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

Optimal. Leaf size=263 \[ -\frac {2 (d+e x)^{3/2} \left (2 A c e (2 c d-b e)-B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )\right )}{3 e^6}+\frac {2 \sqrt {d+e x} \left (A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )\right )}{e^6}+\frac {2 d^2 (B d-A e) (c d-b e)^2}{3 e^6 (d+e x)^{3/2}}-\frac {2 c (d+e x)^{5/2} (-A c e-2 b B e+5 B c d)}{5 e^6}-\frac {2 d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{e^6 \sqrt {d+e x}}+\frac {2 B c^2 (d+e x)^{7/2}}{7 e^6} \]

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Rubi [A]  time = 0.15, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {771} \begin {gather*} -\frac {2 (d+e x)^{3/2} \left (2 A c e (2 c d-b e)-B \left (b^2 e^2-8 b c d e+10 c^2 d^2\right )\right )}{3 e^6}+\frac {2 \sqrt {d+e x} \left (A e \left (b^2 e^2-6 b c d e+6 c^2 d^2\right )-B d \left (3 b^2 e^2-12 b c d e+10 c^2 d^2\right )\right )}{e^6}+\frac {2 d^2 (B d-A e) (c d-b e)^2}{3 e^6 (d+e x)^{3/2}}-\frac {2 c (d+e x)^{5/2} (-A c e-2 b B e+5 B c d)}{5 e^6}-\frac {2 d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{e^6 \sqrt {d+e x}}+\frac {2 B c^2 (d+e x)^{7/2}}{7 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^2*(B*d - A*e)*(c*d - b*e)^2)/(3*e^6*(d + e*x)^(3/2)) - (2*d*(c*d - b*e)*(B*d*(5*c*d - 3*b*e) - 2*A*e*(2*c
*d - b*e)))/(e^6*Sqrt[d + e*x]) + (2*(A*e*(6*c^2*d^2 - 6*b*c*d*e + b^2*e^2) - B*d*(10*c^2*d^2 - 12*b*c*d*e + 3
*b^2*e^2))*Sqrt[d + e*x])/e^6 - (2*(2*A*c*e*(2*c*d - b*e) - B*(10*c^2*d^2 - 8*b*c*d*e + b^2*e^2))*(d + e*x)^(3
/2))/(3*e^6) - (2*c*(5*B*c*d - 2*b*B*e - A*c*e)*(d + e*x)^(5/2))/(5*e^6) + (2*B*c^2*(d + e*x)^(7/2))/(7*e^6)

