\(\int (d+e x)^{3/2} (a^2+2 a b x+b^2 x^2) \, dx\) [1623]

Optimal result

Integrand size = 26, antiderivative size = 71 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 (b d-a e)^2 (d+e x)^{5/2}}{5 e^3}-\frac {4 b (b d-a e) (d+e x)^{7/2}}{7 e^3}+\frac {2 b^2 (d+e x)^{9/2}}{9 e^3} \]

[Out]

2/5*(-a*e+b*d)^2*(e*x+d)^(5/2)/e^3-4/7*b*(-a*e+b*d)*(e*x+d)^(7/2)/e^3+2/9*b^2*(e*x+d)^(9/2)/e^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 45} \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=-\frac {4 b (d+e x)^{7/2} (b d-a e)}{7 e^3}+\frac {2 (d+e x)^{5/2} (b d-a e)^2}{5 e^3}+\frac {2 b^2 (d+e x)^{9/2}}{9 e^3} \]

[In]

Int[(d + e*x)^(3/2)*(a^2 + 2*a*b*x + b^2*x^2),x]

[Out]

(2*(b*d - a*e)^2*(d + e*x)^(5/2))/(5*e^3) - (4*b*(b*d - a*e)*(d + e*x)^(7/2))/(7*e^3) + (2*b^2*(d + e*x)^(9/2)
)/(9*e^3)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 (a+b x)^2 (d+e x)^{3/2} \, dx \\ & = \int \left (\frac {(-b d+a e)^2 (d+e x)^{3/2}}{e^2}-\frac {2 b (b d-a e) (d+e x)^{5/2}}{e^2}+\frac {b^2 (d+e x)^{7/2}}{e^2}\right ) \, dx \\ & = \frac {2 (b d-a e)^2 (d+e x)^{5/2}}{5 e^3}-\frac {4 b (b d-a e) (d+e x)^{7/2}}{7 e^3}+\frac {2 b^2 (d+e x)^{9/2}}{9 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 (d+e x)^{5/2} \left (63 a^2 e^2+18 a b e (-2 d+5 e x)+b^2 \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3} \]

[In]

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

[Out]

(2*(d + e*x)^(5/2)*(63*a^2*e^2 + 18*a*b*e*(-2*d + 5*e*x) + b^2*(8*d^2 - 20*d*e*x + 35*e^2*x^2)))/(315*e^3)

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(\frac {2 \left (\left (\frac {5}{9} b^{2} x^{2}+\frac {10}{7} a b x +a^{2}\right ) e^{2}-\frac {4 b \left (\frac {5 b x}{9}+a \right ) d e}{7}+\frac {8 b^{2} d^{2}}{63}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5 e^{3}}\)	\(54\)
gosper	\(\frac {2 \left (e x +d \right )^{\frac {5}{2}} \left (35 x^{2} b^{2} e^{2}+90 x a b \,e^{2}-20 b^{2} d e x +63 a^{2} e^{2}-36 a b d e +8 b^{2} d^{2}\right )}{315 e^{3}}\)	\(63\)
derivativedivides	\(\frac {\frac {2 b^{2} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (2 a e b -2 b^{2} d \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{3}}\)	\(70\)
default	\(\frac {\frac {2 b^{2} \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (2 a e b -2 b^{2} d \right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}}{e^{3}}\)	\(70\)
trager	\(\frac {2 \left (35 b^{2} e^{4} x^{4}+90 a b \,e^{4} x^{3}+50 b^{2} d \,e^{3} x^{3}+63 a^{2} e^{4} x^{2}+144 a b d \,e^{3} x^{2}+3 b^{2} d^{2} e^{2} x^{2}+126 a^{2} d \,e^{3} x +18 a b \,d^{2} e^{2} x -4 b^{2} d^{3} e x +63 a^{2} d^{2} e^{2}-36 a b \,d^{3} e +8 b^{2} d^{4}\right ) \sqrt {e x +d}}{315 e^{3}}\)	\(141\)
risch	\(\frac {2 \left (35 b^{2} e^{4} x^{4}+90 a b \,e^{4} x^{3}+50 b^{2} d \,e^{3} x^{3}+63 a^{2} e^{4} x^{2}+144 a b d \,e^{3} x^{2}+3 b^{2} d^{2} e^{2} x^{2}+126 a^{2} d \,e^{3} x +18 a b \,d^{2} e^{2} x -4 b^{2} d^{3} e x +63 a^{2} d^{2} e^{2}-36 a b \,d^{3} e +8 b^{2} d^{4}\right ) \sqrt {e x +d}}{315 e^{3}}\)	\(141\)

[In]

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

[Out]

2/5*((5/9*b^2*x^2+10/7*a*b*x+a^2)*e^2-4/7*b*(5/9*b*x+a)*d*e+8/63*b^2*d^2)*(e*x+d)^(5/2)/e^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (59) = 118\).

