3.18.54 \(\int \frac {(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=238 \[ -\frac {6 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^5 (a+b x) (d+e x)}+\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^5 (a+b x) (d+e x)^2}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{3 e^5 (a+b x) (d+e x)^3}+\frac {b^4 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}-\frac {4 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^5 (a+b x)} \]

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Rubi [A]  time = 0.14, antiderivative size = 238, 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 = {770, 21, 43} \begin {gather*} -\frac {6 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^5 (a+b x) (d+e x)}+\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^5 (a+b x) (d+e x)^2}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{3 e^5 (a+b x) (d+e x)^3}-\frac {4 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^5 (a+b x)}+\frac {b^4 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^4,x]

[Out]

(b^4*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^4*(a + b*x)) - ((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^5*(
a + b*x)*(d + e*x)^3) + (2*b*(b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*(d + e*x)^2) - (6*b^2
*(b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*(d + e*x)) - (4*b^3*(b*d - a*e)*Sqrt[a^2 + 2*a*b*
x + b^2*x^2]*Log[d + e*x])/(e^5*(a + b*x))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 181, normalized size = 0.76 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (a^4 e^4+2 a^3 b e^3 (d+3 e x)+6 a^2 b^2 e^2 \left (d^2+3 d e x+3 e^2 x^2\right )-2 a b^3 d e \left (11 d^2+27 d e x+18 e^2 x^2\right )+12 b^3 (d+e x)^3 (b d-a e) \log (d+e x)+b^4 \left (13 d^4+27 d^3 e x+9 d^2 e^2 x^2-9 d e^3 x^3-3 e^4 x^4\right )\right )}{3 e^5 (a+b x) (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^4,x]

[Out]

-1/3*(Sqrt[(a + b*x)^2]*(a^4*e^4 + 2*a^3*b*e^3*(d + 3*e*x) + 6*a^2*b^2*e^2*(d^2 + 3*d*e*x + 3*e^2*x^2) - 2*a*b
^3*d*e*(11*d^2 + 27*d*e*x + 18*e^2*x^2) + b^4*(13*d^4 + 27*d^3*e*x + 9*d^2*e^2*x^2 - 9*d*e^3*x^3 - 3*e^4*x^4)
+ 12*b^3*(b*d - a*e)*(d + e*x)^3*Log[d + e*x]))/(e^5*(a + b*x)*(d + e*x)^3)

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IntegrateAlgebraic [F]  time = 6.61, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^4,x]

[Out]

Defer[IntegrateAlgebraic][((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^4, x]

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fricas [A]  time = 0.46, size = 292, normalized size = 1.23 \begin {gather*} \frac {3 \, b^{4} e^{4} x^{4} + 9 \, b^{4} d e^{3} x^{3} - 13 \, b^{4} d^{4} + 22 \, a b^{3} d^{3} e - 6 \, a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} b d e^{3} - a^{4} e^{4} - 9 \, {\left (b^{4} d^{2} e^{2} - 4 \, a b^{3} d e^{3} + 2 \, a^{2} b^{2} e^{4}\right )} x^{2} - 3 \, {\left (9 \, b^{4} d^{3} e - 18 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3} + 2 \, a^{3} b e^{4}\right )} x - 12 \, {\left (b^{4} d^{4} - a b^{3} d^{3} e + {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (b^{4} d^{2} e^{2} - a b^{3} d e^{3}\right )} x^{2} + 3 \, {\left (b^{4} d^{3} e - a b^{3} d^{2} e^{2}\right )} x\right )} \log \left (e x + d\right )}{3 \, {\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 260, normalized size = 1.09 \begin {gather*} b^{4} x e^{\left (-4\right )} \mathrm {sgn}\left (b x + a\right ) - 4 \, {\left (b^{4} d \mathrm {sgn}\left (b x + a\right ) - a b^{3} e \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) - \frac {{\left (13 \, b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 22 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) + 18 \, {\left (b^{4} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{2} b^{2} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 6 \, {\left (5 \, b^{4} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 9 \, a b^{3} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{2} d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{3} b e^{4} \mathrm {sgn}\left (b x + a\right )\right )} x\right )} e^{\left (-5\right )}}{3 \, {\left (x e + d\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 330, normalized size = 1.39 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (12 a \,b^{3} e^{4} x^{3} \ln \left (e x +d \right )-12 b^{4} d \,e^{3} x^{3} \ln \left (e x +d \right )+3 b^{4} e^{4} x^{4}+36 a \,b^{3} d \,e^{3} x^{2} \ln \left (e x +d \right )-36 b^{4} d^{2} e^{2} x^{2} \ln \left (e x +d \right )+9 b^{4} d \,e^{3} x^{3}-18 a^{2} b^{2} e^{4} x^{2}+36 a \,b^{3} d^{2} e^{2} x \ln \left (e x +d \right )+36 a \,b^{3} d \,e^{3} x^{2}-36 b^{4} d^{3} e x \ln \left (e x +d \right )-9 b^{4} d^{2} e^{2} x^{2}-6 a^{3} b \,e^{4} x -18 a^{2} b^{2} d \,e^{3} x +12 a \,b^{3} d^{3} e \ln \left (e x +d \right )+54 a \,b^{3} d^{2} e^{2} x -12 b^{4} d^{4} \ln \left (e x +d \right )-27 b^{4} d^{3} e x -a^{4} e^{4}-2 a^{3} b d \,e^{3}-6 a^{2} b^{2} d^{2} e^{2}+22 a \,b^{3} d^{3} e -13 b^{4} d^{4}\right )}{3 \left (b x +a \right )^{3} \left (e x +d \right )^{3} e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^4,x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^4,x)

[Out]

int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^4, x)

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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 (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**4,x)

[Out]

Integral((a + b*x)*((a + b*x)**2)**(3/2)/(d + e*x)**4, x)

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