3.20.80 \(\int \frac {(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^5} \, dx\) [1980]

Optimal. Leaf size=246 \[ -\frac {(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^5 (a+b x) (d+e x)^4}+\frac {4 b (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^3}-\frac {3 b^2 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)^2}+\frac {4 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)}+\frac {b^4 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 246, 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 = {784, 21, 45} \begin {gather*} -\frac {3 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)^2}+\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^5 (a+b x) (d+e x)^3}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{4 e^5 (a+b x) (d+e x)^4}+\frac {b^4 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)}+\frac {4 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^5 (a+b x) (d+e 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)^5,x]

[Out]

-1/4*((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*(d + e*x)^4) + (4*b*(b*d - a*e)^3*Sqrt[a^2 +
2*a*b*x + b^2*x^2])/(3*e^5*(a + b*x)*(d + e*x)^3) - (3*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)^2) + (4*b^3*(b*d - a*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*(d + e*x)) + (b^4*Sq
rt[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 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 784

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)^5} \, 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)^5} \, 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)^5} \, 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 d+a e)^4}{e^4 (d+e x)^5}-\frac {4 b (b d-a e)^3}{e^4 (d+e x)^4}+\frac {6 b^2 (b d-a e)^2}{e^4 (d+e x)^3}-\frac {4 b^3 (b d-a e)}{e^4 (d+e x)^2}+\frac {b^4}{e^4 (d+e x)}\right ) \, dx}{a b+b^2 x}\\ &=-\frac {(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^5 (a+b x) (d+e x)^4}+\frac {4 b (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^5 (a+b x) (d+e x)^3}-\frac {3 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)^2}+\frac {4 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) (d+e x)}+\frac {b^4 \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 = 0.07, size = 144, normalized size = 0.59 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left ((b d-a e) \left (3 a^3 e^3+a^2 b e^2 (7 d+16 e x)+a b^2 e \left (13 d^2+40 d e x+36 e^2 x^2\right )+b^3 \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )\right )+12 b^4 (d+e x)^4 \log (d+e x)\right )}{12 e^5 (a+b x) (d+e x)^4} \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)^5,x]

[Out]

(Sqrt[(a + b*x)^2]*((b*d - a*e)*(3*a^3*e^3 + a^2*b*e^2*(7*d + 16*e*x) + a*b^2*e*(13*d^2 + 40*d*e*x + 36*e^2*x^
2) + b^3*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 + 48*e^3*x^3)) + 12*b^4*(d + e*x)^4*Log[d + e*x]))/(12*e^5*(a +
b*x)*(d + e*x)^4)

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Maple [A]
time = 0.07, size = 276, normalized size = 1.12

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {4 b^{3} \left (a e -b d \right ) x^{3}}{e^{2}}-\frac {3 b^{2} \left (a^{2} e^{2}+2 a b d e -3 b^{2} d^{2}\right ) x^{2}}{e^{3}}-\frac {2 b \left (2 a^{3} e^{3}+3 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -11 b^{3} d^{3}\right ) x}{3 e^{4}}-\frac {3 a^{4} e^{4}+4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}+12 a \,b^{3} d^{3} e -25 b^{4} d^{4}}{12 e^{5}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{4}}+\frac {b^{4} \ln \left (e x +d \right ) \sqrt {\left (b x +a \right )^{2}}}{e^{5} \left (b x +a \right )}\)	\(208\)
default	\(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (12 \ln \left (e x +d \right ) b^{4} e^{4} x^{4}+48 \ln \left (e x +d \right ) b^{4} d \,e^{3} x^{3}+72 \ln \left (e x +d \right ) b^{4} d^{2} e^{2} x^{2}-48 a \,b^{3} e^{4} x^{3}+48 b^{4} d \,e^{3} x^{3}+48 \ln \left (e x +d \right ) b^{4} d^{3} e x -36 a^{2} b^{2} e^{4} x^{2}-72 a \,b^{3} d \,e^{3} x^{2}+108 b^{4} d^{2} e^{2} x^{2}+12 \ln \left (e x +d \right ) b^{4} d^{4}-16 a^{3} b \,e^{4} x -24 a^{2} b^{2} d \,e^{3} x -48 a \,b^{3} d^{2} e^{2} x +88 b^{4} d^{3} e x -3 a^{4} e^{4}-4 a^{3} b d \,e^{3}-6 a^{2} b^{2} d^{2} e^{2}-12 a \,b^{3} d^{3} e +25 b^{4} d^{4}\right )}{12 \left (b x +a \right )^{3} e^{5} \left (e x +d \right )^{4}}\)	\(276\)

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)^5,x,method=_RETURNVERBOSE)

[Out]

