\(\int (a+b x)^2 (c+d x)^{3/2} \, dx\) [1390]

Optimal result

Integrand size = 17, antiderivative size = 71 \[ \int (a+b x)^2 (c+d x)^{3/2} \, dx=\frac {2 (b c-a d)^2 (c+d x)^{5/2}}{5 d^3}-\frac {4 b (b c-a d) (c+d x)^{7/2}}{7 d^3}+\frac {2 b^2 (c+d x)^{9/2}}{9 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, 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} \[ \int (a+b x)^2 (c+d x)^{3/2} \, dx=-\frac {4 b (c+d x)^{7/2} (b c-a d)}{7 d^3}+\frac {2 (c+d x)^{5/2} (b c-a d)^2}{5 d^3}+\frac {2 b^2 (c+d x)^{9/2}}{9 d^3} \]

[In]

Int[(a + b*x)^2*(c + d*x)^(3/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.86 \[ \int (a+b x)^2 (c+d x)^{3/2} \, dx=\frac {2 (c+d x)^{5/2} \left (63 a^2 d^2+18 a b d (-2 c+5 d x)+b^2 \left (8 c^2-20 c d x+35 d^2 x^2\right )\right )}{315 d^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(\frac {2 \left (\left (\frac {5}{9} b^{2} x^{2}+\frac {10}{7} a b x +a^{2}\right ) d^{2}-\frac {4 b \left (\frac {5 b x}{9}+a \right ) c d}{7}+\frac {8 b^{2} c^{2}}{63}\right ) \left (d x +c \right )^{\frac {5}{2}}}{5 d^{3}}\)	\(54\)
derivativedivides	\(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {9}{2}}}{9}+\frac {4 \left (a d -b c \right ) b \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}}{d^{3}}\)	\(56\)
default	\(\frac {\frac {2 b^{2} \left (d x +c \right )^{\frac {9}{2}}}{9}+\frac {4 \left (a d -b c \right ) b \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a d -b c \right )^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}}{d^{3}}\)	\(56\)
gosper	\(\frac {2 \left (d x +c \right )^{\frac {5}{2}} \left (35 d^{2} x^{2} b^{2}+90 x a b \,d^{2}-20 x \,b^{2} c d +63 a^{2} d^{2}-36 a b c d +8 b^{2} c^{2}\right )}{315 d^{3}}\)	\(63\)
trager	\(\frac {2 \left (35 b^{2} d^{4} x^{4}+90 a b \,d^{4} x^{3}+50 b^{2} c \,d^{3} x^{3}+63 a^{2} d^{4} x^{2}+144 a b c \,d^{3} x^{2}+3 b^{2} c^{2} d^{2} x^{2}+126 a^{2} c \,d^{3} x +18 a b \,c^{2} d^{2} x -4 b^{2} c^{3} d x +63 a^{2} c^{2} d^{2}-36 a b \,c^{3} d +8 b^{2} c^{4}\right ) \sqrt {d x +c}}{315 d^{3}}\)	\(141\)
risch	\(\frac {2 \left (35 b^{2} d^{4} x^{4}+90 a b \,d^{4} x^{3}+50 b^{2} c \,d^{3} x^{3}+63 a^{2} d^{4} x^{2}+144 a b c \,d^{3} x^{2}+3 b^{2} c^{2} d^{2} x^{2}+126 a^{2} c \,d^{3} x +18 a b \,c^{2} d^{2} x -4 b^{2} c^{3} d x +63 a^{2} c^{2} d^{2}-36 a b \,c^{3} d +8 b^{2} c^{4}\right ) \sqrt {d x +c}}{315 d^{3}}\)	\(141\)

[In]

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

[Out]

2/5*((5/9*b^2*x^2+10/7*a*b*x+a^2)*d^2-4/7*b*(5/9*b*x+a)*c*d+8/63*b^2*c^2)*(d*x+c)^(5/2)/d^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 = 0.22 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.93 \[ \int (a+b x)^2 (c+d x)^{3/2} \, dx=\frac {2 \, {\left (35 \, b^{2} d^{4} x^{4} + 8 \, b^{2} c^{4} - 36 \, a b c^{3} d + 63 \, a^{2} c^{2} d^{2} + 10 \, {\left (5 \, b^{2} c d^{3} + 9 \, a b d^{4}\right )} x^{3} + 3 \, {\left (b^{2} c^{2} d^{2} + 48 \, a b c d^{3} + 21 \, a^{2} d^{4}\right )} x^{2} - 2 \, {\left (2 \, b^{2} c^{3} d - 9 \, a b c^{2} d^{2} - 63 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x + c}}{315 \, d^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.48 \[ \int (a+b x)^2 (c+d x)^{3/2} \, dx=\begin {cases} \frac {2 \left (\frac {b^{2} \left (c + d x\right )^{\frac {9}{2}}}{9 d^{2}} + \frac {\left (c + d x\right )^{\frac {7}{2}} \cdot \left (2 a b d - 2 b^{2} c\right )}{7 d^{2}} + \frac {\left (c + d x\right )^{\frac {5}{2}} \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{5 d^{2}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (\begin {cases} a^{2} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{3}}{3 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(b**2*(c + d*x)**(9/2)/(9*d**2) + (c + d*x)**(7/2)*(2*a*b*d - 2*b**2*c)/(7*d**2) + (c + d*x)**(5/
2)*(a**2*d**2 - 2*a*b*c*d + b**2*c**2)/(5*d**2))/d, Ne(d, 0)), (c**(3/2)*Piecewise((a**2*x, Eq(b, 0)), ((a + b
*x)**3/(3*b), True)), True))

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

2/315*(35*(d*x + c)^(9/2)*b^2 - 90*(b^2*c - a*b*d)*(d*x + c)^(7/2) + 63*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(d*x +
c)^(5/2))/d^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 = 0.32 (sec) , antiderivative size = 360, normalized size of antiderivative = 5.07 \[ \int (a+b x)^2 (c+d x)^{3/2} \, dx=\frac {2 \, {\left (315 \, \sqrt {d x + c} a^{2} c^{2} + 210 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a^{2} c + \frac {210 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a b c^{2}}{d} + 21 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a^{2} + \frac {21 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} b^{2} c^{2}}{d^{2}} + \frac {84 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a b c}{d} + \frac {18 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} b^{2} c}{d^{2}} + \frac {18 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} a b}{d} + \frac {{\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} - 180 \, {\left (d x + c\right )}^{\frac {7}{2}} c + 378 \, {\left (d x + c\right )}^{\frac {5}{2}} c^{2} - 420 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {d x + c} c^{4}\right )} b^{2}}{d^{2}}\right )}}{315 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int (a+b x)^2 (c+d x)^{3/2} \, dx=\frac {2\,{\left (c+d\,x\right )}^{5/2}\,\left (35\,b^2\,{\left (c+d\,x\right )}^2+63\,a^2\,d^2+63\,b^2\,c^2-90\,b^2\,c\,\left (c+d\,x\right )+90\,a\,b\,d\,\left (c+d\,x\right )-126\,a\,b\,c\,d\right )}{315\,d^3} \]

[In]

int((a + b*x)^2*(c + d*x)^(3/2),x)

[Out]

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