3.1587 \(\int \frac {(d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=222 \[ \frac {(a+b x) (d+e x)^3 (b d-a e)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^4}{4 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (b d-a e)^4 \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e x (a+b x) (b d-a e)^3}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^2 (b d-a e)^2}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 222, 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} \[ \frac {e x (a+b x) (b d-a e)^3}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^2 (b d-a e)^2}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^3 (b d-a e)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^4}{4 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (b d-a e)^4 \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {(d+e x)^4}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \left (\frac {e (b d-a e)^3}{b^5}+\frac {(b d-a e)^4}{b^4 \left (a b+b^2 x\right )}+\frac {e (b d-a e)^2 (d+e x)}{b^4}+\frac {e (b d-a e) (d+e x)^2}{b^3}+\frac {e (d+e x)^3}{b^2}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {e (b d-a e)^3 x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(b d-a e)^2 (a+b x) (d+e x)^2}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(b d-a e) (a+b x) (d+e x)^3}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^4}{4 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(b d-a e)^4 (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 = 130, normalized size = 0.59 \[ \frac {(a+b x) \left (b e x \left (-12 a^3 e^3+6 a^2 b e^2 (8 d+e x)-4 a b^2 e \left (18 d^2+6 d e x+e^2 x^2\right )+b^3 \left (48 d^3+36 d^2 e x+16 d e^2 x^2+3 e^3 x^3\right )\right )+12 (b d-a e)^4 \log (a+b x)\right )}{12 b^5 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(b*e*x*(-12*a^3*e^3 + 6*a^2*b*e^2*(8*d + e*x) - 4*a*b^2*e*(18*d^2 + 6*d*e*x + e^2*x^2) + b^3*(48*d^
3 + 36*d^2*e*x + 16*d*e^2*x^2 + 3*e^3*x^3)) + 12*(b*d - a*e)^4*Log[a + b*x]))/(12*b^5*Sqrt[(a + b*x)^2])

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fricas [A]  time = 0.91, size = 181, normalized size = 0.82 \[ \frac {3 \, b^{4} e^{4} x^{4} + 4 \, {\left (4 \, b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (6 \, b^{4} d^{2} e^{2} - 4 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 12 \, {\left (4 \, b^{4} d^{3} e - 6 \, a b^{3} d^{2} e^{2} + 4 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x + 12 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \log \left (b x + a\right )}{12 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/12*(3*b^4*e^4*x^4 + 4*(4*b^4*d*e^3 - a*b^3*e^4)*x^3 + 6*(6*b^4*d^2*e^2 - 4*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 +
12*(4*b^4*d^3*e - 6*a*b^3*d^2*e^2 + 4*a^2*b^2*d*e^3 - a^3*b*e^4)*x + 12*(b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d
^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*log(b*x + a))/b^5

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giac [A]  time = 0.18, size = 264, normalized size = 1.19 \[ \frac {3 \, b^{3} x^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) + 16 \, b^{3} d x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 36 \, b^{3} d^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 48 \, b^{3} d^{3} x e \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{2} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) - 24 \, a b^{2} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 72 \, a b^{2} d^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 48 \, a^{2} b d x e^{3} \mathrm {sgn}\left (b x + a\right ) - 12 \, a^{3} x e^{4} \mathrm {sgn}\left (b x + a\right )}{12 \, b^{4}} + \frac {{\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/12*(3*b^3*x^4*e^4*sgn(b*x + a) + 16*b^3*d*x^3*e^3*sgn(b*x + a) + 36*b^3*d^2*x^2*e^2*sgn(b*x + a) + 48*b^3*d^
3*x*e*sgn(b*x + a) - 4*a*b^2*x^3*e^4*sgn(b*x + a) - 24*a*b^2*d*x^2*e^3*sgn(b*x + a) - 72*a*b^2*d^2*x*e^2*sgn(b
*x + a) + 6*a^2*b*x^2*e^4*sgn(b*x + a) + 48*a^2*b*d*x*e^3*sgn(b*x + a) - 12*a^3*x*e^4*sgn(b*x + a))/b^4 + (b^4
*d^4*sgn(b*x + a) - 4*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) +
a^4*e^4*sgn(b*x + a))*log(abs(b*x + a))/b^5