1/12*((b*x+a)^2)^(3/2)*(12*ln(e*x+d)*b^4*e^4*x^4+48*ln(e*x+d)*b^4*d*e^3*x^3+72*ln(e*x+d)*b^4*d^2*e^2*x^2-48*a*
b^3*e^4*x^3+48*b^4*d*e^3*x^3+48*ln(e*x+d)*b^4*d^3*e*x-36*a^2*b^2*e^4*x^2-72*a*b^3*d*e^3*x^2+108*b^4*d^2*e^2*x^
2+12*ln(e*x+d)*b^4*d^4-16*a^3*b*e^4*x-24*a^2*b^2*d*e^3*x-48*a*b^3*d^2*e^2*x+88*b^4*d^3*e*x-3*a^4*e^4-4*a^3*b*d
*e^3-6*a^2*b^2*d^2*e^2-12*a*b^3*d^3*e+25*b^4*d^4)/(b*x+a)^3/e^5/(e*x+d)^4

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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)^5,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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 2.04, size = 251, normalized size = 1.02 \begin {gather*} \frac {25 \, b^{4} d^{4} - {\left (48 \, a b^{3} x^{3} + 36 \, a^{2} b^{2} x^{2} + 16 \, a^{3} b x + 3 \, a^{4}\right )} e^{4} + 4 \, {\left (12 \, b^{4} d x^{3} - 18 \, a b^{3} d x^{2} - 6 \, a^{2} b^{2} d x - a^{3} b d\right )} e^{3} + 6 \, {\left (18 \, b^{4} d^{2} x^{2} - 8 \, a b^{3} d^{2} x - a^{2} b^{2} d^{2}\right )} e^{2} + 4 \, {\left (22 \, b^{4} d^{3} x - 3 \, a b^{3} d^{3}\right )} e + 12 \, {\left (b^{4} x^{4} e^{4} + 4 \, b^{4} d x^{3} e^{3} + 6 \, b^{4} d^{2} x^{2} e^{2} + 4 \, b^{4} d^{3} x e + b^{4} d^{4}\right )} \log \left (x e + d\right )}{12 \, {\left (x^{4} e^{9} + 4 \, d x^{3} e^{8} + 6 \, d^{2} x^{2} e^{7} + 4 \, d^{3} x e^{6} + d^{4} 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)^5,x, algorithm="fricas")

[Out]

1/12*(25*b^4*d^4 - (48*a*b^3*x^3 + 36*a^2*b^2*x^2 + 16*a^3*b*x + 3*a^4)*e^4 + 4*(12*b^4*d*x^3 - 18*a*b^3*d*x^2
- 6*a^2*b^2*d*x - a^3*b*d)*e^3 + 6*(18*b^4*d^2*x^2 - 8*a*b^3*d^2*x - a^2*b^2*d^2)*e^2 + 4*(22*b^4*d^3*x - 3*a
*b^3*d^3)*e + 12*(b^4*x^4*e^4 + 4*b^4*d*x^3*e^3 + 6*b^4*d^2*x^2*e^2 + 4*b^4*d^3*x*e + b^4*d^4)*log(x*e + d))/(
x^4*e^9 + 4*d*x^3*e^8 + 6*d^2*x^2*e^7 + 4*d^3*x*e^6 + d^4*e^5)

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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 )^{5}}\, 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)**5,x)

[Out]

Integral((a + b*x)*((a + b*x)**2)**(3/2)/(d + e*x)**5, x)

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Giac [A]
time = 0.74, size = 268, normalized size = 1.09 \begin {gather*} b^{4} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (48 \, {\left (b^{4} d e^{2} \mathrm {sgn}\left (b x + a\right ) - a b^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} x^{3} + 36 \, {\left (3 \, b^{4} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{3} d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{2} b^{2} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 8 \, {\left (11 \, b^{4} d^{3} \mathrm {sgn}\left (b x + a\right ) - 6 \, a b^{3} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 3 \, a^{2} b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a^{3} b e^{3} \mathrm {sgn}\left (b x + a\right )\right )} x + {\left (25 \, b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 12 \, 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 ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\right )}\right )} e^{\left (-4\right )}}{12 \, {\left (x e + d\right )}^{4}} \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)^5,x, algorithm="giac")

[Out]

b^4*e^(-5)*log(abs(x*e + d))*sgn(b*x + a) + 1/12*(48*(b^4*d*e^2*sgn(b*x + a) - a*b^3*e^3*sgn(b*x + a))*x^3 + 3
6*(3*b^4*d^2*e*sgn(b*x + a) - 2*a*b^3*d*e^2*sgn(b*x + a) - a^2*b^2*e^3*sgn(b*x + a))*x^2 + 8*(11*b^4*d^3*sgn(b
*x + a) - 6*a*b^3*d^2*e*sgn(b*x + a) - 3*a^2*b^2*d*e^2*sgn(b*x + a) - 2*a^3*b*e^3*sgn(b*x + a))*x + (25*b^4*d^
4*sgn(b*x + a) - 12*a*b^3*d^3*e*sgn(b*x + a) - 6*a^2*b^2*d^2*e^2*sgn(b*x + a) - 4*a^3*b*d*e^3*sgn(b*x + a) - 3
*a^4*e^4*sgn(b*x + a))*e^(-1))*e^(-4)/(x*e + d)^4

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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 )}^5} \,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)^5,x)

[Out]

int(((a + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x)^5, x)

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