3.1740 \(\int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=169 \[ -\frac {2 b^2 (d+e x)^{3/2} (-3 a B e-A b e+4 b B d)}{3 e^5}+\frac {6 b \sqrt {d+e x} (b d-a e) (-a B e-A b e+2 b B d)}{e^5}+\frac {2 (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 \sqrt {d+e x}}-\frac {2 (b d-a e)^3 (B d-A e)}{3 e^5 (d+e x)^{3/2}}+\frac {2 b^3 B (d+e x)^{5/2}}{5 e^5} \]

[Out]

-2/3*(-a*e+b*d)^3*(-A*e+B*d)/e^5/(e*x+d)^(3/2)-2/3*b^2*(-A*b*e-3*B*a*e+4*B*b*d)*(e*x+d)^(3/2)/e^5+2/5*b^3*B*(e
*x+d)^(5/2)/e^5+2*(-a*e+b*d)^2*(-3*A*b*e-B*a*e+4*B*b*d)/e^5/(e*x+d)^(1/2)+6*b*(-a*e+b*d)*(-A*b*e-B*a*e+2*B*b*d
)*(e*x+d)^(1/2)/e^5

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Rubi [A]  time = 0.07, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {77} \[ -\frac {2 b^2 (d+e x)^{3/2} (-3 a B e-A b e+4 b B d)}{3 e^5}+\frac {6 b \sqrt {d+e x} (b d-a e) (-a B e-A b e+2 b B d)}{e^5}+\frac {2 (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 \sqrt {d+e x}}-\frac {2 (b d-a e)^3 (B d-A e)}{3 e^5 (d+e x)^{3/2}}+\frac {2 b^3 B (d+e x)^{5/2}}{5 e^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^3*(A + B*x))/(d + e*x)^(5/2),x]

[Out]

(-2*(b*d - a*e)^3*(B*d - A*e))/(3*e^5*(d + e*x)^(3/2)) + (2*(b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e))/(e^5*Sq
rt[d + e*x]) + (6*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e)*Sqrt[d + e*x])/e^5 - (2*b^2*(4*b*B*d - A*b*e - 3*a*B
*e)*(d + e*x)^(3/2))/(3*e^5) + (2*b^3*B*(d + e*x)^(5/2))/(5*e^5)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin {align*} \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx &=\int \left (\frac {(-b d+a e)^3 (-B d+A e)}{e^4 (d+e x)^{5/2}}+\frac {(-b d+a e)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^{3/2}}-\frac {3 b (b d-a e) (-2 b B d+A b e+a B e)}{e^4 \sqrt {d+e x}}+\frac {b^2 (-4 b B d+A b e+3 a B e) \sqrt {d+e x}}{e^4}+\frac {b^3 B (d+e x)^{3/2}}{e^4}\right ) \, dx\\ &=-\frac {2 (b d-a e)^3 (B d-A e)}{3 e^5 (d+e x)^{3/2}}+\frac {2 (b d-a e)^2 (4 b B d-3 A b e-a B e)}{e^5 \sqrt {d+e x}}+\frac {6 b (b d-a e) (2 b B d-A b e-a B e) \sqrt {d+e x}}{e^5}-\frac {2 b^2 (4 b B d-A b e-3 a B e) (d+e x)^{3/2}}{3 e^5}+\frac {2 b^3 B (d+e x)^{5/2}}{5 e^5}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 145, normalized size = 0.86 \[ \frac {2 \left (-5 b^2 (d+e x)^3 (-3 a B e-A b e+4 b B d)+45 b (d+e x)^2 (b d-a e) (-a B e-A b e+2 b B d)+15 (d+e x) (b d-a e)^2 (-a B e-3 A b e+4 b B d)-5 (b d-a e)^3 (B d-A e)+3 b^3 B (d+e x)^4\right )}{15 e^5 (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^3*(A + B*x))/(d + e*x)^(5/2),x]

[Out]

(2*(-5*(b*d - a*e)^3*(B*d - A*e) + 15*(b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e)*(d + e*x) + 45*b*(b*d - a*e)*(
2*b*B*d - A*b*e - a*B*e)*(d + e*x)^2 - 5*b^2*(4*b*B*d - A*b*e - 3*a*B*e)*(d + e*x)^3 + 3*b^3*B*(d + e*x)^4))/(
15*e^5*(d + e*x)^(3/2))

