\(\int \frac {(a+b x)^4}{(c+d x)^{5/2}} \, dx\) [1435]

Optimal result

Integrand size = 17, antiderivative size = 125 \[ \int \frac {(a+b x)^4}{(c+d x)^{5/2}} \, dx=-\frac {2 (b c-a d)^4}{3 d^5 (c+d x)^{3/2}}+\frac {8 b (b c-a d)^3}{d^5 \sqrt {c+d x}}+\frac {12 b^2 (b c-a d)^2 \sqrt {c+d x}}{d^5}-\frac {8 b^3 (b c-a d) (c+d x)^{3/2}}{3 d^5}+\frac {2 b^4 (c+d x)^{5/2}}{5 d^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 125, 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} \[ \int \frac {(a+b x)^4}{(c+d x)^{5/2}} \, dx=-\frac {8 b^3 (c+d x)^{3/2} (b c-a d)}{3 d^5}+\frac {12 b^2 \sqrt {c+d x} (b c-a d)^2}{d^5}+\frac {8 b (b c-a d)^3}{d^5 \sqrt {c+d x}}-\frac {2 (b c-a d)^4}{3 d^5 (c+d x)^{3/2}}+\frac {2 b^4 (c+d x)^{5/2}}{5 d^5} \]

[In]

Int[(a + b*x)^4/(c + d*x)^(5/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.22 \[ \int \frac {(a+b x)^4}{(c+d x)^{5/2}} \, dx=\frac {2 \left (-5 a^4 d^4-20 a^3 b d^3 (2 c+3 d x)+30 a^2 b^2 d^2 \left (8 c^2+12 c d x+3 d^2 x^2\right )+20 a b^3 d \left (-16 c^3-24 c^2 d x-6 c d^2 x^2+d^3 x^3\right )+b^4 \left (128 c^4+192 c^3 d x+48 c^2 d^2 x^2-8 c d^3 x^3+3 d^4 x^4\right )\right )}{15 d^5 (c+d x)^{3/2}} \]

[In]

Integrate[(a + b*x)^4/(c + d*x)^(5/2),x]

[Out]

(2*(-5*a^4*d^4 - 20*a^3*b*d^3*(2*c + 3*d*x) + 30*a^2*b^2*d^2*(8*c^2 + 12*c*d*x + 3*d^2*x^2) + 20*a*b^3*d*(-16*
c^3 - 24*c^2*d*x - 6*c*d^2*x^2 + d^3*x^3) + b^4*(128*c^4 + 192*c^3*d*x + 48*c^2*d^2*x^2 - 8*c*d^3*x^3 + 3*d^4*
x^4)))/(15*d^5*(c + d*x)^(3/2))

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.02

method	result	size
risch	\(\frac {2 b^{2} \left (3 d^{2} x^{2} b^{2}+20 x a b \,d^{2}-14 x \,b^{2} c d +90 a^{2} d^{2}-160 a b c d +73 b^{2} c^{2}\right ) \sqrt {d x +c}}{15 d^{5}}-\frac {2 \left (12 b d x +a d +11 b c \right ) \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{3 d^{5} \left (d x +c \right )^{\frac {3}{2}}}\)	\(128\)
pseudoelliptic	\(-\frac {2 \left (\left (-\frac {3}{5} b^{4} x^{4}-4 a \,b^{3} x^{3}-18 a^{2} b^{2} x^{2}+12 a^{3} b x +a^{4}\right ) d^{4}+8 b c \left (\frac {1}{5} b^{3} x^{3}+3 a \,b^{2} x^{2}-9 a^{2} b x +a^{3}\right ) d^{3}-48 \left (\frac {1}{5} b^{2} x^{2}-2 a b x +a^{2}\right ) b^{2} c^{2} d^{2}+64 b^{3} \left (-\frac {3 b x}{5}+a \right ) c^{3} d -\frac {128 b^{4} c^{4}}{5}\right )}{3 \left (d x +c \right )^{\frac {3}{2}} d^{5}}\)	\(143\)
gosper	\(-\frac {2 \left (-3 d^{4} x^{4} b^{4}-20 a \,b^{3} d^{4} x^{3}+8 b^{4} c \,d^{3} x^{3}-90 a^{2} b^{2} d^{4} x^{2}+120 a \,b^{3} c \,d^{3} x^{2}-48 b^{4} c^{2} d^{2} x^{2}+60 a^{3} b \,d^{4} x -360 a^{2} b^{2} c \,d^{3} x +480 a \,b^{3} c^{2} d^{2} x -192 b^{4} c^{3} d x +5 a^{4} d^{4}+40 a^{3} b c \,d^{3}-240 a^{2} b^{2} c^{2} d^{2}+320 a \,b^{3} c^{3} d -128 b^{4} c^{4}\right )}{15 \left (d x +c \right )^{\frac {3}{2}} d^{5}}\)	\(186\)
trager	\(-\frac {2 \left (-3 d^{4} x^{4} b^{4}-20 a \,b^{3} d^{4} x^{3}+8 b^{4} c \,d^{3} x^{3}-90 a^{2} b^{2} d^{4} x^{2}+120 a \,b^{3} c \,d^{3} x^{2}-48 b^{4} c^{2} d^{2} x^{2}+60 a^{3} b \,d^{4} x -360 a^{2} b^{2} c \,d^{3} x +480 a \,b^{3} c^{2} d^{2} x -192 b^{4} c^{3} d x +5 a^{4} d^{4}+40 a^{3} b c \,d^{3}-240 a^{2} b^{2} c^{2} d^{2}+320 a \,b^{3} c^{3} d -128 b^{4} c^{4}\right )}{15 \left (d x +c \right )^{\frac {3}{2}} d^{5}}\)	\(186\)
derivativedivides	\(\frac {\frac {2 b^{4} \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {8 a \,b^{3} d \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {8 b^{4} c \left (d x +c \right )^{\frac {3}{2}}}{3}+12 a^{2} b^{2} d^{2} \sqrt {d x +c}-24 a \,b^{3} c d \sqrt {d x +c}+12 b^{4} c^{2} \sqrt {d x +c}-\frac {8 b \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{\sqrt {d x +c}}-\frac {2 \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}}{d^{5}}\)	\(198\)
default	\(\frac {\frac {2 b^{4} \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {8 a \,b^{3} d \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {8 b^{4} c \left (d x +c \right )^{\frac {3}{2}}}{3}+12 a^{2} b^{2} d^{2} \sqrt {d x +c}-24 a \,b^{3} c d \sqrt {d x +c}+12 b^{4} c^{2} \sqrt {d x +c}-\frac {8 b \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{\sqrt {d x +c}}-\frac {2 \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}{3 \left (d x +c \right )^{\frac {3}{2}}}}{d^{5}}\)	\(198\)

