\(\int (a+b x)^3 \sqrt {c+d x} \, dx\) [1377]

Optimal result

Integrand size = 17, antiderivative size = 100 \[ \int (a+b x)^3 \sqrt {c+d x} \, dx=-\frac {2 (b c-a d)^3 (c+d x)^{3/2}}{3 d^4}+\frac {6 b (b c-a d)^2 (c+d x)^{5/2}}{5 d^4}-\frac {6 b^2 (b c-a d) (c+d x)^{7/2}}{7 d^4}+\frac {2 b^3 (c+d x)^{9/2}}{9 d^4} \]

[Out]

-2/3*(-a*d+b*c)^3*(d*x+c)^(3/2)/d^4+6/5*b*(-a*d+b*c)^2*(d*x+c)^(5/2)/d^4-6/7*b^2*(-a*d+b*c)*(d*x+c)^(7/2)/d^4+
2/9*b^3*(d*x+c)^(9/2)/d^4

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 100, 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} \[ \int (a+b x)^3 \sqrt {c+d x} \, dx=-\frac {6 b^2 (c+d x)^{7/2} (b c-a d)}{7 d^4}+\frac {6 b (c+d x)^{5/2} (b c-a d)^2}{5 d^4}-\frac {2 (c+d x)^{3/2} (b c-a d)^3}{3 d^4}+\frac {2 b^3 (c+d x)^{9/2}}{9 d^4} \]

[In]

Int[(a + b*x)^3*Sqrt[c + d*x],x]

[Out]

(-2*(b*c - a*d)^3*(c + d*x)^(3/2))/(3*d^4) + (6*b*(b*c - a*d)^2*(c + d*x)^(5/2))/(5*d^4) - (6*b^2*(b*c - a*d)*
(c + d*x)^(7/2))/(7*d^4) + (2*b^3*(c + d*x)^(9/2))/(9*d^4)

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)^3 \sqrt {c+d x}}{d^3}+\frac {3 b (b c-a d)^2 (c+d x)^{3/2}}{d^3}-\frac {3 b^2 (b c-a d) (c+d x)^{5/2}}{d^3}+\frac {b^3 (c+d x)^{7/2}}{d^3}\right ) \, dx \\ & = -\frac {2 (b c-a d)^3 (c+d x)^{3/2}}{3 d^4}+\frac {6 b (b c-a d)^2 (c+d x)^{5/2}}{5 d^4}-\frac {6 b^2 (b c-a d) (c+d x)^{7/2}}{7 d^4}+\frac {2 b^3 (c+d x)^{9/2}}{9 d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.02 \[ \int (a+b x)^3 \sqrt {c+d x} \, dx=\frac {2 (c+d x)^{3/2} \left (105 a^3 d^3+63 a^2 b d^2 (-2 c+3 d x)+9 a b^2 d \left (8 c^2-12 c d x+15 d^2 x^2\right )+b^3 \left (-16 c^3+24 c^2 d x-30 c d^2 x^2+35 d^3 x^3\right )\right )}{315 d^4} \]

[In]

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

[Out]

(2*(c + d*x)^(3/2)*(105*a^3*d^3 + 63*a^2*b*d^2*(-2*c + 3*d*x) + 9*a*b^2*d*(8*c^2 - 12*c*d*x + 15*d^2*x^2) + b^
3*(-16*c^3 + 24*c^2*d*x - 30*c*d^2*x^2 + 35*d^3*x^3)))/(315*d^4)

