\(\int \frac {(a+b x)^4}{\sqrt {c+d x}} \, dx\) [1414]

Optimal result

Integrand size = 17, antiderivative size = 127 \[ \int \frac {(a+b x)^4}{\sqrt {c+d x}} \, dx=\frac {2 (b c-a d)^4 \sqrt {c+d x}}{d^5}-\frac {8 b (b c-a d)^3 (c+d x)^{3/2}}{3 d^5}+\frac {12 b^2 (b c-a d)^2 (c+d x)^{5/2}}{5 d^5}-\frac {8 b^3 (b c-a d) (c+d x)^{7/2}}{7 d^5}+\frac {2 b^4 (c+d x)^{9/2}}{9 d^5} \]

[Out]

-8/3*b*(-a*d+b*c)^3*(d*x+c)^(3/2)/d^5+12/5*b^2*(-a*d+b*c)^2*(d*x+c)^(5/2)/d^5-8/7*b^3*(-a*d+b*c)*(d*x+c)^(7/2)
/d^5+2/9*b^4*(d*x+c)^(9/2)/d^5+2*(-a*d+b*c)^4*(d*x+c)^(1/2)/d^5

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 127, 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} \[ \int \frac {(a+b x)^4}{\sqrt {c+d x}} \, dx=-\frac {8 b^3 (c+d x)^{7/2} (b c-a d)}{7 d^5}+\frac {12 b^2 (c+d x)^{5/2} (b c-a d)^2}{5 d^5}-\frac {8 b (c+d x)^{3/2} (b c-a d)^3}{3 d^5}+\frac {2 \sqrt {c+d x} (b c-a d)^4}{d^5}+\frac {2 b^4 (c+d x)^{9/2}}{9 d^5} \]

[In]

Int[(a + b*x)^4/Sqrt[c + d*x],x]

[Out]

(2*(b*c - a*d)^4*Sqrt[c + d*x])/d^5 - (8*b*(b*c - a*d)^3*(c + d*x)^(3/2))/(3*d^5) + (12*b^2*(b*c - a*d)^2*(c +
d*x)^(5/2))/(5*d^5) - (8*b^3*(b*c - a*d)*(c + d*x)^(7/2))/(7*d^5) + (2*b^4*(c + d*x)^(9/2))/(9*d^5)

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)^4}{d^4 \sqrt {c+d x}}-\frac {4 b (b c-a d)^3 \sqrt {c+d x}}{d^4}+\frac {6 b^2 (b c-a d)^2 (c+d x)^{3/2}}{d^4}-\frac {4 b^3 (b c-a d) (c+d x)^{5/2}}{d^4}+\frac {b^4 (c+d x)^{7/2}}{d^4}\right ) \, dx \\ & = \frac {2 (b c-a d)^4 \sqrt {c+d x}}{d^5}-\frac {8 b (b c-a d)^3 (c+d x)^{3/2}}{3 d^5}+\frac {12 b^2 (b c-a d)^2 (c+d x)^{5/2}}{5 d^5}-\frac {8 b^3 (b c-a d) (c+d x)^{7/2}}{7 d^5}+\frac {2 b^4 (c+d x)^{9/2}}{9 d^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.20 \[ \int \frac {(a+b x)^4}{\sqrt {c+d x}} \, dx=\frac {2 \sqrt {c+d x} \left (315 a^4 d^4+420 a^3 b d^3 (-2 c+d x)+126 a^2 b^2 d^2 \left (8 c^2-4 c d x+3 d^2 x^2\right )+36 a b^3 d \left (-16 c^3+8 c^2 d x-6 c d^2 x^2+5 d^3 x^3\right )+b^4 \left (128 c^4-64 c^3 d x+48 c^2 d^2 x^2-40 c d^3 x^3+35 d^4 x^4\right )\right )}{315 d^5} \]

[In]

Integrate[(a + b*x)^4/Sqrt[c + d*x],x]

[Out]