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fricas [A]  time = 0.76, size = 284, normalized size = 1.68 \[ \frac {2 \, {\left (3 \, B b^{3} e^{4} x^{4} + 128 \, B b^{3} d^{4} - 5 \, A a^{3} e^{4} - 80 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 120 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - 10 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - {\left (8 \, B b^{3} d e^{3} - 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 3 \, {\left (16 \, B b^{3} d^{2} e^{2} - 10 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 15 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 3 \, {\left (64 \, B b^{3} d^{3} e - 40 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 60 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*B*b^3*e^4*x^4 + 128*B*b^3*d^4 - 5*A*a^3*e^4 - 80*(3*B*a*b^2 + A*b^3)*d^3*e + 120*(B*a^2*b + A*a*b^2)*d
^2*e^2 - 10*(B*a^3 + 3*A*a^2*b)*d*e^3 - (8*B*b^3*d*e^3 - 5*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 3*(16*B*b^3*d^2*e^2
- 10*(3*B*a*b^2 + A*b^3)*d*e^3 + 15*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 3*(64*B*b^3*d^3*e - 40*(3*B*a*b^2 + A*b^3)*
d^2*e^2 + 60*(B*a^2*b + A*a*b^2)*d*e^3 - 5*(B*a^3 + 3*A*a^2*b)*e^4)*x)*sqrt(e*x + d)/(e^7*x^2 + 2*d*e^6*x + d^
2*e^5)

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giac [B]  time = 1.34, size = 365, normalized size = 2.16 \[ \frac {2}{15} \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{3} e^{20} - 20 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{3} d e^{20} + 90 \, \sqrt {x e + d} B b^{3} d^{2} e^{20} + 15 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{2} e^{21} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{3} e^{21} - 135 \, \sqrt {x e + d} B a b^{2} d e^{21} - 45 \, \sqrt {x e + d} A b^{3} d e^{21} + 45 \, \sqrt {x e + d} B a^{2} b e^{22} + 45 \, \sqrt {x e + d} A a b^{2} e^{22}\right )} e^{\left (-25\right )} + \frac {2 \, {\left (12 \, {\left (x e + d\right )} B b^{3} d^{3} - B b^{3} d^{4} - 27 \, {\left (x e + d\right )} B a b^{2} d^{2} e - 9 \, {\left (x e + d\right )} A b^{3} d^{2} e + 3 \, B a b^{2} d^{3} e + A b^{3} d^{3} e + 18 \, {\left (x e + d\right )} B a^{2} b d e^{2} + 18 \, {\left (x e + d\right )} A a b^{2} d e^{2} - 3 \, B a^{2} b d^{2} e^{2} - 3 \, A a b^{2} d^{2} e^{2} - 3 \, {\left (x e + d\right )} B a^{3} e^{3} - 9 \, {\left (x e + d\right )} A a^{2} b e^{3} + B a^{3} d e^{3} + 3 \, A a^{2} b d e^{3} - A a^{3} e^{4}\right )} e^{\left (-5\right )}}{3 \, {\left (x e + d\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

2/15*(3*(x*e + d)^(5/2)*B*b^3*e^20 - 20*(x*e + d)^(3/2)*B*b^3*d*e^20 + 90*sqrt(x*e + d)*B*b^3*d^2*e^20 + 15*(x
*e + d)^(3/2)*B*a*b^2*e^21 + 5*(x*e + d)^(3/2)*A*b^3*e^21 - 135*sqrt(x*e + d)*B*a*b^2*d*e^21 - 45*sqrt(x*e + d
)*A*b^3*d*e^21 + 45*sqrt(x*e + d)*B*a^2*b*e^22 + 45*sqrt(x*e + d)*A*a*b^2*e^22)*e^(-25) + 2/3*(12*(x*e + d)*B*
b^3*d^3 - B*b^3*d^4 - 27*(x*e + d)*B*a*b^2*d^2*e - 9*(x*e + d)*A*b^3*d^2*e + 3*B*a*b^2*d^3*e + A*b^3*d^3*e + 1
8*(x*e + d)*B*a^2*b*d*e^2 + 18*(x*e + d)*A*a*b^2*d*e^2 - 3*B*a^2*b*d^2*e^2 - 3*A*a*b^2*d^2*e^2 - 3*(x*e + d)*B
*a^3*e^3 - 9*(x*e + d)*A*a^2*b*e^3 + B*a^3*d*e^3 + 3*A*a^2*b*d*e^3 - A*a^3*e^4)*e^(-5)/(x*e + d)^(3/2)

