\(\int \frac {(a+b x)^2}{(c+d x)^{3/2}} \, dx\) [1427]

Optimal result

Integrand size = 17, antiderivative size = 67 \[ \int \frac {(a+b x)^2}{(c+d x)^{3/2}} \, dx=-\frac {2 (b c-a d)^2}{d^3 \sqrt {c+d x}}-\frac {4 b (b c-a d) \sqrt {c+d x}}{d^3}+\frac {2 b^2 (c+d x)^{3/2}}{3 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {45} \[ \int \frac {(a+b x)^2}{(c+d x)^{3/2}} \, dx=-\frac {4 b \sqrt {c+d x} (b c-a d)}{d^3}-\frac {2 (b c-a d)^2}{d^3 \sqrt {c+d x}}+\frac {2 b^2 (c+d x)^{3/2}}{3 d^3} \]

[In]

Int[(a + b*x)^2/(c + d*x)^(3/2),x]

[Out]

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

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])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(-b c+a d)^2}{d^2 (c+d x)^{3/2}}-\frac {2 b (b c-a d)}{d^2 \sqrt {c+d x}}+\frac {b^2 \sqrt {c+d x}}{d^2}\right ) \, dx \\ & = -\frac {2 (b c-a d)^2}{d^3 \sqrt {c+d x}}-\frac {4 b (b c-a d) \sqrt {c+d x}}{d^3}+\frac {2 b^2 (c+d x)^{3/2}}{3 d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^2}{(c+d x)^{3/2}} \, dx=\frac {2 \left (-3 a^2 d^2+6 a b d (2 c+d x)+b^2 \left (-8 c^2-4 c d x+d^2 x^2\right )\right )}{3 d^3 \sqrt {c+d x}} \]

[In]

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

[Out]

(2*(-3*a^2*d^2 + 6*a*b*d*(2*c + d*x) + b^2*(-8*c^2 - 4*c*d*x + d^2*x^2)))/(3*d^3*Sqrt[c + d*x])

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.82

method	result	size
pseudoelliptic	\(\frac {\frac {2 \left (d^{2} x^{2}-4 c d x -8 c^{2}\right ) b^{2}}{3}+8 \left (\frac {d x}{2}+c \right ) d a b -2 a^{2} d^{2}}{d^{3} \sqrt {d x +c}}\)	\(55\)
risch	\(\frac {2 b \left (b d x +6 a d -5 b c \right ) \sqrt {d x +c}}{3 d^{3}}-\frac {2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{d^{3} \sqrt {d x +c}}\)	\(61\)
gosper	\(-\frac {2 \left (-d^{2} x^{2} b^{2}-6 x a b \,d^{2}+4 x \,b^{2} c d +3 a^{2} d^{2}-12 a b c d +8 b^{2} c^{2}\right )}{3 \sqrt {d x +c}\, d^{3}}\)	\(63\)
trager	\(-\frac {2 \left (-d^{2} x^{2} b^{2}-6 x a b \,d^{2}+4 x \,b^{2} c d +3 a^{2} d^{2}-12 a b c d +8 b^{2} c^{2}\right )}{3 \sqrt {d x +c}\, d^{3}}\)	\(63\)
derivativedivides	\(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+4 a b d \sqrt {d x +c}-4 b^{2} c \sqrt {d x +c}-\frac {2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{\sqrt {d x +c}}}{d^{3}}\)	\(74\)
default	\(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {3}{2}}}{3}+4 a b d \sqrt {d x +c}-4 b^{2} c \sqrt {d x +c}-\frac {2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{\sqrt {d x +c}}}{d^{3}}\)	\(74\)

[In]

int((b*x+a)^2/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/3*((d^2*x^2-4*c*d*x-8*c^2)*b^2+12*(1/2*d*x+c)*d*a*b-3*a^2*d^2)/(d*x+c)^(1/2)/d^3

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b x)^2}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (b^{2} d^{2} x^{2} - 8 \, b^{2} c^{2} + 12 \, a b c d - 3 \, a^{2} d^{2} - 2 \, {\left (2 \, b^{2} c d - 3 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (d^{4} x + c d^{3}\right )}} \]

[In]

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

[Out]

2/3*(b^2*d^2*x^2 - 8*b^2*c^2 + 12*a*b*c*d - 3*a^2*d^2 - 2*(2*b^2*c*d - 3*a*b*d^2)*x)*sqrt(d*x + c)/(d^4*x + c*
d^3)

Sympy [A] (verification not implemented)

Time = 1.48 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31 \[ \int \frac {(a+b x)^2}{(c+d x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{2} \left (c + d x\right )^{\frac {3}{2}}}{3 d^{2}} + \frac {\sqrt {c + d x} \left (2 a b d - 2 b^{2} c\right )}{d^{2}} - \frac {\left (a d - b c\right )^{2}}{d^{2} \sqrt {c + d x}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\begin {cases} a^{2} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{3}}{3 b} & \text {otherwise} \end {cases}}{c^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(b**2*(c + d*x)**(3/2)/(3*d**2) + sqrt(c + d*x)*(2*a*b*d - 2*b**2*c)/d**2 - (a*d - b*c)**2/(d**2*
sqrt(c + d*x)))/d, Ne(d, 0)), (Piecewise((a**2*x, Eq(b, 0)), ((a + b*x)**3/(3*b), True))/c**(3/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12 \[ \int \frac {(a+b x)^2}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {{\left (d x + c\right )}^{\frac {3}{2}} b^{2} - 6 \, {\left (b^{2} c - a b d\right )} \sqrt {d x + c}}{d^{2}} - \frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}}{\sqrt {d x + c} d^{2}}\right )}}{3 \, d} \]

[In]

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

[Out]

2/3*(((d*x + c)^(3/2)*b^2 - 6*(b^2*c - a*b*d)*sqrt(d*x + c))/d^2 - 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/(sqrt(d*x
+ c)*d^2))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.25 \[ \int \frac {(a+b x)^2}{(c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}}{\sqrt {d x + c} d^{3}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b^{2} d^{6} - 6 \, \sqrt {d x + c} b^{2} c d^{6} + 6 \, \sqrt {d x + c} a b d^{7}\right )}}{3 \, d^{9}} \]

[In]

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

[Out]

-2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/(sqrt(d*x + c)*d^3) + 2/3*((d*x + c)^(3/2)*b^2*d^6 - 6*sqrt(d*x + c)*b^2*c*
d^6 + 6*sqrt(d*x + c)*a*b*d^7)/d^9

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2}{(c+d x)^{3/2}} \, dx=\frac {\frac {2\,b^2\,{\left (c+d\,x\right )}^2}{3}-2\,a^2\,d^2-2\,b^2\,c^2-4\,b^2\,c\,\left (c+d\,x\right )+4\,a\,b\,d\,\left (c+d\,x\right )+4\,a\,b\,c\,d}{d^3\,\sqrt {c+d\,x}} \]

[In]

int((a + b*x)^2/(c + d*x)^(3/2),x)

[Out]

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