[In]

int((b*x+a)^4/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/15*b^2*(3*b^2*d^2*x^2+20*a*b*d^2*x-14*b^2*c*d*x+90*a^2*d^2-160*a*b*c*d+73*b^2*c^2)*(d*x+c)^(1/2)/d^5-2/3*(12
*b*d*x+a*d+11*b*c)*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/d^5/(d*x+c)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b x)^4}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (3 \, b^{4} d^{4} x^{4} + 128 \, b^{4} c^{4} - 320 \, a b^{3} c^{3} d + 240 \, a^{2} b^{2} c^{2} d^{2} - 40 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4} - 4 \, {\left (2 \, b^{4} c d^{3} - 5 \, a b^{3} d^{4}\right )} x^{3} + 6 \, {\left (8 \, b^{4} c^{2} d^{2} - 20 \, a b^{3} c d^{3} + 15 \, a^{2} b^{2} d^{4}\right )} x^{2} + 12 \, {\left (16 \, b^{4} c^{3} d - 40 \, a b^{3} c^{2} d^{2} + 30 \, a^{2} b^{2} c d^{3} - 5 \, a^{3} b d^{4}\right )} x\right )} \sqrt {d x + c}}{15 \, {\left (d^{7} x^{2} + 2 \, c d^{6} x + c^{2} d^{5}\right )}} \]

[In]

integrate((b*x+a)^4/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*b^4*d^4*x^4 + 128*b^4*c^4 - 320*a*b^3*c^3*d + 240*a^2*b^2*c^2*d^2 - 40*a^3*b*c*d^3 - 5*a^4*d^4 - 4*(2*
b^4*c*d^3 - 5*a*b^3*d^4)*x^3 + 6*(8*b^4*c^2*d^2 - 20*a*b^3*c*d^3 + 15*a^2*b^2*d^4)*x^2 + 12*(16*b^4*c^3*d - 40
*a*b^3*c^2*d^2 + 30*a^2*b^2*c*d^3 - 5*a^3*b*d^4)*x)*sqrt(d*x + c)/(d^7*x^2 + 2*c*d^6*x + c^2*d^5)

Sympy [A] (verification not implemented)

Time = 4.57 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.30 \[ \int \frac {(a+b x)^4}{(c+d x)^{5/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{4} \left (c + d x\right )^{\frac {5}{2}}}{5 d^{4}} - \frac {4 b \left (a d - b c\right )^{3}}{d^{4} \sqrt {c + d x}} + \frac {\left (c + d x\right )^{\frac {3}{2}} \cdot \left (4 a b^{3} d - 4 b^{4} c\right )}{3 d^{4}} + \frac {\sqrt {c + d x} \left (6 a^{2} b^{2} d^{2} - 12 a b^{3} c d + 6 b^{4} c^{2}\right )}{d^{4}} - \frac {\left (a d - b c\right )^{4}}{3 d^{4} \left (c + d x\right )^{\frac {3}{2}}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\begin {cases} a^{4} x & \text {for}\: b = 0 \\\frac {\left (a + b x\right )^{5}}{5 b} & \text {otherwise} \end {cases}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((b*x+a)**4/(d*x+c)**(5/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.50 \[ \int \frac {(a+b x)^4}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} - 20 \, {\left (b^{4} c - a b^{3} d\right )} {\left (d x + c\right )}^{\frac {3}{2}} + 90 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \sqrt {d x + c}}{d^{4}} - \frac {5 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4} - 12 \, {\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} {\left (d x + c\right )}\right )}}{{\left (d x + c\right )}^{\frac {3}{2}} d^{4}}\right )}}{15 \, d} \]

