36 a^{2} b^{2} e^{4} x^{2} \ln \left (b x +a \right )-36 a \,b^{3} d \,e^{3} x^{2} \ln \left (b x +a \right )-9 a \,b^{3} e^{4} x^{3}+36 a^{3} b \,e^{4} x \ln \left (b x +a \right )-36 a^{2} b^{2} d \,e^{3} x \ln \left (b x +a \right )+9 a^{2} b^{2} e^{4} x^{2}-36 a \,b^{3} d \,e^{3} x^{2}+18 b^{4} d^{2} e^{2} x^{2}+12 a^{4} e^{4} \ln \left (b x +a \right )-12 a^{3} b d \,e^{3} \ln \left (b x +a \right )+27 a^{3} b \,e^{4} x -54 a^{2} b^{2} d \,e^{3} x +18 a \,b^{3} d^{2} e^{2} x +6 b^{4} d^{3} e x +13 a^{4} e^{4}-22 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}+2 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \left (b x +a \right )^{2}}{3 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{5}} \]

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*b^4*e^4*x^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*a*b^3*e^4*x^3+36*a^3*b*e^4*x*ln(b*x+a)-36*a^2*b^2*d*e^3*x*ln(b*x+a)+9*a^2*b^2*e^4*x^2-
36*a*b^3*d*e^3*x^2+18*b^4*d^2*e^2*x^2+12*a^4*e^4*ln(b*x+a)-12*a^3*b*d*e^3*ln(b*x+a)+27*a^3*b*e^4*x-54*a^2*b^2*
d*e^3*x+18*a*b^3*d^2*e^2*x+6*b^4*d^3*e*x+13*a^4*e^4-22*a^3*b*d*e^3+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 = 0.99, size = 755, normalized size = 3.76 \[ \frac {1}{12} \, b e^{4} {\left (\frac {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}} - \frac {60 \, a \log \left (b x + a\right )}{b^{6}}\right )} + \frac {1}{3} \, b d e^{3} {\left (\frac {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}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} + \frac {1}{12} \, a e^{4} {\left (\frac {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}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} - \frac {1}{2} \, b d^{2} e^{2} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{3} \, a d e^{3} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{12} \, b d^{4} {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{3} \, a d^{3} e {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{3} \, b d^{3} e {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{2} \, a d^{2} e^{2} {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {a d^{4}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} \]

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) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b)^3) -
3*a^3/(b^8*(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) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b)^3) - 3*a^3/(b^8*(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^6*(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^6*(x + a/b)^4)) - 1/3*b*d^3*e*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3*a^2/(b^7*
(x + a/b)^4)) - 1/2*a*d^2*e^2*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3*a^2/(b^7*(x + a/b)^4)) - 1/4*a*
d^4/(b^5*(x + a/b)^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x\right ) \left (d + e x\right )^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

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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