Time = 0.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.93 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (35 \, b^{2} e^{4} x^{4} + 8 \, b^{2} d^{4} - 36 \, a b d^{3} e + 63 \, a^{2} d^{2} e^{2} + 10 \, {\left (5 \, b^{2} d e^{3} + 9 \, a b e^{4}\right )} x^{3} + 3 \, {\left (b^{2} d^{2} e^{2} + 48 \, a b d e^{3} + 21 \, a^{2} e^{4}\right )} x^{2} - 2 \, {\left (2 \, b^{2} d^{3} e - 9 \, a b d^{2} e^{2} - 63 \, a^{2} d e^{3}\right )} x\right )} \sqrt {e x + d}}{315 \, e^{3}} \]

[In]

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

[Out]

2/315*(35*b^2*e^4*x^4 + 8*b^2*d^4 - 36*a*b*d^3*e + 63*a^2*d^2*e^2 + 10*(5*b^2*d*e^3 + 9*a*b*e^4)*x^3 + 3*(b^2*
d^2*e^2 + 48*a*b*d*e^3 + 21*a^2*e^4)*x^2 - 2*(2*b^2*d^3*e - 9*a*b*d^2*e^2 - 63*a^2*d*e^3)*x)*sqrt(e*x + d)/e^3

Sympy [A] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.55 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\begin {cases} \frac {2 \left (\frac {b^{2} \left (d + e x\right )^{\frac {9}{2}}}{9 e^{2}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \cdot \left (2 a b e - 2 b^{2} d\right )}{7 e^{2}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (a^{2} e^{2} - 2 a b d e + b^{2} d^{2}\right )}{5 e^{2}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {3}{2}} \left (a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(b**2*(d + e*x)**(9/2)/(9*e**2) + (d + e*x)**(7/2)*(2*a*b*e - 2*b**2*d)/(7*e**2) + (d + e*x)**(5/
2)*(a**2*e**2 - 2*a*b*d*e + b**2*d**2)/(5*e**2))/e, Ne(e, 0)), (d**(3/2)*(a**2*x + a*b*x**2 + b**2*x**3/3), Tr
ue))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} b^{2} - 90 \, {\left (b^{2} d - a b e\right )} {\left (e x + d\right )}^{\frac {7}{2}} + 63 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} {\left (e x + d\right )}^{\frac {5}{2}}\right )}}{315 \, e^{3}} \]

[In]

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

[Out]

2/315*(35*(e*x + d)^(9/2)*b^2 - 90*(b^2*d - a*b*e)*(e*x + d)^(7/2) + 63*(b^2*d^2 - 2*a*b*d*e + a^2*e^2)*(e*x +
d)^(5/2))/e^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (59) = 118\).

Time = 0.29 (sec) , antiderivative size = 360, normalized size of antiderivative = 5.07 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2 \, {\left (315 \, \sqrt {e x + d} a^{2} d^{2} + 210 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{2} d + \frac {210 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a b d^{2}}{e} + 21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{2} + \frac {21 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} b^{2} d^{2}}{e^{2}} + \frac {84 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a b d}{e} + \frac {18 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} b^{2} d}{e^{2}} + \frac {18 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a b}{e} + \frac {{\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} b^{2}}{e^{2}}\right )}}{315 \, e} \]

[In]

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

[Out]

2/315*(315*sqrt(e*x + d)*a^2*d^2 + 210*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^2*d + 210*((e*x + d)^(3/2) - 3*
sqrt(e*x + d)*d)*a*b*d^2/e + 21*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a^2 + 21*(3*
(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*b^2*d^2/e^2 + 84*(3*(e*x + d)^(5/2) - 10*(e*x +
d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*b*d/e + 18*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2
)*d^2 - 35*sqrt(e*x + d)*d^3)*b^2*d/e^2 + 18*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^
2 - 35*sqrt(e*x + d)*d^3)*a*b/e + (35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*
(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*b^2/e^2)/e

Mupad [B] (verification not implemented)

Time = 9.58 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int (d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right ) \, dx=\frac {2\,{\left (d+e\,x\right )}^{5/2}\,\left (35\,b^2\,{\left (d+e\,x\right )}^2+63\,a^2\,e^2+63\,b^2\,d^2-90\,b^2\,d\,\left (d+e\,x\right )+90\,a\,b\,e\,\left (d+e\,x\right )-126\,a\,b\,d\,e\right )}{315\,e^3} \]

[In]

int((d + e*x)^(3/2)*(a^2 + b^2*x^2 + 2*a*b*x),x)

[Out]

(2*(d + e*x)^(5/2)*(35*b^2*(d + e*x)^2 + 63*a^2*e^2 + 63*b^2*d^2 - 90*b^2*d*(d + e*x) + 90*a*b*e*(d + e*x) - 1
26*a*b*d*e))/(315*e^3)