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

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

[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])

________________________________________________________________________________________

Rubi [A]  time = 0.0803956, antiderivative size = 222, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 0.0806139, size = 130, normalized size = 0.59 $\frac{(a+b x) \left (b e x \left (6 a^2 b e^2 (8 d+e x)-12 a^3 e^3-4 a b^2 e \left (18 d^2+6 d e x+e^2 x^2\right )+b^3 \left (36 d^2 e x+48 d^3+16 d e^2 x^2+3 e^3 x^3\right )\right )+12 (b d-a e)^4 \log (a+b x)\right )}{12 b^5 \sqrt{(a+b x)^2}}$

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])

________________________________________________________________________________________

Maple [A]  time = 0.158, size = 223, normalized size = 1. \begin{align*}{\frac{ \left ( bx+a \right ) \left ( 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 \left ( bx+a \right ){a}^{4}{e}^{4}-48\,\ln \left ( bx+a \right ){a}^{3}bd{e}^{3}+72\,\ln \left ( bx+a \right ){a}^{2}{b}^{2}{d}^{2}{e}^{2}-48\,\ln \left ( bx+a \right ) a{b}^{3}{d}^{3}e+12\,\ln \left ( bx+a \right ){b}^{4}{d}^{4}-12\,x{a}^{3}b{e}^{4}+48\,x{a}^{2}{b}^{2}d{e}^{3}-72\,xa{b}^{3}{d}^{2}{e}^{2}+48\,x{b}^{4}{d}^{3}e \right ) }{12\,{b}^{5}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

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

________________________________________________________________________________________

Maxima [B]  time = 1.1538, size = 590, normalized size = 2.66 \begin{align*} \frac{6 \, a^{2} b^{2} d^{2} e^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{20 \, a^{3} b d e^{3} \log \left (x + \frac{a}{b}\right )}{3 \,{\left (b^{2}\right )}^{\frac{5}{2}}} + \frac{13 \, a^{4} e^{4} \log \left (x + \frac{a}{b}\right )}{6 \,{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{6 \, a b d^{2} e^{2} x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{20 \, a^{2} d e^{3} x}{3 \,{\left (b^{2}\right )}^{\frac{3}{2}}} - \frac{13 \, a^{3} e^{4} x}{6 \,{\left (b^{2}\right )}^{\frac{3}{2}} b} + \frac{3 \, d^{2} e^{2} x^{2}}{\sqrt{b^{2}}} - \frac{10 \, a d e^{3} x^{2}}{3 \, \sqrt{b^{2}} b} + \frac{13 \, a^{2} e^{4} x^{2}}{12 \, \sqrt{b^{2}} b^{2}} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} e^{4} x^{3}}{4 \, b^{2}} + \sqrt{\frac{1}{b^{2}}} d^{4} \log \left (x + \frac{a}{b}\right ) - \frac{4 \, a \sqrt{\frac{1}{b^{2}}} d^{3} e \log \left (x + \frac{a}{b}\right )}{b} + \frac{8 \, a^{3} \sqrt{\frac{1}{b^{2}}} d e^{3} \log \left (x + \frac{a}{b}\right )}{3 \, b^{3}} - \frac{7 \, a^{4} \sqrt{\frac{1}{b^{2}}} e^{4} \log \left (x + \frac{a}{b}\right )}{6 \, b^{4}} + \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{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}} \end{align*}

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]

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

________________________________________________________________________________________

Fricas [A]  time = 1.5845, size = 369, normalized size = 1.66 \begin{align*} \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}} \end{align*}

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

________________________________________________________________________________________

Sympy [A]  time = 0.647496, size = 134, normalized size = 0.6 \begin{align*} \frac{e^{4} x^{4}}{4 b} - \frac{x^{3} \left (a e^{4} - 4 b d e^{3}\right )}{3 b^{2}} + \frac{x^{2} \left (a^{2} e^{4} - 4 a b d e^{3} + 6 b^{2} d^{2} e^{2}\right )}{2 b^{3}} - \frac{x \left (a^{3} e^{4} - 4 a^{2} b d e^{3} + 6 a b^{2} d^{2} e^{2} - 4 b^{3} d^{3} e\right )}{b^{4}} + \frac{\left (a e - b d\right )^{4} \log{\left (a + b x \right )}}{b^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.22037, size = 356, normalized size = 1.6 \begin{align*} \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}} \end{align*}

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