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^2}{(d+e x)^{5/2}} \, dx &=\int \left (-\frac {d^2 (B d-A e) (c d-b e)^2}{e^5 (d+e x)^{5/2}}+\frac {d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{e^5 (d+e x)^{3/2}}+\frac {A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )-B d \left (10 c^2 d^2-12 b c d e+3 b^2 e^2\right )}{e^5 \sqrt {d+e x}}+\frac {\left (-2 A c e (2 c d-b e)+B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )\right ) \sqrt {d+e x}}{e^5}+\frac {c (-5 B c d+2 b B e+A c e) (d+e x)^{3/2}}{e^5}+\frac {B c^2 (d+e x)^{5/2}}{e^5}\right ) \, dx\\ &=\frac {2 d^2 (B d-A e) (c d-b e)^2}{3 e^6 (d+e x)^{3/2}}-\frac {2 d (c d-b e) (B d (5 c d-3 b e)-2 A e (2 c d-b e))}{e^6 \sqrt {d+e x}}+\frac {2 \left (A e \left (6 c^2 d^2-6 b c d e+b^2 e^2\right )-B d \left (10 c^2 d^2-12 b c d e+3 b^2 e^2\right )\right ) \sqrt {d+e x}}{e^6}-\frac {2 \left (2 A c e (2 c d-b e)-B \left (10 c^2 d^2-8 b c d e+b^2 e^2\right )\right ) (d+e x)^{3/2}}{3 e^6}-\frac {2 c (5 B c d-2 b B e-A c e) (d+e x)^{5/2}}{5 e^6}+\frac {2 B c^2 (d+e x)^{7/2}}{7 e^6}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 271, normalized size = 1.03 \begin {gather*} \frac {2 \left (7 A e \left (5 b^2 e^2 \left (8 d^2+12 d e x+3 e^2 x^2\right )+10 b c e \left (-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3\right )+c^2 \left (128 d^4+192 d^3 e x+48 d^2 e^2 x^2-8 d e^3 x^3+3 e^4 x^4\right )\right )+B \left (35 b^2 e^2 \left (-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3\right )+14 b c e \left (128 d^4+192 d^3 e x+48 d^2 e^2 x^2-8 d e^3 x^3+3 e^4 x^4\right )-5 c^2 \left (256 d^5+384 d^4 e x+96 d^3 e^2 x^2-16 d^2 e^3 x^3+6 d e^4 x^4-3 e^5 x^5\right )\right )\right )}{105 e^6 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(7*A*e*(5*b^2*e^2*(8*d^2 + 12*d*e*x + 3*e^2*x^2) + 10*b*c*e*(-16*d^3 - 24*d^2*e*x - 6*d*e^2*x^2 + e^3*x^3)
+ c^2*(128*d^4 + 192*d^3*e*x + 48*d^2*e^2*x^2 - 8*d*e^3*x^3 + 3*e^4*x^4)) + B*(35*b^2*e^2*(-16*d^3 - 24*d^2*e*
x - 6*d*e^2*x^2 + e^3*x^3) + 14*b*c*e*(128*d^4 + 192*d^3*e*x + 48*d^2*e^2*x^2 - 8*d*e^3*x^3 + 3*e^4*x^4) - 5*c
^2*(256*d^5 + 384*d^4*e*x + 96*d^3*e^2*x^2 - 16*d^2*e^3*x^3 + 6*d*e^4*x^4 - 3*e^5*x^5))))/(105*e^6*(d + e*x)^(
3/2))

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IntegrateAlgebraic [A]  time = 0.17, size = 399, normalized size = 1.52 \begin {gather*} \frac {2 \left (-35 A b^2 d^2 e^3+210 A b^2 d e^3 (d+e x)+105 A b^2 e^3 (d+e x)^2+70 A b c d^3 e^2-630 A b c d^2 e^2 (d+e x)-630 A b c d e^2 (d+e x)^2+70 A b c e^2 (d+e x)^3-35 A c^2 d^4 e+420 A c^2 d^3 e (d+e x)+630 A c^2 d^2 e (d+e x)^2-140 A c^2 d e (d+e x)^3+21 A c^2 e (d+e x)^4+35 b^2 B d^3 e^2-315 b^2 B d^2 e^2 (d+e x)-315 b^2 B d e^2 (d+e x)^2+35 b^2 B e^2 (d+e x)^3-70 b B c d^4 e+840 b B c d^3 e (d+e x)+1260 b B c d^2 e (d+e x)^2-280 b B c d e (d+e x)^3+42 b B c e (d+e x)^4+35 B c^2 d^5-525 B c^2 d^4 (d+e x)-1050 B c^2 d^3 (d+e x)^2+350 B c^2 d^2 (d+e x)^3-105 B c^2 d (d+e x)^4+15 B c^2 (d+e x)^5\right )}{105 e^6 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^2)/(d + e*x)^(5/2),x]

[Out]

(2*(35*B*c^2*d^5 - 70*b*B*c*d^4*e - 35*A*c^2*d^4*e + 35*b^2*B*d^3*e^2 + 70*A*b*c*d^3*e^2 - 35*A*b^2*d^2*e^3 -
525*B*c^2*d^4*(d + e*x) + 840*b*B*c*d^3*e*(d + e*x) + 420*A*c^2*d^3*e*(d + e*x) - 315*b^2*B*d^2*e^2*(d + e*x)
- 630*A*b*c*d^2*e^2*(d + e*x) + 210*A*b^2*d*e^3*(d + e*x) - 1050*B*c^2*d^3*(d + e*x)^2 + 1260*b*B*c*d^2*e*(d +
e*x)^2 + 630*A*c^2*d^2*e*(d + e*x)^2 - 315*b^2*B*d*e^2*(d + e*x)^2 - 630*A*b*c*d*e^2*(d + e*x)^2 + 105*A*b^2*
e^3*(d + e*x)^2 + 350*B*c^2*d^2*(d + e*x)^3 - 280*b*B*c*d*e*(d + e*x)^3 - 140*A*c^2*d*e*(d + e*x)^3 + 35*b^2*B
*e^2*(d + e*x)^3 + 70*A*b*c*e^2*(d + e*x)^3 - 105*B*c^2*d*(d + e*x)^4 + 42*b*B*c*e*(d + e*x)^4 + 21*A*c^2*e*(d
+ e*x)^4 + 15*B*c^2*(d + e*x)^5))/(105*e^6*(d + e*x)^(3/2))