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {\frac {2 b^{3} \left (d x +c \right )^{\frac {9}{2}}}{9}+\frac {6 \left (a d -b c \right ) b^{2} \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {6 \left (a d -b c \right )^{2} b \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a d -b c \right )^{3} \left (d x +c \right )^{\frac {3}{2}}}{3}}{d^{4}}\)	\(78\)
default	\(\frac {\frac {2 b^{3} \left (d x +c \right )^{\frac {9}{2}}}{9}+\frac {6 \left (a d -b c \right ) b^{2} \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {6 \left (a d -b c \right )^{2} b \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 \left (a d -b c \right )^{3} \left (d x +c \right )^{\frac {3}{2}}}{3}}{d^{4}}\)	\(78\)
pseudoelliptic	\(\frac {2 \left (\left (\frac {1}{3} b^{3} x^{3}+\frac {9}{7} a \,b^{2} x^{2}+\frac {9}{5} a^{2} b x +a^{3}\right ) d^{3}-\frac {6 \left (\frac {5}{21} b^{2} x^{2}+\frac {6}{7} a b x +a^{2}\right ) b c \,d^{2}}{5}+\frac {24 \left (\frac {b x}{3}+a \right ) b^{2} c^{2} d}{35}-\frac {16 b^{3} c^{3}}{105}\right ) \left (d x +c \right )^{\frac {3}{2}}}{3 d^{4}}\)	\(93\)
gosper	\(\frac {2 \left (d x +c \right )^{\frac {3}{2}} \left (35 d^{3} x^{3} b^{3}+135 x^{2} a \,b^{2} d^{3}-30 x^{2} b^{3} c \,d^{2}+189 x \,a^{2} b \,d^{3}-108 x a \,b^{2} c \,d^{2}+24 x \,b^{3} c^{2} d +105 a^{3} d^{3}-126 a^{2} b c \,d^{2}+72 a \,b^{2} c^{2} d -16 b^{3} c^{3}\right )}{315 d^{4}}\)	\(116\)
trager	\(\frac {2 \left (35 b^{3} d^{4} x^{4}+135 a \,b^{2} d^{4} x^{3}+5 b^{3} c \,d^{3} x^{3}+189 a^{2} b \,d^{4} x^{2}+27 a \,b^{2} c \,d^{3} x^{2}-6 b^{3} c^{2} d^{2} x^{2}+105 a^{3} d^{4} x +63 a^{2} b c \,d^{3} x -36 a \,b^{2} c^{2} d^{2} x +8 b^{3} c^{3} d x +105 a^{3} c \,d^{3}-126 a^{2} b \,c^{2} d^{2}+72 a \,b^{2} c^{3} d -16 b^{3} c^{4}\right ) \sqrt {d x +c}}{315 d^{4}}\)	\(170\)
risch	\(\frac {2 \left (35 b^{3} d^{4} x^{4}+135 a \,b^{2} d^{4} x^{3}+5 b^{3} c \,d^{3} x^{3}+189 a^{2} b \,d^{4} x^{2}+27 a \,b^{2} c \,d^{3} x^{2}-6 b^{3} c^{2} d^{2} x^{2}+105 a^{3} d^{4} x +63 a^{2} b c \,d^{3} x -36 a \,b^{2} c^{2} d^{2} x +8 b^{3} c^{3} d x +105 a^{3} c \,d^{3}-126 a^{2} b \,c^{2} d^{2}+72 a \,b^{2} c^{3} d -16 b^{3} c^{4}\right ) \sqrt {d x +c}}{315 d^{4}}\)	\(170\)

[In]

int((b*x+a)^3*(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/d^4*(1/9*b^3*(d*x+c)^(9/2)+3/7*(a*d-b*c)*b^2*(d*x+c)^(7/2)+3/5*(a*d-b*c)^2*b*(d*x+c)^(5/2)+1/3*(a*d-b*c)^3*(
d*x+c)^(3/2))

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.64 \[ \int (a+b x)^3 \sqrt {c+d x} \, dx=\frac {2 \, {\left (35 \, b^{3} d^{4} x^{4} - 16 \, b^{3} c^{4} + 72 \, a b^{2} c^{3} d - 126 \, a^{2} b c^{2} d^{2} + 105 \, a^{3} c d^{3} + 5 \, {\left (b^{3} c d^{3} + 27 \, a b^{2} d^{4}\right )} x^{3} - 3 \, {\left (2 \, b^{3} c^{2} d^{2} - 9 \, a b^{2} c d^{3} - 63 \, a^{2} b d^{4}\right )} x^{2} + {\left (8 \, b^{3} c^{3} d - 36 \, a b^{2} c^{2} d^{2} + 63 \, a^{2} b c d^{3} + 105 \, a^{3} d^{4}\right )} x\right )} \sqrt {d x + c}}{315 \, d^{4}} \]

[In]

integrate((b*x+a)^3*(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*b^3*d^4*x^4 - 16*b^3*c^4 + 72*a*b^2*c^3*d - 126*a^2*b*c^2*d^2 + 105*a^3*c*d^3 + 5*(b^3*c*d^3 + 27*a*
b^2*d^4)*x^3 - 3*(2*b^3*c^2*d^2 - 9*a*b^2*c*d^3 - 63*a^2*b*d^4)*x^2 + (8*b^3*c^3*d - 36*a*b^2*c^2*d^2 + 63*a^2
*b*c*d^3 + 105*a^3*d^4)*x)*sqrt(d*x + c)/d^4