[In]

integrate((b*x+a)^4/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

2/15*((3*(d*x + c)^(5/2)*b^4 - 20*(b^4*c - a*b^3*d)*(d*x + c)^(3/2) + 90*(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)
*sqrt(d*x + c))/d^4 - 5*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4 - 12*(b^4*c^3 -
3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*(d*x + c))/((d*x + c)^(3/2)*d^4))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (109) = 218\).

Time = 0.33 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.83 \[ \int \frac {(a+b x)^4}{(c+d x)^{5/2}} \, dx=\frac {2 \, {\left (12 \, {\left (d x + c\right )} b^{4} c^{3} - b^{4} c^{4} - 36 \, {\left (d x + c\right )} a b^{3} c^{2} d + 4 \, a b^{3} c^{3} d + 36 \, {\left (d x + c\right )} a^{2} b^{2} c d^{2} - 6 \, a^{2} b^{2} c^{2} d^{2} - 12 \, {\left (d x + c\right )} a^{3} b d^{3} + 4 \, a^{3} b c d^{3} - a^{4} d^{4}\right )}}{3 \, {\left (d x + c\right )}^{\frac {3}{2}} d^{5}} + \frac {2 \, {\left (3 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4} d^{20} - 20 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{4} c d^{20} + 90 \, \sqrt {d x + c} b^{4} c^{2} d^{20} + 20 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{3} d^{21} - 180 \, \sqrt {d x + c} a b^{3} c d^{21} + 90 \, \sqrt {d x + c} a^{2} b^{2} d^{22}\right )}}{15 \, d^{25}} \]

[In]

integrate((b*x+a)^4/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

2/3*(12*(d*x + c)*b^4*c^3 - b^4*c^4 - 36*(d*x + c)*a*b^3*c^2*d + 4*a*b^3*c^3*d + 36*(d*x + c)*a^2*b^2*c*d^2 -
6*a^2*b^2*c^2*d^2 - 12*(d*x + c)*a^3*b*d^3 + 4*a^3*b*c*d^3 - a^4*d^4)/((d*x + c)^(3/2)*d^5) + 2/15*(3*(d*x + c
)^(5/2)*b^4*d^20 - 20*(d*x + c)^(3/2)*b^4*c*d^20 + 90*sqrt(d*x + c)*b^4*c^2*d^20 + 20*(d*x + c)^(3/2)*a*b^3*d^
21 - 180*sqrt(d*x + c)*a*b^3*c*d^21 + 90*sqrt(d*x + c)*a^2*b^2*d^22)/d^25

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.40 \[ \int \frac {(a+b x)^4}{(c+d x)^{5/2}} \, dx=\frac {2\,b^4\,{\left (c+d\,x\right )}^{5/2}}{5\,d^5}-\frac {\left (8\,b^4\,c-8\,a\,b^3\,d\right )\,{\left (c+d\,x\right )}^{3/2}}{3\,d^5}+\frac {\left (c+d\,x\right )\,\left (-8\,a^3\,b\,d^3+24\,a^2\,b^2\,c\,d^2-24\,a\,b^3\,c^2\,d+8\,b^4\,c^3\right )-\frac {2\,a^4\,d^4}{3}-\frac {2\,b^4\,c^4}{3}-4\,a^2\,b^2\,c^2\,d^2+\frac {8\,a\,b^3\,c^3\,d}{3}+\frac {8\,a^3\,b\,c\,d^3}{3}}{d^5\,{\left (c+d\,x\right )}^{3/2}}+\frac {12\,b^2\,{\left (a\,d-b\,c\right )}^2\,\sqrt {c+d\,x}}{d^5} \]

[In]

int((a + b*x)^4/(c + d*x)^(5/2),x)

[Out]

(2*b^4*(c + d*x)^(5/2))/(5*d^5) - ((8*b^4*c - 8*a*b^3*d)*(c + d*x)^(3/2))/(3*d^5) + ((c + d*x)*(8*b^4*c^3 - 8*
a^3*b*d^3 + 24*a^2*b^2*c*d^2 - 24*a*b^3*c^2*d) - (2*a^4*d^4)/3 - (2*b^4*c^4)/3 - 4*a^2*b^2*c^2*d^2 + (8*a*b^3*
c^3*d)/3 + (8*a^3*b*c*d^3)/3)/(d^5*(c + d*x)^(3/2)) + (12*b^2*(a*d - b*c)^2*(c + d*x)^(1/2))/d^5