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fricas [A]  time = 0.42, size = 310, normalized size = 1.18 \begin {gather*} \frac {2 \, {\left (15 \, B c^{2} e^{5} x^{5} - 1280 \, B c^{2} d^{5} + 280 \, A b^{2} d^{2} e^{3} + 896 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e - 560 \, {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} - 3 \, {\left (10 \, B c^{2} d e^{4} - 7 \, {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + {\left (80 \, B c^{2} d^{2} e^{3} - 56 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} + 35 \, {\left (B b^{2} + 2 \, A b c\right )} e^{5}\right )} x^{3} - 3 \, {\left (160 \, B c^{2} d^{3} e^{2} - 35 \, A b^{2} e^{5} - 112 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} + 70 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{4}\right )} x^{2} - 12 \, {\left (160 \, B c^{2} d^{4} e - 35 \, A b^{2} d e^{4} - 112 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + 70 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{105 \, {\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*B*c^2*e^5*x^5 - 1280*B*c^2*d^5 + 280*A*b^2*d^2*e^3 + 896*(2*B*b*c + A*c^2)*d^4*e - 560*(B*b^2 + 2*A*
b*c)*d^3*e^2 - 3*(10*B*c^2*d*e^4 - 7*(2*B*b*c + A*c^2)*e^5)*x^4 + (80*B*c^2*d^2*e^3 - 56*(2*B*b*c + A*c^2)*d*e
^4 + 35*(B*b^2 + 2*A*b*c)*e^5)*x^3 - 3*(160*B*c^2*d^3*e^2 - 35*A*b^2*e^5 - 112*(2*B*b*c + A*c^2)*d^2*e^3 + 70*
(B*b^2 + 2*A*b*c)*d*e^4)*x^2 - 12*(160*B*c^2*d^4*e - 35*A*b^2*d*e^4 - 112*(2*B*b*c + A*c^2)*d^3*e^2 + 70*(B*b^
2 + 2*A*b*c)*d^2*e^3)*x)*sqrt(e*x + d)/(e^8*x^2 + 2*d*e^7*x + d^2*e^6)