Sympy [A] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.67 \[ \int (a+b x)^3 \sqrt {c+d x} \, dx=\begin {cases} \frac {2 \left (\frac {b^{3} \left (c + d x\right )^{\frac {9}{2}}}{9 d^{3}} + \frac {\left (c + d x\right )^{\frac {7}{2}} \cdot \left (3 a b^{2} d - 3 b^{3} c\right )}{7 d^{3}} + \frac {\left (c + d x\right )^{\frac {5}{2}} \cdot \left (3 a^{2} b d^{2} - 6 a b^{2} c d + 3 b^{3} c^{2}\right )}{5 d^{3}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \left (a^{3} d^{3} - 3 a^{2} b c d^{2} + 3 a b^{2} c^{2} d - b^{3} c^{3}\right )}{3 d^{3}}\right )}{d} & \text {for}\: d \neq 0 \\\sqrt {c} \left (\begin {cases} a^{3} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{4}}{4 b} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((b*x+a)**3*(d*x+c)**(1/2),x)

[Out]

Piecewise((2*(b**3*(c + d*x)**(9/2)/(9*d**3) + (c + d*x)**(7/2)*(3*a*b**2*d - 3*b**3*c)/(7*d**3) + (c + d*x)**
(5/2)*(3*a**2*b*d**2 - 6*a*b**2*c*d + 3*b**3*c**2)/(5*d**3) + (c + d*x)**(3/2)*(a**3*d**3 - 3*a**2*b*c*d**2 +
3*a*b**2*c**2*d - b**3*c**3)/(3*d**3))/d, Ne(d, 0)), (sqrt(c)*Piecewise((a**3*x, Eq(b, 0)), ((a + b*x)**4/(4*b
), True)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18 \[ \int (a+b x)^3 \sqrt {c+d x} \, dx=\frac {2 \, {\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} b^{3} - 135 \, {\left (b^{3} c - a b^{2} d\right )} {\left (d x + c\right )}^{\frac {7}{2}} + 189 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} {\left (d x + c\right )}^{\frac {5}{2}} - 105 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left (d x + c\right )}^{\frac {3}{2}}\right )}}{315 \, d^{4}} \]

[In]

integrate((b*x+a)^3*(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

2/315*(35*(d*x + c)^(9/2)*b^3 - 135*(b^3*c - a*b^2*d)*(d*x + c)^(7/2) + 189*(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2
)*(d*x + c)^(5/2) - 105*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(d*x + c)^(3/2))/d^4

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (84) = 168\).

Time = 0.34 (sec) , antiderivative size = 322, normalized size of antiderivative = 3.22 \[ \int (a+b x)^3 \sqrt {c+d x} \, dx=\frac {2 \, {\left (315 \, \sqrt {d x + c} a^{3} c + 105 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a^{3} + \frac {315 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a^{2} b c}{d} + \frac {63 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a b^{2} c}{d^{2}} + \frac {63 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a^{2} b}{d} + \frac {9 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} b^{3} c}{d^{3}} + \frac {27 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} a b^{2}}{d^{2}} + \frac {{\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} - 180 \, {\left (d x + c\right )}^{\frac {7}{2}} c + 378 \, {\left (d x + c\right )}^{\frac {5}{2}} c^{2} - 420 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {d x + c} c^{4}\right )} b^{3}}{d^{3}}\right )}}{315 \, d} \]

[In]

integrate((b*x+a)^3*(d*x+c)^(1/2),x, algorithm="giac")

[Out]

2/315*(315*sqrt(d*x + c)*a^3*c + 105*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a^3 + 315*((d*x + c)^(3/2) - 3*sqrt
(d*x + c)*c)*a^2*b*c/d + 63*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a*b^2*c/d^2 + 63
*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a^2*b/d + 9*(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^3*c/d^3 + 27*(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^2/d^2 + (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^3/d^3)/d

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int (a+b x)^3 \sqrt {c+d x} \, dx=\frac {2\,b^3\,{\left (c+d\,x\right )}^{9/2}}{9\,d^4}-\frac {\left (6\,b^3\,c-6\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^{7/2}}{7\,d^4}+\frac {2\,{\left (a\,d-b\,c\right )}^3\,{\left (c+d\,x\right )}^{3/2}}{3\,d^4}+\frac {6\,b\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^{5/2}}{5\,d^4} \]

[In]

int((a + b*x)^3*(c + d*x)^(1/2),x)

[Out]

(2*b^3*(c + d*x)^(9/2))/(9*d^4) - ((6*b^3*c - 6*a*b^2*d)*(c + d*x)^(7/2))/(7*d^4) + (2*(a*d - b*c)^3*(c + d*x)
^(3/2))/(3*d^4) + (6*b*(a*d - b*c)^2*(c + d*x)^(5/2))/(5*d^4)