(2*Sqrt[c + d*x]*(315*a^4*d^4 + 420*a^3*b*d^3*(-2*c + d*x) + 126*a^2*b^2*d^2*(8*c^2 - 4*c*d*x + 3*d^2*x^2) + 3
6*a*b^3*d*(-16*c^3 + 8*c^2*d*x - 6*c*d^2*x^2 + 5*d^3*x^3) + b^4*(128*c^4 - 64*c^3*d*x + 48*c^2*d^2*x^2 - 40*c*
d^3*x^3 + 35*d^4*x^4)))/(315*d^5)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {\frac {2 b^{4} \left (d x +c \right )^{\frac {9}{2}}}{9}+\frac {8 \left (a d -b c \right ) b^{3} \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {12 \left (a d -b c \right )^{2} b^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {8 \left (a d -b c \right )^{3} b \left (d x +c \right )^{\frac {3}{2}}}{3}+2 \left (a d -b c \right )^{4} \sqrt {d x +c}}{d^{5}}\)	\(99\)
default	\(\frac {\frac {2 b^{4} \left (d x +c \right )^{\frac {9}{2}}}{9}+\frac {8 \left (a d -b c \right ) b^{3} \left (d x +c \right )^{\frac {7}{2}}}{7}+\frac {12 \left (a d -b c \right )^{2} b^{2} \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {8 \left (a d -b c \right )^{3} b \left (d x +c \right )^{\frac {3}{2}}}{3}+2 \left (a d -b c \right )^{4} \sqrt {d x +c}}{d^{5}}\)	\(99\)
pseudoelliptic	\(\frac {2 \sqrt {d x +c}\, \left (\left (\frac {1}{9} b^{4} x^{4}+\frac {4}{7} a \,b^{3} x^{3}+\frac {6}{5} a^{2} b^{2} x^{2}+\frac {4}{3} a^{3} b x +a^{4}\right ) d^{4}-\frac {8 \left (\frac {1}{21} b^{3} x^{3}+\frac {9}{35} a \,b^{2} x^{2}+\frac {3}{5} a^{2} b x +a^{3}\right ) b c \,d^{3}}{3}+\frac {16 b^{2} \left (\frac {1}{21} b^{2} x^{2}+\frac {2}{7} a b x +a^{2}\right ) c^{2} d^{2}}{5}-\frac {64 b^{3} \left (\frac {b x}{9}+a \right ) c^{3} d}{35}+\frac {128 b^{4} c^{4}}{315}\right )}{d^{5}}\)	\(143\)
gosper	\(\frac {2 \sqrt {d x +c}\, \left (35 d^{4} x^{4} b^{4}+180 a \,b^{3} d^{4} x^{3}-40 b^{4} c \,d^{3} x^{3}+378 a^{2} b^{2} d^{4} x^{2}-216 a \,b^{3} c \,d^{3} x^{2}+48 b^{4} c^{2} d^{2} x^{2}+420 a^{3} b \,d^{4} x -504 a^{2} b^{2} c \,d^{3} x +288 a \,b^{3} c^{2} d^{2} x -64 b^{4} c^{3} d x +315 a^{4} d^{4}-840 a^{3} b c \,d^{3}+1008 a^{2} b^{2} c^{2} d^{2}-576 a \,b^{3} c^{3} d +128 b^{4} c^{4}\right )}{315 d^{5}}\)	\(186\)
trager	\(\frac {2 \sqrt {d x +c}\, \left (35 d^{4} x^{4} b^{4}+180 a \,b^{3} d^{4} x^{3}-40 b^{4} c \,d^{3} x^{3}+378 a^{2} b^{2} d^{4} x^{2}-216 a \,b^{3} c \,d^{3} x^{2}+48 b^{4} c^{2} d^{2} x^{2}+420 a^{3} b \,d^{4} x -504 a^{2} b^{2} c \,d^{3} x +288 a \,b^{3} c^{2} d^{2} x -64 b^{4} c^{3} d x +315 a^{4} d^{4}-840 a^{3} b c \,d^{3}+1008 a^{2} b^{2} c^{2} d^{2}-576 a \,b^{3} c^{3} d +128 b^{4} c^{4}\right )}{315 d^{5}}\)	\(186\)
risch	\(\frac {2 \sqrt {d x +c}\, \left (35 d^{4} x^{4} b^{4}+180 a \,b^{3} d^{4} x^{3}-40 b^{4} c \,d^{3} x^{3}+378 a^{2} b^{2} d^{4} x^{2}-216 a \,b^{3} c \,d^{3} x^{2}+48 b^{4} c^{2} d^{2} x^{2}+420 a^{3} b \,d^{4} x -504 a^{2} b^{2} c \,d^{3} x +288 a \,b^{3} c^{2} d^{2} x -64 b^{4} c^{3} d x +315 a^{4} d^{4}-840 a^{3} b c \,d^{3}+1008 a^{2} b^{2} c^{2} d^{2}-576 a \,b^{3} c^{3} d +128 b^{4} c^{4}\right )}{315 d^{5}}\)	\(186\)