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giac [A]  time = 0.26, size = 427, normalized size = 1.62 \begin {gather*} \frac {2}{105} \, {\left (15 \, {\left (x e + d\right )}^{\frac {7}{2}} B c^{2} e^{36} - 105 \, {\left (x e + d\right )}^{\frac {5}{2}} B c^{2} d e^{36} + 350 \, {\left (x e + d\right )}^{\frac {3}{2}} B c^{2} d^{2} e^{36} - 1050 \, \sqrt {x e + d} B c^{2} d^{3} e^{36} + 42 \, {\left (x e + d\right )}^{\frac {5}{2}} B b c e^{37} + 21 \, {\left (x e + d\right )}^{\frac {5}{2}} A c^{2} e^{37} - 280 \, {\left (x e + d\right )}^{\frac {3}{2}} B b c d e^{37} - 140 \, {\left (x e + d\right )}^{\frac {3}{2}} A c^{2} d e^{37} + 1260 \, \sqrt {x e + d} B b c d^{2} e^{37} + 630 \, \sqrt {x e + d} A c^{2} d^{2} e^{37} + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{2} e^{38} + 70 \, {\left (x e + d\right )}^{\frac {3}{2}} A b c e^{38} - 315 \, \sqrt {x e + d} B b^{2} d e^{38} - 630 \, \sqrt {x e + d} A b c d e^{38} + 105 \, \sqrt {x e + d} A b^{2} e^{39}\right )} e^{\left (-42\right )} - \frac {2 \, {\left (15 \, {\left (x e + d\right )} B c^{2} d^{4} - B c^{2} d^{5} - 24 \, {\left (x e + d\right )} B b c d^{3} e - 12 \, {\left (x e + d\right )} A c^{2} d^{3} e + 2 \, B b c d^{4} e + A c^{2} d^{4} e + 9 \, {\left (x e + d\right )} B b^{2} d^{2} e^{2} + 18 \, {\left (x e + d\right )} A b c d^{2} e^{2} - B b^{2} d^{3} e^{2} - 2 \, A b c d^{3} e^{2} - 6 \, {\left (x e + d\right )} A b^{2} d e^{3} + A b^{2} d^{2} e^{3}\right )} e^{\left (-6\right )}}{3 \, {\left (x e + d\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*(x*e + d)^(7/2)*B*c^2*e^36 - 105*(x*e + d)^(5/2)*B*c^2*d*e^36 + 350*(x*e + d)^(3/2)*B*c^2*d^2*e^36 -
1050*sqrt(x*e + d)*B*c^2*d^3*e^36 + 42*(x*e + d)^(5/2)*B*b*c*e^37 + 21*(x*e + d)^(5/2)*A*c^2*e^37 - 280*(x*e
+ d)^(3/2)*B*b*c*d*e^37 - 140*(x*e + d)^(3/2)*A*c^2*d*e^37 + 1260*sqrt(x*e + d)*B*b*c*d^2*e^37 + 630*sqrt(x*e
+ d)*A*c^2*d^2*e^37 + 35*(x*e + d)^(3/2)*B*b^2*e^38 + 70*(x*e + d)^(3/2)*A*b*c*e^38 - 315*sqrt(x*e + d)*B*b^2*
d*e^38 - 630*sqrt(x*e + d)*A*b*c*d*e^38 + 105*sqrt(x*e + d)*A*b^2*e^39)*e^(-42) - 2/3*(15*(x*e + d)*B*c^2*d^4
- B*c^2*d^5 - 24*(x*e + d)*B*b*c*d^3*e - 12*(x*e + d)*A*c^2*d^3*e + 2*B*b*c*d^4*e + A*c^2*d^4*e + 9*(x*e + d)*
B*b^2*d^2*e^2 + 18*(x*e + d)*A*b*c*d^2*e^2 - B*b^2*d^3*e^2 - 2*A*b*c*d^3*e^2 - 6*(x*e + d)*A*b^2*d*e^3 + A*b^2
*d^2*e^3)*e^(-6)/(x*e + d)^(3/2)

