\(\int \frac {(a+b x)^2}{(c+d x)^{5/2}} \, dx\) [1437]

Optimal result

Integrand size = 17, antiderivative size = 67 \[ \int \frac {(a+b x)^2}{(c+d x)^{5/2}} \, dx=-\frac {2 (b c-a d)^2}{3 d^3 (c+d x)^{3/2}}+\frac {4 b (b c-a d)}{d^3 \sqrt {c+d x}}+\frac {2 b^2 \sqrt {c+d x}}{d^3} \]

[Out]

-2/3*(-a*d+b*c)^2/d^3/(d*x+c)^(3/2)+4*b*(-a*d+b*c)/d^3/(d*x+c)^(1/2)+2*b^2*(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} \[ \int \frac {(a+b x)^2}{(c+d x)^{5/2}} \, dx=\frac {4 b (b c-a d)}{d^3 \sqrt {c+d x}}-\frac {2 (b c-a d)^2}{3 d^3 (c+d x)^{3/2}}+\frac {2 b^2 \sqrt {c+d x}}{d^3} \]

[In]

Int[(a + b*x)^2/(c + d*x)^(5/2),x]

[Out]

(-2*(b*c - a*d)^2)/(3*d^3*(c + d*x)^(3/2)) + (4*b*(b*c - a*d))/(d^3*Sqrt[c + d*x]) + (2*b^2*Sqrt[c + d*x])/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)^{5/2}}-\frac {2 b (b c-a d)}{d^2 (c+d x)^{3/2}}+\frac {b^2}{d^2 \sqrt {c+d x}}\right ) \, dx \\ & = -\frac {2 (b c-a d)^2}{3 d^3 (c+d x)^{3/2}}+\frac {4 b (b c-a d)}{d^3 \sqrt {c+d x}}+\frac {2 b^2 \sqrt {c+d x}}{d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x)^2}{(c+d x)^{5/2}} \, dx=\frac {-2 a^2 d^2-4 a b d (2 c+3 d x)+2 b^2 \left (8 c^2+12 c d x+3 d^2 x^2\right )}{3 d^3 (c+d x)^{3/2}} \]

[In]

Integrate[(a + b*x)^2/(c + d*x)^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {2 b^{2} \sqrt {d x +c}}{d^{3}}-\frac {2 \left (6 b d x +a d +5 b c \right ) \left (a d -b c \right )}{3 d^{3} \left (d x +c \right )^{\frac {3}{2}}}\)	\(50\)
pseudoelliptic	\(-\frac {2 \left (\left (-3 b^{2} x^{2}+6 a b x +a^{2}\right ) d^{2}+4 b c \left (-3 b x +a \right ) d -8 b^{2} c^{2}\right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{3}}\)	\(54\)
gosper	\(-\frac {2 \left (-3 d^{2} x^{2} b^{2}+6 x a b \,d^{2}-12 x \,b^{2} c d +a^{2} d^{2}+4 a b c d -8 b^{2} c^{2}\right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{3}}\)	\(62\)
trager	\(-\frac {2 \left (-3 d^{2} x^{2} b^{2}+6 x a b \,d^{2}-12 x \,b^{2} c d +a^{2} d^{2}+4 a b c d -8 b^{2} c^{2}\right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{3}}\)	\(62\)
derivativedivides	\(\frac {2 b^{2} \sqrt {d x +c}-\frac {4 \left (a d -b c \right ) b}{\sqrt {d x +c}}-\frac {2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}}{d^{3}}\)	\(66\)
default	\(\frac {2 b^{2} \sqrt {d x +c}-\frac {4 \left (a d -b c \right ) b}{\sqrt {d x +c}}-\frac {2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}}{d^{3}}\)	\(66\)

[In]

int((b*x+a)^2/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2*b^2*(d*x+c)^(1/2)/d^3-2/3*(6*b*d*x+a*d+5*b*c)*(a*d-b*c)/d^3/(d*x+c)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.27 \[ \int \frac {(a+b x)^2}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, b^{2} d^{2} x^{2} + 8 \, b^{2} c^{2} - 4 \, a b c d - a^{2} d^{2} + 6 \, {\left (2 \, b^{2} c d - a b d^{2}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

[In]

integrate((b*x+a)^2/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

2/3*(3*b^2*d^2*x^2 + 8*b^2*c^2 - 4*a*b*c*d - a^2*d^2 + 6*(2*b^2*c*d - a*b*d^2)*x)*sqrt(d*x + c)/(d^5*x^2 + 2*c
*d^4*x + c^2*d^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (61) = 122\).

