\(\int \frac {(a+b x)^2}{\sqrt {c+d x}} \, dx\) [1416]

Optimal result

Integrand size = 17, antiderivative size = 69 \[ \int \frac {(a+b x)^2}{\sqrt {c+d x}} \, dx=\frac {2 (b c-a d)^2 \sqrt {c+d x}}{d^3}-\frac {4 b (b c-a d) (c+d x)^{3/2}}{3 d^3}+\frac {2 b^2 (c+d x)^{5/2}}{5 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 69, 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} \[ \int \frac {(a+b x)^2}{\sqrt {c+d x}} \, dx=-\frac {4 b (c+d x)^{3/2} (b c-a d)}{3 d^3}+\frac {2 \sqrt {c+d x} (b c-a d)^2}{d^3}+\frac {2 b^2 (c+d x)^{5/2}}{5 d^3} \]

[In]

Int[(a + b*x)^2/Sqrt[c + d*x],x]

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b x)^2}{\sqrt {c+d x}} \, dx=\frac {2 \sqrt {c+d x} \left (15 a^2 d^2+10 a b d (-2 c+d x)+b^2 \left (8 c^2-4 c d x+3 d^2 x^2\right )\right )}{15 d^3} \]

[In]

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

[Out]

(2*Sqrt[c + d*x]*(15*a^2*d^2 + 10*a*b*d*(-2*c + d*x) + b^2*(8*c^2 - 4*c*d*x + 3*d^2*x^2)))/(15*d^3)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(\frac {2 \sqrt {d x +c}\, \left (\left (\frac {1}{5} b^{2} x^{2}+\frac {2}{3} a b x +a^{2}\right ) d^{2}-\frac {4 b \left (\frac {b x}{5}+a \right ) c d}{3}+\frac {8 b^{2} c^{2}}{15}\right )}{d^{3}}\)	\(54\)
derivativedivides	\(\frac {\frac {2 \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}+\frac {4 \left (a d -b c \right ) b \left (d x +c \right )^{\frac {3}{2}}}{3}+2 \left (a d -b c \right )^{2} \sqrt {d x +c}}{d^{3}}\)	\(55\)
default	\(\frac {\frac {2 \left (d x +c \right )^{\frac {5}{2}} b^{2}}{5}+\frac {4 \left (a d -b c \right ) b \left (d x +c \right )^{\frac {3}{2}}}{3}+2 \left (a d -b c \right )^{2} \sqrt {d x +c}}{d^{3}}\)	\(55\)
gosper	\(\frac {2 \sqrt {d x +c}\, \left (3 d^{2} x^{2} b^{2}+10 x a b \,d^{2}-4 x \,b^{2} c d +15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right )}{15 d^{3}}\)	\(63\)
trager	\(\frac {2 \sqrt {d x +c}\, \left (3 d^{2} x^{2} b^{2}+10 x a b \,d^{2}-4 x \,b^{2} c d +15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right )}{15 d^{3}}\)	\(63\)
risch	\(\frac {2 \sqrt {d x +c}\, \left (3 d^{2} x^{2} b^{2}+10 x a b \,d^{2}-4 x \,b^{2} c d +15 a^{2} d^{2}-20 a b c d +8 b^{2} c^{2}\right )}{15 d^{3}}\)	\(63\)

[In]

int((b*x+a)^2/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*(d*x+c)^(1/2)*((1/5*b^2*x^2+2/3*a*b*x+a^2)*d^2-4/3*b*(1/5*b*x+a)*c*d+8/15*b^2*c^2)/d^3

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x)^2}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (3 \, b^{2} d^{2} x^{2} + 8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2} - 2 \, {\left (2 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{15 \, d^{3}} \]

[In]

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

[Out]

2/15*(3*b^2*d^2*x^2 + 8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2 - 2*(2*b^2*c*d - 5*a*b*d^2)*x)*sqrt(d*x + c)/d^3

Sympy [A] (verification not implemented)

Time = 0.83 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.51 \[ \int \frac {(a+b x)^2}{\sqrt {c+d x}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{2} \left (c + d x\right )^{\frac {5}{2}}}{5 d^{2}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \cdot \left (2 a b d - 2 b^{2} c\right )}{3 d^{2}} + \frac {\sqrt {c + d x} \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{d^{2}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\begin {cases} a^{2} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{3}}{3 b} & \text {otherwise} \end {cases}}{\sqrt {c}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(b**2*(c + d*x)**(5/2)/(5*d**2) + (c + d*x)**(3/2)*(2*a*b*d - 2*b**2*c)/(3*d**2) + sqrt(c + d*x)*
(a**2*d**2 - 2*a*b*c*d + b**2*c**2)/d**2)/d, Ne(d, 0)), (Piecewise((a**2*x, Eq(b, 0)), ((a + b*x)**3/(3*b), Tr
ue))/sqrt(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b x)^2}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {d x + c} a^{2} + \frac {10 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a b}{d} + \frac {{\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} b^{2}}{d^{2}}\right )}}{15 \, d} \]

[In]

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

[Out]

2/15*(15*sqrt(d*x + c)*a^2 + 10*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a*b/d + (3*(d*x + c)^(5/2) - 10*(d*x + c
)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*b^2/d^2)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.19 \[ \int \frac {(a+b x)^2}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (15 \, \sqrt {d x + c} a^{2} + \frac {10 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a b}{d} + \frac {{\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} b^{2}}{d^{2}}\right )}}{15 \, d} \]

[In]

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

[Out]

2/15*(15*sqrt(d*x + c)*a^2 + 10*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a*b/d + (3*(d*x + c)^(5/2) - 10*(d*x + c
)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*b^2/d^2)/d

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b x)^2}{\sqrt {c+d x}} \, dx=\frac {2\,\sqrt {c+d\,x}\,\left (3\,b^2\,{\left (c+d\,x\right )}^2+15\,a^2\,d^2+15\,b^2\,c^2-10\,b^2\,c\,\left (c+d\,x\right )+10\,a\,b\,d\,\left (c+d\,x\right )-30\,a\,b\,c\,d\right )}{15\,d^3} \]

[In]

int((a + b*x)^2/(c + d*x)^(1/2),x)

[Out]

(2*(c + d*x)^(1/2)*(3*b^2*(c + d*x)^2 + 15*a^2*d^2 + 15*b^2*c^2 - 10*b^2*c*(c + d*x) + 10*a*b*d*(c + d*x) - 30
*a*b*c*d))/(15*d^3)