[In]

int((b*x+a)^4/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/d^5*(1/9*b^4*(d*x+c)^(9/2)+4/7*(a*d-b*c)*b^3*(d*x+c)^(7/2)+6/5*(a*d-b*c)^2*b^2*(d*x+c)^(5/2)+4/3*(a*d-b*c)^3
*b*(d*x+c)^(3/2)+(a*d-b*c)^4*(d*x+c)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.43 \[ \int \frac {(a+b x)^4}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (35 \, b^{4} d^{4} x^{4} + 128 \, b^{4} c^{4} - 576 \, a b^{3} c^{3} d + 1008 \, a^{2} b^{2} c^{2} d^{2} - 840 \, a^{3} b c d^{3} + 315 \, a^{4} d^{4} - 20 \, {\left (2 \, b^{4} c d^{3} - 9 \, a b^{3} d^{4}\right )} x^{3} + 6 \, {\left (8 \, b^{4} c^{2} d^{2} - 36 \, a b^{3} c d^{3} + 63 \, a^{2} b^{2} d^{4}\right )} x^{2} - 4 \, {\left (16 \, b^{4} c^{3} d - 72 \, a b^{3} c^{2} d^{2} + 126 \, a^{2} b^{2} c d^{3} - 105 \, a^{3} b d^{4}\right )} x\right )} \sqrt {d x + c}}{315 \, d^{5}} \]

[In]

integrate((b*x+a)^4/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*b^4*d^4*x^4 + 128*b^4*c^4 - 576*a*b^3*c^3*d + 1008*a^2*b^2*c^2*d^2 - 840*a^3*b*c*d^3 + 315*a^4*d^4 -
20*(2*b^4*c*d^3 - 9*a*b^3*d^4)*x^3 + 6*(8*b^4*c^2*d^2 - 36*a*b^3*c*d^3 + 63*a^2*b^2*d^4)*x^2 - 4*(16*b^4*c^3*
d - 72*a*b^3*c^2*d^2 + 126*a^2*b^2*c*d^3 - 105*a^3*b*d^4)*x)*sqrt(d*x + c)/d^5

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (117) = 234\).

Time = 1.02 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.90 \[ \int \frac {(a+b x)^4}{\sqrt {c+d x}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{4} \left (c + d x\right )^{\frac {9}{2}}}{9 d^{4}} + \frac {\left (c + d x\right )^{\frac {7}{2}} \cdot \left (4 a b^{3} d - 4 b^{4} c\right )}{7 d^{4}} + \frac {\left (c + d x\right )^{\frac {5}{2}} \cdot \left (6 a^{2} b^{2} d^{2} - 12 a b^{3} c d + 6 b^{4} c^{2}\right )}{5 d^{4}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \cdot \left (4 a^{3} b d^{3} - 12 a^{2} b^{2} c d^{2} + 12 a b^{3} c^{2} d - 4 b^{4} c^{3}\right )}{3 d^{4}} + \frac {\sqrt {c + d x} \left (a^{4} d^{4} - 4 a^{3} b c d^{3} + 6 a^{2} b^{2} c^{2} d^{2} - 4 a b^{3} c^{3} d + b^{4} c^{4}\right )}{d^{4}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\begin {cases} a^{4} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{5}}{5 b} & \text {otherwise} \end {cases}}{\sqrt {c}} & \text {otherwise} \end {cases} \]

[In]

integrate((b*x+a)**4/(d*x+c)**(1/2),x)

[Out]