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maple [A]  time = 0.01, size = 301, normalized size = 1.78 \[ -\frac {2 \left (-3 B \,b^{3} x^{4} e^{4}-5 A \,b^{3} e^{4} x^{3}-15 B a \,b^{2} e^{4} x^{3}+8 B \,b^{3} d \,e^{3} x^{3}-45 A a \,b^{2} e^{4} x^{2}+30 A \,b^{3} d \,e^{3} x^{2}-45 B \,a^{2} b \,e^{4} x^{2}+90 B a \,b^{2} d \,e^{3} x^{2}-48 B \,b^{3} d^{2} e^{2} x^{2}+45 A \,a^{2} b \,e^{4} x -180 A a \,b^{2} d \,e^{3} x +120 A \,b^{3} d^{2} e^{2} x +15 B \,a^{3} e^{4} x -180 B \,a^{2} b d \,e^{3} x +360 B a \,b^{2} d^{2} e^{2} x -192 B \,b^{3} d^{3} e x +5 a^{3} A \,e^{4}+30 A \,a^{2} b d \,e^{3}-120 A a \,b^{2} d^{2} e^{2}+80 A \,b^{3} d^{3} e +10 B \,a^{3} d \,e^{3}-120 B \,a^{2} b \,d^{2} e^{2}+240 B a \,b^{2} d^{3} e -128 B \,b^{3} d^{4}\right )}{15 \left (e x +d \right )^{\frac {3}{2}} e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x)

[Out]

-2/15/(e*x+d)^(3/2)*(-3*B*b^3*e^4*x^4-5*A*b^3*e^4*x^3-15*B*a*b^2*e^4*x^3+8*B*b^3*d*e^3*x^3-45*A*a*b^2*e^4*x^2+
30*A*b^3*d*e^3*x^2-45*B*a^2*b*e^4*x^2+90*B*a*b^2*d*e^3*x^2-48*B*b^3*d^2*e^2*x^2+45*A*a^2*b*e^4*x-180*A*a*b^2*d
*e^3*x+120*A*b^3*d^2*e^2*x+15*B*a^3*e^4*x-180*B*a^2*b*d*e^3*x+360*B*a*b^2*d^2*e^2*x-192*B*b^3*d^3*e*x+5*A*a^3*
e^4+30*A*a^2*b*d*e^3-120*A*a*b^2*d^2*e^2+80*A*b^3*d^3*e+10*B*a^3*d*e^3-120*B*a^2*b*d^2*e^2+240*B*a*b^2*d^3*e-1
28*B*b^3*d^4)/e^5

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maxima [A]  time = 0.61, size = 271, normalized size = 1.60 \[ \frac {2 \, {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} B b^{3} - 5 \, {\left (4 \, B b^{3} d - {\left (3 \, B a b^{2} + A b^{3}\right )} e\right )} {\left (e x + d\right )}^{\frac {3}{2}} + 45 \, {\left (2 \, B b^{3} d^{2} - {\left (3 \, B a b^{2} + A b^{3}\right )} d e + {\left (B a^{2} b + A a b^{2}\right )} e^{2}\right )} \sqrt {e x + d}}{e^{4}} - \frac {5 \, {\left (B b^{3} d^{4} + A a^{3} e^{4} - {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - 3 \, {\left (4 \, B b^{3} d^{3} - 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e + 6 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{2} - {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{3}\right )} {\left (e x + d\right )}\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} e^{4}}\right )}}{15 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

2/15*((3*(e*x + d)^(5/2)*B*b^3 - 5*(4*B*b^3*d - (3*B*a*b^2 + A*b^3)*e)*(e*x + d)^(3/2) + 45*(2*B*b^3*d^2 - (3*
B*a*b^2 + A*b^3)*d*e + (B*a^2*b + A*a*b^2)*e^2)*sqrt(e*x + d))/e^4 - 5*(B*b^3*d^4 + A*a^3*e^4 - (3*B*a*b^2 + A
*b^3)*d^3*e + 3*(B*a^2*b + A*a*b^2)*d^2*e^2 - (B*a^3 + 3*A*a^2*b)*d*e^3 - 3*(4*B*b^3*d^3 - 3*(3*B*a*b^2 + A*b^
3)*d^2*e + 6*(B*a^2*b + A*a*b^2)*d*e^2 - (B*a^3 + 3*A*a^2*b)*e^3)*(e*x + d))/((e*x + d)^(3/2)*e^4))/e