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maple [A]  time = 0.06, size = 341, normalized size = 1.30 \begin {gather*} \frac {\frac {512}{15} B b c \,d^{4} e -\frac {4}{7} B \,c^{2} d \,e^{4} x^{4}+2 A \,b^{2} e^{5} x^{2}+\frac {16}{3} A \,b^{2} d^{2} e^{3}+\frac {256}{15} A \,c^{2} d^{4} e -\frac {32}{3} B \,b^{2} d^{3} e^{2}+\frac {2}{3} B \,b^{2} e^{5} x^{3}+\frac {2}{5} A \,c^{2} e^{5} x^{4}+\frac {256}{5} B b c \,d^{3} e^{2} x -32 A b c \,d^{2} e^{3} x -8 A b c d \,e^{4} x^{2}+\frac {64}{5} B b c \,d^{2} e^{3} x^{2}-\frac {32}{15} B b c d \,e^{4} x^{3}+\frac {2}{7} B \,c^{2} x^{5} e^{5}-\frac {512}{21} B \,c^{2} d^{5}+\frac {32}{21} B \,c^{2} d^{2} e^{3} x^{3}+\frac {128}{5} A \,c^{2} d^{3} e^{2} x -16 B \,b^{2} d^{2} e^{3} x -\frac {16}{15} A \,c^{2} d \,e^{4} x^{3}+\frac {32}{5} A \,c^{2} d^{2} e^{3} x^{2}-4 B \,b^{2} d \,e^{4} x^{2}-\frac {64}{7} B \,c^{2} d^{3} e^{2} x^{2}+8 A \,b^{2} d \,e^{4} x +\frac {4}{5} B b c \,e^{5} x^{4}+\frac {4}{3} A b c \,e^{5} x^{3}-\frac {64}{3} A b c \,d^{3} e^{2}-\frac {256}{7} B \,c^{2} d^{4} e x}{\left (e x +d \right )^{\frac {3}{2}} e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*B*c^2*e^5*x^5+21*A*c^2*e^5*x^4+42*B*b*c*e^5*x^4-30*B*c^2*d*e^4*x^4+70*A*b*c*e^5*x^3-56*A*c^2*d*e^4*x
^3+35*B*b^2*e^5*x^3-112*B*b*c*d*e^4*x^3+80*B*c^2*d^2*e^3*x^3+105*A*b^2*e^5*x^2-420*A*b*c*d*e^4*x^2+336*A*c^2*d
^2*e^3*x^2-210*B*b^2*d*e^4*x^2+672*B*b*c*d^2*e^3*x^2-480*B*c^2*d^3*e^2*x^2+420*A*b^2*d*e^4*x-1680*A*b*c*d^2*e^
3*x+1344*A*c^2*d^3*e^2*x-840*B*b^2*d^2*e^3*x+2688*B*b*c*d^3*e^2*x-1920*B*c^2*d^4*e*x+280*A*b^2*d^2*e^3-1120*A*
b*c*d^3*e^2+896*A*c^2*d^4*e-560*B*b^2*d^3*e^2+1792*B*b*c*d^4*e-1280*B*c^2*d^5)/(e*x+d)^(3/2)/e^6

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maxima [A]  time = 0.54, size = 297, normalized size = 1.13 \begin {gather*} \frac {2 \, {\left (\frac {15 \, {\left (e x + d\right )}^{\frac {7}{2}} B c^{2} - 21 \, {\left (5 \, B c^{2} d - {\left (2 \, B b c + A c^{2}\right )} e\right )} {\left (e x + d\right )}^{\frac {5}{2}} + 35 \, {\left (10 \, B c^{2} d^{2} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e + {\left (B b^{2} + 2 \, A b c\right )} e^{2}\right )} {\left (e x + d\right )}^{\frac {3}{2}} - 105 \, {\left (10 \, B c^{2} d^{3} - A b^{2} e^{3} - 6 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e + 3 \, {\left (B b^{2} + 2 \, A b c\right )} d e^{2}\right )} \sqrt {e x + d}}{e^{5}} + \frac {35 \, {\left (B c^{2} d^{5} - A b^{2} d^{2} e^{3} - {\left (2 \, B b c + A c^{2}\right )} d^{4} e + {\left (B b^{2} + 2 \, A b c\right )} d^{3} e^{2} - 3 \, {\left (5 \, B c^{2} d^{4} - 2 \, A b^{2} d e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e + 3 \, {\left (B b^{2} + 2 \, A b c\right )} d^{2} e^{2}\right )} {\left (e x + d\right )}\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} e^{5}}\right )}}{105 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*((15*(e*x + d)^(7/2)*B*c^2 - 21*(5*B*c^2*d - (2*B*b*c + A*c^2)*e)*(e*x + d)^(5/2) + 35*(10*B*c^2*d^2 - 4
*(2*B*b*c + A*c^2)*d*e + (B*b^2 + 2*A*b*c)*e^2)*(e*x + d)^(3/2) - 105*(10*B*c^2*d^3 - A*b^2*e^3 - 6*(2*B*b*c +
A*c^2)*d^2*e + 3*(B*b^2 + 2*A*b*c)*d*e^2)*sqrt(e*x + d))/e^5 + 35*(B*c^2*d^5 - A*b^2*d^2*e^3 - (2*B*b*c + A*c
^2)*d^4*e + (B*b^2 + 2*A*b*c)*d^3*e^2 - 3*(5*B*c^2*d^4 - 2*A*b^2*d*e^3 - 4*(2*B*b*c + A*c^2)*d^3*e + 3*(B*b^2
+ 2*A*b*c)*d^2*e^2)*(e*x + d))/((e*x + d)^(3/2)*e^5))/e