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maple [A]  time = 0.06, size = 223, normalized size = 1.00 \[ \frac {\left (b x +a \right ) \left (3 b^{4} e^{4} x^{4}-4 a \,b^{3} e^{4} x^{3}+16 b^{4} d \,e^{3} x^{3}+6 a^{2} b^{2} e^{4} x^{2}-24 a \,b^{3} d \,e^{3} x^{2}+36 b^{4} d^{2} e^{2} x^{2}+12 a^{4} e^{4} \ln \left (b x +a \right )-48 a^{3} b d \,e^{3} \ln \left (b x +a \right )-12 a^{3} b \,e^{4} x +72 a^{2} b^{2} d^{2} e^{2} \ln \left (b x +a \right )+48 a^{2} b^{2} d \,e^{3} x -48 a \,b^{3} d^{3} e \ln \left (b x +a \right )-72 a \,b^{3} d^{2} e^{2} x +12 b^{4} d^{4} \ln \left (b x +a \right )+48 b^{4} d^{3} e x \right )}{12 \sqrt {\left (b x +a \right )^{2}}\, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/((b*x+a)^2)^(1/2),x)

[Out]

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

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maxima [B]  time = 1.07, size = 348, normalized size = 1.57 \[ \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} e^{4} x^{3}}{4 \, b^{2}} + \frac {3 \, d^{2} e^{2} x^{2}}{b} - \frac {10 \, a d e^{3} x^{2}}{3 \, b^{2}} + \frac {13 \, a^{2} e^{4} x^{2}}{12 \, b^{3}} + \frac {4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d e^{3} x^{2}}{3 \, b^{2}} - \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a e^{4} x^{2}}{12 \, b^{3}} - \frac {6 \, a d^{2} e^{2} x}{b^{2}} + \frac {20 \, a^{2} d e^{3} x}{3 \, b^{3}} - \frac {13 \, a^{3} e^{4} x}{6 \, b^{4}} + \frac {d^{4} \log \left (x + \frac {a}{b}\right )}{b} - \frac {4 \, a d^{3} e \log \left (x + \frac {a}{b}\right )}{b^{2}} + \frac {6 \, a^{2} d^{2} e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {4 \, a^{3} d e^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {a^{4} e^{4} \log \left (x + \frac {a}{b}\right )}{b^{5}} + \frac {4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d^{3} e}{b^{2}} - \frac {8 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d e^{3}}{3 \, b^{4}} + \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} e^{4}}{6 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/4*sqrt(b^2*x^2 + 2*a*b*x + a^2)*e^4*x^3/b^2 + 3*d^2*e^2*x^2/b - 10/3*a*d*e^3*x^2/b^2 + 13/12*a^2*e^4*x^2/b^3
+ 4/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*d*e^3*x^2/b^2 - 7/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*e^4*x^2/b^3 - 6*a*d^
2*e^2*x/b^2 + 20/3*a^2*d*e^3*x/b^3 - 13/6*a^3*e^4*x/b^4 + d^4*log(x + a/b)/b - 4*a*d^3*e*log(x + a/b)/b^2 + 6*
a^2*d^2*e^2*log(x + a/b)/b^3 - 4*a^3*d*e^3*log(x + a/b)/b^4 + a^4*e^4*log(x + a/b)/b^5 + 4*sqrt(b^2*x^2 + 2*a*
b*x + a^2)*d^3*e/b^2 - 8/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*d*e^3/b^4 + 7/6*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3
*e^4/b^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^4/((a + b*x)^2)^(1/2),x)

[Out]

int((d + e*x)^4/((a + b*x)^2)^(1/2), x)

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sympy [A]  time = 0.43, size = 136, normalized size = 0.61 \[ x^{3} \left (- \frac {a e^{4}}{3 b^{2}} + \frac {4 d e^{3}}{3 b}\right ) + x^{2} \left (\frac {a^{2} e^{4}}{2 b^{3}} - \frac {2 a d e^{3}}{b^{2}} + \frac {3 d^{2} e^{2}}{b}\right ) + x \left (- \frac {a^{3} e^{4}}{b^{4}} + \frac {4 a^{2} d e^{3}}{b^{3}} - \frac {6 a d^{2} e^{2}}{b^{2}} + \frac {4 d^{3} e}{b}\right ) + \frac {e^{4} x^{4}}{4 b} + \frac {\left (a e - b d\right )^{4} \log {\left (a + b x \right )}}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/((b*x+a)**2)**(1/2),x)

[Out]

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

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