Time = 0.46 (sec) , antiderivative size = 265, normalized size of antiderivative = 3.96 \[ \int \frac {(a+b x)^2}{(c+d x)^{5/2}} \, dx=\begin {cases} - \frac {2 a^{2} d^{2}}{3 c d^{3} \sqrt {c + d x} + 3 d^{4} x \sqrt {c + d x}} - \frac {8 a b c d}{3 c d^{3} \sqrt {c + d x} + 3 d^{4} x \sqrt {c + d x}} - \frac {12 a b d^{2} x}{3 c d^{3} \sqrt {c + d x} + 3 d^{4} x \sqrt {c + d x}} + \frac {16 b^{2} c^{2}}{3 c d^{3} \sqrt {c + d x} + 3 d^{4} x \sqrt {c + d x}} + \frac {24 b^{2} c d x}{3 c d^{3} \sqrt {c + d x} + 3 d^{4} x \sqrt {c + d x}} + \frac {6 b^{2} d^{2} x^{2}}{3 c d^{3} \sqrt {c + d x} + 3 d^{4} x \sqrt {c + d x}} & \text {for}\: d \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((b*x+a)**2/(d*x+c)**(5/2),x)

[Out]

Piecewise((-2*a**2*d**2/(3*c*d**3*sqrt(c + d*x) + 3*d**4*x*sqrt(c + d*x)) - 8*a*b*c*d/(3*c*d**3*sqrt(c + d*x)
+ 3*d**4*x*sqrt(c + d*x)) - 12*a*b*d**2*x/(3*c*d**3*sqrt(c + d*x) + 3*d**4*x*sqrt(c + d*x)) + 16*b**2*c**2/(3*
c*d**3*sqrt(c + d*x) + 3*d**4*x*sqrt(c + d*x)) + 24*b**2*c*d*x/(3*c*d**3*sqrt(c + d*x) + 3*d**4*x*sqrt(c + d*x
)) + 6*b**2*d**2*x**2/(3*c*d**3*sqrt(c + d*x) + 3*d**4*x*sqrt(c + d*x)), Ne(d, 0)), ((a**2*x + a*b*x**2 + b**2
*x**3/3)/c**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b x)^2}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, \sqrt {d x + c} b^{2}}{d^{2}} - \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2} - 6 \, {\left (b^{2} c - a b d\right )} {\left (d x + c\right )}}{{\left (d x + c\right )}^{\frac {3}{2}} d^{2}}\right )}}{3 \, d} \]

[In]

integrate((b*x+a)^2/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b x)^2}{(c+d x)^{5/2}} \, dx=\frac {2 \, \sqrt {d x + c} b^{2}}{d^{3}} + \frac {2 \, {\left (6 \, {\left (d x + c\right )} b^{2} c - b^{2} c^{2} - 6 \, {\left (d x + c\right )} a b d + 2 \, a b c d - a^{2} d^{2}\right )}}{3 \, {\left (d x + c\right )}^{\frac {3}{2}} d^{3}} \]

[In]

integrate((b*x+a)^2/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b x)^2}{(c+d x)^{5/2}} \, dx=\frac {6\,b^2\,{\left (c+d\,x\right )}^2-2\,a^2\,d^2-2\,b^2\,c^2+12\,b^2\,c\,\left (c+d\,x\right )-12\,a\,b\,d\,\left (c+d\,x\right )+4\,a\,b\,c\,d}{3\,d^3\,{\left (c+d\,x\right )}^{3/2}} \]

[In]

int((a + b*x)^2/(c + d*x)^(5/2),x)

[Out]

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