3.21.26 \(\int \frac {(a+b x) (d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [2026]

Optimal. Leaf size=162 \[ \frac {3 e^2 (b d-a e) x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (a+b x) (d+e x)^2}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^3}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (b d-a e)^2 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 162, 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 = {782, 660, 45} \begin {gather*} -\frac {(d+e x)^3}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (a+b x) (d+e x)^2}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (a+b x) (b d-a e)^2 \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e^2 x (a+b x) (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

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 660

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 782

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

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Mathematica [A]
time = 0.04, size = 120, normalized size = 0.74 \begin {gather*} \frac {2 a^3 e^3-2 a^2 b e^2 (3 d+2 e x)+3 a b^2 e \left (2 d^2+2 d e x-e^2 x^2\right )+b^3 \left (-2 d^3+6 d e^2 x^2+e^3 x^3\right )+6 e (b d-a e)^2 (a+b x) \log (a+b x)}{2 b^4 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*e^3 - 2*a^2*b*e^2*(3*d + 2*e*x) + 3*a*b^2*e*(2*d^2 + 2*d*e*x - e^2*x^2) + b^3*(-2*d^3 + 6*d*e^2*x^2 + e
^3*x^3) + 6*e*(b*d - a*e)^2*(a + b*x)*Log[a + b*x])/(2*b^4*Sqrt[(a + b*x)^2])

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Maple [A]
time = 0.08, size = 209, normalized size = 1.29

method	result	size
risch	\(-\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{2} \left (-\frac {1}{2} b e \,x^{2}+2 a e x -3 b d x \right )}{\left (b x +a \right ) b^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{\left (b x +a \right )^{2} b^{4}}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, e \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{4}}\)	\(149\)
default	\(\frac {\left (b^{3} e^{3} x^{3}+6 \ln \left (b x +a \right ) a^{2} b \,e^{3} x -12 \ln \left (b x +a \right ) a \,b^{2} d \,e^{2} x +6 \ln \left (b x +a \right ) b^{3} d^{2} e x -3 a \,b^{2} e^{3} x^{2}+6 b^{3} d \,e^{2} x^{2}+6 \ln \left (b x +a \right ) a^{3} e^{3}-12 \ln \left (b x +a \right ) a^{2} b d \,e^{2}+6 \ln \left (b x +a \right ) a \,b^{2} d^{2} e -4 a^{2} b \,e^{3} x +6 a \,b^{2} d \,e^{2} x +2 a^{3} e^{3}-6 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -2 b^{3} d^{3}\right ) \left (b x +a \right )^{2}}{2 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^3*e^3*x^3+6*ln(b*x+a)*a^2*b*e^3*x-12*ln(b*x+a)*a*b^2*d*e^2*x+6*ln(b*x+a)*b^3*d^2*e*x-3*a*b^2*e^3*x^2+6*
b^3*d*e^2*x^2+6*ln(b*x+a)*a^3*e^3-12*ln(b*x+a)*a^2*b*d*e^2+6*ln(b*x+a)*a*b^2*d^2*e-4*a^2*b*e^3*x+6*a*b^2*d*e^2
*x+2*a^3*e^3-6*a^2*b*d*e^2+6*a*b^2*d^2*e-2*b^3*d^3)*(b*x+a)^2/b^4/((b*x+a)^2)^(3/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 467 vs. \(2 (121) = 242\).
time = 0.27, size = 467, normalized size = 2.88 \begin {gather*} \frac {x^{3} e^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b} - \frac {5 \, a x^{2} e^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {{\left (3 \, b d e^{2} + a e^{3}\right )} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {6 \, a^{2} e^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} - \frac {5 \, a^{3} e^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {a d^{3}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {3 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {3 \, {\left (b d^{2} e + a d e^{2}\right )} \log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {2 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {b d^{3} + 3 \, a d^{2} e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {12 \, a^{3} x e^{3}}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {6 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {6 \, {\left (b d^{2} e + a d e^{2}\right )} a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, a^{4} e^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {9 \, {\left (b d^{2} e + a d e^{2}\right )} a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {{\left (b d^{3} + 3 \, a d^{2} e\right )} a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^3*e^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b) - 5/2*a*x^2*e^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + (3*b*d*e^2
+ a*e^3)*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 6*a^2*e^3*log(x + a/b)/b^4 - 5*a^3*e^3/(sqrt(b^2*x^2 + 2*a
*b*x + a^2)*b^4) - 1/2*a*d^3/(b^3*(x + a/b)^2) - 3*(3*b*d*e^2 + a*e^3)*a*log(x + a/b)/b^4 + 3*(b*d^2*e + a*d*e
^2)*log(x + a/b)/b^3 + 2*(3*b*d*e^2 + a*e^3)*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - (b*d^3 + 3*a*d^2*e)/(sq
rt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 12*a^3*x*e^3/(b^5*(x + a/b)^2) - 6*(3*b*d*e^2 + a*e^3)*a^2*x/(b^5*(x + a/b)
^2) + 6*(b*d^2*e + a*d*e^2)*a*x/(b^4*(x + a/b)^2) + 23/2*a^4*e^3/(b^6*(x + a/b)^2) - 11/2*(3*b*d*e^2 + a*e^3)*
a^3/(b^6*(x + a/b)^2) + 9/2*(b*d^2*e + a*d*e^2)*a^2/(b^5*(x + a/b)^2) + 1/2*(b*d^3 + 3*a*d^2*e)*a/(b^4*(x + a/
b)^2)

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Fricas [A]
time = 2.51, size = 155, normalized size = 0.96 \begin {gather*} -\frac {2 \, b^{3} d^{3} - 6 \, a b^{2} d^{2} e - {\left (b^{3} x^{3} - 3 \, a b^{2} x^{2} - 4 \, a^{2} b x + 2 \, a^{3}\right )} e^{3} - 6 \, {\left (b^{3} d x^{2} + a b^{2} d x - a^{2} b d\right )} e^{2} - 6 \, {\left ({\left (a^{2} b x + a^{3}\right )} e^{3} - 2 \, {\left (a b^{2} d x + a^{2} b d\right )} e^{2} + {\left (b^{3} d^{2} x + a b^{2} d^{2}\right )} e\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x + a b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.17, size = 148, normalized size = 0.91 \begin {gather*} \frac {b^{2} x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, b^{2} d x e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b x e^{3} \mathrm {sgn}\left (b x + a\right )}{2 \, b^{4}} + \frac {3 \, {\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}}{{\left (b x + a\right )} b^{4} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*x^2*e^3*sgn(b*x + a) + 6*b^2*d*x*e^2*sgn(b*x + a) - 4*a*b*x*e^3*sgn(b*x + a))/b^4 + 3*(b^2*d^2*e - 2*
a*b*d*e^2 + a^2*e^3)*log(abs(b*x + a))/(b^4*sgn(b*x + a)) - (b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3
)/((b*x + a)*b^4*sgn(b*x + a))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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