Piecewise((2*(b**4*(c + d*x)**(9/2)/(9*d**4) + (c + d*x)**(7/2)*(4*a*b**3*d - 4*b**4*c)/(7*d**4) + (c + d*x)**
(5/2)*(6*a**2*b**2*d**2 - 12*a*b**3*c*d + 6*b**4*c**2)/(5*d**4) + (c + d*x)**(3/2)*(4*a**3*b*d**3 - 12*a**2*b*
*2*c*d**2 + 12*a*b**3*c**2*d - 4*b**4*c**3)/(3*d**4) + sqrt(c + d*x)*(a**4*d**4 - 4*a**3*b*c*d**3 + 6*a**2*b**
2*c**2*d**2 - 4*a*b**3*c**3*d + b**4*c**4)/d**4)/d, Ne(d, 0)), (Piecewise((a**4*x, Eq(b, 0)), ((a + b*x)**5/(5
*b), True))/sqrt(c), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b x)^4}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (315 \, \sqrt {d x + c} a^{4} + \frac {420 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a^{3} b}{d} + \frac {126 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a^{2} b^{2}}{d^{2}} + \frac {36 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} a b^{3}}{d^{3}} + \frac {{\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} - 180 \, {\left (d x + c\right )}^{\frac {7}{2}} c + 378 \, {\left (d x + c\right )}^{\frac {5}{2}} c^{2} - 420 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {d x + c} c^{4}\right )} b^{4}}{d^{4}}\right )}}{315 \, d} \]

[In]

integrate((b*x+a)^4/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

2/315*(315*sqrt(d*x + c)*a^4 + 420*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a^3*b/d + 126*(3*(d*x + c)^(5/2) - 10
*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a^2*b^2/d^2 + 36*(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^3/d^3 + (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^4/d^4)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.61 \[ \int \frac {(a+b x)^4}{\sqrt {c+d x}} \, dx=\frac {2 \, {\left (315 \, \sqrt {d x + c} a^{4} + \frac {420 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {d x + c} c\right )} a^{3} b}{d} + \frac {126 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {d x + c} c^{2}\right )} a^{2} b^{2}}{d^{2}} + \frac {36 \, {\left (5 \, {\left (d x + c\right )}^{\frac {7}{2}} - 21 \, {\left (d x + c\right )}^{\frac {5}{2}} c + 35 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{2} - 35 \, \sqrt {d x + c} c^{3}\right )} a b^{3}}{d^{3}} + \frac {{\left (35 \, {\left (d x + c\right )}^{\frac {9}{2}} - 180 \, {\left (d x + c\right )}^{\frac {7}{2}} c + 378 \, {\left (d x + c\right )}^{\frac {5}{2}} c^{2} - 420 \, {\left (d x + c\right )}^{\frac {3}{2}} c^{3} + 315 \, \sqrt {d x + c} c^{4}\right )} b^{4}}{d^{4}}\right )}}{315 \, d} \]

[In]

integrate((b*x+a)^4/(d*x+c)^(1/2),x, algorithm="giac")

[Out]

2/315*(315*sqrt(d*x + c)*a^4 + 420*((d*x + c)^(3/2) - 3*sqrt(d*x + c)*c)*a^3*b/d + 126*(3*(d*x + c)^(5/2) - 10
*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)*a^2*b^2/d^2 + 36*(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^3/d^3 + (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^4/d^4)/d

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.88 \[ \int \frac {(a+b x)^4}{\sqrt {c+d x}} \, dx=\frac {2\,b^4\,{\left (c+d\,x\right )}^{9/2}}{9\,d^5}-\frac {\left (8\,b^4\,c-8\,a\,b^3\,d\right )\,{\left (c+d\,x\right )}^{7/2}}{7\,d^5}+\frac {2\,{\left (a\,d-b\,c\right )}^4\,\sqrt {c+d\,x}}{d^5}+\frac {12\,b^2\,{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^{5/2}}{5\,d^5}+\frac {8\,b\,{\left (a\,d-b\,c\right )}^3\,{\left (c+d\,x\right )}^{3/2}}{3\,d^5} \]

[In]

int((a + b*x)^4/(c + d*x)^(1/2),x)

[Out]

(2*b^4*(c + d*x)^(9/2))/(9*d^5) - ((8*b^4*c - 8*a*b^3*d)*(c + d*x)^(7/2))/(7*d^5) + (2*(a*d - b*c)^4*(c + d*x)
^(1/2))/d^5 + (12*b^2*(a*d - b*c)^2*(c + d*x)^(5/2))/(5*d^5) + (8*b*(a*d - b*c)^3*(c + d*x)^(3/2))/(3*d^5)