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mupad [B]  time = 1.27, size = 264, normalized size = 1.56 \[ \frac {{\left (d+e\,x\right )}^{3/2}\,\left (2\,A\,b^3\,e-8\,B\,b^3\,d+6\,B\,a\,b^2\,e\right )}{3\,e^5}-\frac {\left (d+e\,x\right )\,\left (2\,B\,a^3\,e^3-12\,B\,a^2\,b\,d\,e^2+6\,A\,a^2\,b\,e^3+18\,B\,a\,b^2\,d^2\,e-12\,A\,a\,b^2\,d\,e^2-8\,B\,b^3\,d^3+6\,A\,b^3\,d^2\,e\right )+\frac {2\,A\,a^3\,e^4}{3}+\frac {2\,B\,b^3\,d^4}{3}-\frac {2\,A\,b^3\,d^3\,e}{3}-\frac {2\,B\,a^3\,d\,e^3}{3}+2\,A\,a\,b^2\,d^2\,e^2+2\,B\,a^2\,b\,d^2\,e^2-2\,A\,a^2\,b\,d\,e^3-2\,B\,a\,b^2\,d^3\,e}{e^5\,{\left (d+e\,x\right )}^{3/2}}+\frac {2\,B\,b^3\,{\left (d+e\,x\right )}^{5/2}}{5\,e^5}+\frac {6\,b\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{e^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^3)/(d + e*x)^(5/2),x)

[Out]

((d + e*x)^(3/2)*(2*A*b^3*e - 8*B*b^3*d + 6*B*a*b^2*e))/(3*e^5) - ((d + e*x)*(2*B*a^3*e^3 - 8*B*b^3*d^3 + 6*A*
a^2*b*e^3 + 6*A*b^3*d^2*e - 12*A*a*b^2*d*e^2 + 18*B*a*b^2*d^2*e - 12*B*a^2*b*d*e^2) + (2*A*a^3*e^4)/3 + (2*B*b
^3*d^4)/3 - (2*A*b^3*d^3*e)/3 - (2*B*a^3*d*e^3)/3 + 2*A*a*b^2*d^2*e^2 + 2*B*a^2*b*d^2*e^2 - 2*A*a^2*b*d*e^3 -
2*B*a*b^2*d^3*e)/(e^5*(d + e*x)^(3/2)) + (2*B*b^3*(d + e*x)^(5/2))/(5*e^5) + (6*b*(a*e - b*d)*(d + e*x)^(1/2)*
(A*b*e + B*a*e - 2*B*b*d))/e^5

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sympy [A]  time = 84.19, size = 199, normalized size = 1.18 \[ \frac {2 B b^{3} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{5}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (2 A b^{3} e + 6 B a b^{2} e - 8 B b^{3} d\right )}{3 e^{5}} + \frac {\sqrt {d + e x} \left (6 A a b^{2} e^{2} - 6 A b^{3} d e + 6 B a^{2} b e^{2} - 18 B a b^{2} d e + 12 B b^{3} d^{2}\right )}{e^{5}} - \frac {2 \left (a e - b d\right )^{2} \left (3 A b e + B a e - 4 B b d\right )}{e^{5} \sqrt {d + e x}} + \frac {2 \left (- A e + B d\right ) \left (a e - b d\right )^{3}}{3 e^{5} \left (d + e x\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(B*x+A)/(e*x+d)**(5/2),x)

[Out]

2*B*b**3*(d + e*x)**(5/2)/(5*e**5) + (d + e*x)**(3/2)*(2*A*b**3*e + 6*B*a*b**2*e - 8*B*b**3*d)/(3*e**5) + sqrt
(d + e*x)*(6*A*a*b**2*e**2 - 6*A*b**3*d*e + 6*B*a**2*b*e**2 - 18*B*a*b**2*d*e + 12*B*b**3*d**2)/e**5 - 2*(a*e
- b*d)**2*(3*A*b*e + B*a*e - 4*B*b*d)/(e**5*sqrt(d + e*x)) + 2*(-A*e + B*d)*(a*e - b*d)**3/(3*e**5*(d + e*x)**
(3/2))

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