3.1563 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^4} \, dx\)

Optimal. Leaf size=194 \[ \frac {3 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^4 (a+b x) (d+e x)}-\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{2 e^4 (a+b x) (d+e x)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^4 (a+b x) (d+e x)^3}+\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^4 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {646, 43} \[ \frac {3 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^4 (a+b x) (d+e x)}-\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{2 e^4 (a+b x) (d+e x)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^4 (a+b x) (d+e x)^3}+\frac {b^3 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 104, normalized size = 0.54 \[ \frac {\sqrt {(a+b x)^2} \left ((b d-a e) \left (2 a^2 e^2+a b e (5 d+9 e x)+b^2 \left (11 d^2+27 d e x+18 e^2 x^2\right )\right )+6 b^3 (d+e x)^3 \log (d+e x)\right )}{6 e^4 (a+b x) (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.92, size = 177, normalized size = 0.91 \[ \frac {11 \, b^{3} d^{3} - 6 \, a b^{2} d^{2} e - 3 \, a^{2} b d e^{2} - 2 \, a^{3} e^{3} + 18 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 9 \, {\left (3 \, b^{3} d^{2} e - 2 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x + 6 \, {\left (b^{3} e^{3} x^{3} + 3 \, b^{3} d e^{2} x^{2} + 3 \, b^{3} d^{2} e x + b^{3} d^{3}\right )} \log \left (e x + d\right )}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(11*b^3*d^3 - 6*a*b^2*d^2*e - 3*a^2*b*d*e^2 - 2*a^3*e^3 + 18*(b^3*d*e^2 - a*b^2*e^3)*x^2 + 9*(3*b^3*d^2*e
- 2*a*b^2*d*e^2 - a^2*b*e^3)*x + 6*(b^3*e^3*x^3 + 3*b^3*d*e^2*x^2 + 3*b^3*d^2*e*x + b^3*d^3)*log(e*x + d))/(e^
7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)

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giac [A]  time = 0.17, size = 177, normalized size = 0.91 \[ b^{3} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (18 \, {\left (b^{3} d e \mathrm {sgn}\left (b x + a\right ) - a b^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 9 \, {\left (3 \, b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{2} d e \mathrm {sgn}\left (b x + a\right ) - a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right )\right )} x + {\left (11 \, b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 6 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\right )}\right )} e^{\left (-3\right )}}{6 \, {\left (x e + d\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^3*e^(-4)*log(abs(x*e + d))*sgn(b*x + a) + 1/6*(18*(b^3*d*e*sgn(b*x + a) - a*b^2*e^2*sgn(b*x + a))*x^2 + 9*(3
*b^3*d^2*sgn(b*x + a) - 2*a*b^2*d*e*sgn(b*x + a) - a^2*b*e^2*sgn(b*x + a))*x + (11*b^3*d^3*sgn(b*x + a) - 6*a*
b^2*d^2*e*sgn(b*x + a) - 3*a^2*b*d*e^2*sgn(b*x + a) - 2*a^3*e^3*sgn(b*x + a))*e^(-1))*e^(-3)/(x*e + d)^3

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maple [A]  time = 0.05, size = 186, normalized size = 0.96 \[ \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (6 b^{3} e^{3} x^{3} \ln \left (e x +d \right )+18 b^{3} d \,e^{2} x^{2} \ln \left (e x +d \right )-18 a \,b^{2} e^{3} x^{2}+18 b^{3} d^{2} e x \ln \left (e x +d \right )+18 b^{3} d \,e^{2} x^{2}-9 a^{2} b \,e^{3} x -18 a \,b^{2} d \,e^{2} x +6 b^{3} d^{3} \ln \left (e x +d \right )+27 b^{3} d^{2} e x -2 a^{3} e^{3}-3 a^{2} b d \,e^{2}-6 a \,b^{2} d^{2} e +11 b^{3} d^{3}\right )}{6 \left (b x +a \right )^{3} \left (e x +d \right )^{3} e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((b*x+a)^2)^(3/2)*(6*ln(e*x+d)*x^3*b^3*e^3+18*b^3*d*e^2*x^2*ln(e*x+d)+18*b^3*d^2*e*x*ln(e*x+d)-18*a*b^2*e^
3*x^2+18*b^3*d*e^2*x^2+6*b^3*d^3*ln(e*x+d)-9*a^2*b*e^3*x-18*a*b^2*d*e^2*x+27*b^3*d^2*e*x-2*a^3*e^3-3*a^2*b*d*e
^2-6*a*b^2*d^2*e+11*b^3*d^3)/(b*x+a)^3/e^4/(e*x+d)^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = -0.01 \[ \int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \[ \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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