\(\int (a+b x) (c+d x)^{3/2} \, dx\) [1391]

Optimal result

Integrand size = 15, antiderivative size = 42 \[ \int (a+b x) (c+d x)^{3/2} \, dx=-\frac {2 (b c-a d) (c+d x)^{5/2}}{5 d^2}+\frac {2 b (c+d x)^{7/2}}{7 d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int (a+b x) (c+d x)^{3/2} \, dx=\frac {2 b (c+d x)^{7/2}}{7 d^2}-\frac {2 (c+d x)^{5/2} (b c-a d)}{5 d^2} \]

[In]

Int[(a + b*x)*(c + d*x)^(3/2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71 \[ \int (a+b x) (c+d x)^{3/2} \, dx=\frac {2 (c+d x)^{5/2} (-2 b c+7 a d+5 b d x)}{35 d^2} \]

[In]

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

[Out]

(2*(c + d*x)^(5/2)*(-2*b*c + 7*a*d + 5*b*d*x))/(35*d^2)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.64

method	result	size
gosper	\(\frac {2 \left (d x +c \right )^{\frac {5}{2}} \left (5 b d x +7 a d -2 b c \right )}{35 d^{2}}\)	\(27\)
pseudoelliptic	\(\frac {2 \left (\left (5 b x +7 a \right ) d -2 b c \right ) \left (d x +c \right )^{\frac {5}{2}}}{35 d^{2}}\)	\(28\)
derivativedivides	\(\frac {\frac {2 b \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a d -b c \right ) \left (d x +c \right )^{\frac {5}{2}}}{5}}{d^{2}}\)	\(34\)
default	\(\frac {\frac {2 b \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {2 \left (a d -b c \right ) \left (d x +c \right )^{\frac {5}{2}}}{5}}{d^{2}}\)	\(34\)
trager	\(\frac {2 \left (5 b \,d^{3} x^{3}+7 a \,d^{3} x^{2}+8 b c \,d^{2} x^{2}+14 a c \,d^{2} x +b \,c^{2} d x +7 a \,c^{2} d -2 b \,c^{3}\right ) \sqrt {d x +c}}{35 d^{2}}\)	\(70\)
risch	\(\frac {2 \left (5 b \,d^{3} x^{3}+7 a \,d^{3} x^{2}+8 b c \,d^{2} x^{2}+14 a c \,d^{2} x +b \,c^{2} d x +7 a \,c^{2} d -2 b \,c^{3}\right ) \sqrt {d x +c}}{35 d^{2}}\)	\(70\)

[In]

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

[Out]

2/35*(d*x+c)^(5/2)*(5*b*d*x+7*a*d-2*b*c)/d^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (34) = 68\).

Time = 0.22 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.64 \[ \int (a+b x) (c+d x)^{3/2} \, dx=\frac {2 \, {\left (5 \, b d^{3} x^{3} - 2 \, b c^{3} + 7 \, a c^{2} d + {\left (8 \, b c d^{2} + 7 \, a d^{3}\right )} x^{2} + {\left (b c^{2} d + 14 \, a c d^{2}\right )} x\right )} \sqrt {d x + c}}{35 \, d^{2}} \]

[In]

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

[Out]

2/35*(5*b*d^3*x^3 - 2*b*c^3 + 7*a*c^2*d + (8*b*c*d^2 + 7*a*d^3)*x^2 + (b*c^2*d + 14*a*c*d^2)*x)*sqrt(d*x + c)/
d^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (41) = 82\).

Time = 0.22 (sec) , antiderivative size = 146, normalized size of antiderivative = 3.48 \[ \int (a+b x) (c+d x)^{3/2} \, dx=\begin {cases} \frac {2 a c^{2} \sqrt {c + d x}}{5 d} + \frac {4 a c x \sqrt {c + d x}}{5} + \frac {2 a d x^{2} \sqrt {c + d x}}{5} - \frac {4 b c^{3} \sqrt {c + d x}}{35 d^{2}} + \frac {2 b c^{2} x \sqrt {c + d x}}{35 d} + \frac {16 b c x^{2} \sqrt {c + d x}}{35} + \frac {2 b d x^{3} \sqrt {c + d x}}{7} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (a x + \frac {b x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*a*c**2*sqrt(c + d*x)/(5*d) + 4*a*c*x*sqrt(c + d*x)/5 + 2*a*d*x**2*sqrt(c + d*x)/5 - 4*b*c**3*sqrt
(c + d*x)/(35*d**2) + 2*b*c**2*x*sqrt(c + d*x)/(35*d) + 16*b*c*x**2*sqrt(c + d*x)/35 + 2*b*d*x**3*sqrt(c + d*x
)/7, Ne(d, 0)), (c**(3/2)*(a*x + b*x**2/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79 \[ \int (a+b x) (c+d x)^{3/2} \, dx=\frac {2 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} b - 7 \, {\left (b c - a d\right )} {\left (d x + c\right )}^{\frac {5}{2}}\right )}}{35 \, d^{2}} \]

[In]

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

[Out]

2/35*(5*(d*x + c)^(7/2)*b - 7*(b*c - a*d)*(d*x + c)^(5/2))/d^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (34) = 68\).

Time = 0.30 (sec) , antiderivative size = 192, normalized size of antiderivative = 4.57 \[ \int (a+b x) (c+d x)^{3/2} \, dx=\frac {2 \, {\left (105 \, \sqrt {d x + c} a c^{2} + 70 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a c + \frac {35 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} b c^{2}}{d} + 7 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a + \frac {14 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} b c}{d} + \frac {3 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} b}{d}\right )}}{105 \, d} \]

[In]

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

[Out]

2/105*(105*sqrt(d*x + c)*a*c^2 + 70*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a*c + 35*((d*x + c)^(3/2) - 3*sqrt(d
*x + c)*c)*b*c^2/d + 7*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a + 14*(3*(d*x + c)^(
5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*b*c/d + 3*(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/d)/d

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.69 \[ \int (a+b x) (c+d x)^{3/2} \, dx=\frac {2\,{\left (c+d\,x\right )}^{5/2}\,\left (7\,a\,d-7\,b\,c+5\,b\,\left (c+d\,x\right )\right )}{35\,d^2} \]

[In]

int((a + b*x)*(c + d*x)^(3/2),x)

[Out]

(2*(c + d*x)^(5/2)*(7*a*d - 7*b*c + 5*b*(c + d*x)))/(35*d^2)