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mupad [B]  time = 0.07, size = 316, normalized size = 1.20 \begin {gather*} \frac {{\left (d+e\,x\right )}^{5/2}\,\left (2\,A\,c^2\,e-10\,B\,c^2\,d+4\,B\,b\,c\,e\right )}{5\,e^6}-\frac {\left (d+e\,x\right )\,\left (6\,B\,b^2\,d^2\,e^2-4\,A\,b^2\,d\,e^3-16\,B\,b\,c\,d^3\,e+12\,A\,b\,c\,d^2\,e^2+10\,B\,c^2\,d^4-8\,A\,c^2\,d^3\,e\right )-\frac {2\,B\,c^2\,d^5}{3}+\frac {2\,A\,c^2\,d^4\,e}{3}+\frac {2\,A\,b^2\,d^2\,e^3}{3}-\frac {2\,B\,b^2\,d^3\,e^2}{3}+\frac {4\,B\,b\,c\,d^4\,e}{3}-\frac {4\,A\,b\,c\,d^3\,e^2}{3}}{e^6\,{\left (d+e\,x\right )}^{3/2}}+\frac {\sqrt {d+e\,x}\,\left (-6\,B\,b^2\,d\,e^2+2\,A\,b^2\,e^3+24\,B\,b\,c\,d^2\,e-12\,A\,b\,c\,d\,e^2-20\,B\,c^2\,d^3+12\,A\,c^2\,d^2\,e\right )}{e^6}+\frac {{\left (d+e\,x\right )}^{3/2}\,\left (2\,B\,b^2\,e^2-16\,B\,b\,c\,d\,e+4\,A\,b\,c\,e^2+20\,B\,c^2\,d^2-8\,A\,c^2\,d\,e\right )}{3\,e^6}+\frac {2\,B\,c^2\,{\left (d+e\,x\right )}^{7/2}}{7\,e^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d + e*x)^(5/2)*(2*A*c^2*e - 10*B*c^2*d + 4*B*b*c*e))/(5*e^6) - ((d + e*x)*(10*B*c^2*d^4 - 4*A*b^2*d*e^3 - 8*
A*c^2*d^3*e + 6*B*b^2*d^2*e^2 - 16*B*b*c*d^3*e + 12*A*b*c*d^2*e^2) - (2*B*c^2*d^5)/3 + (2*A*c^2*d^4*e)/3 + (2*
A*b^2*d^2*e^3)/3 - (2*B*b^2*d^3*e^2)/3 + (4*B*b*c*d^4*e)/3 - (4*A*b*c*d^3*e^2)/3)/(e^6*(d + e*x)^(3/2)) + ((d
+ e*x)^(1/2)*(2*A*b^2*e^3 - 20*B*c^2*d^3 + 12*A*c^2*d^2*e - 6*B*b^2*d*e^2 - 12*A*b*c*d*e^2 + 24*B*b*c*d^2*e))/
e^6 + ((d + e*x)^(3/2)*(2*B*b^2*e^2 + 20*B*c^2*d^2 + 4*A*b*c*e^2 - 8*A*c^2*d*e - 16*B*b*c*d*e))/(3*e^6) + (2*B
*c^2*(d + e*x)^(7/2))/(7*e^6)

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sympy [A]  time = 75.51, size = 292, normalized size = 1.11 \begin {gather*} \frac {2 B c^{2} \left (d + e x\right )^{\frac {7}{2}}}{7 e^{6}} + \frac {2 d^{2} \left (- A e + B d\right ) \left (b e - c d\right )^{2}}{3 e^{6} \left (d + e x\right )^{\frac {3}{2}}} - \frac {2 d \left (b e - c d\right ) \left (- 2 A b e^{2} + 4 A c d e + 3 B b d e - 5 B c d^{2}\right )}{e^{6} \sqrt {d + e x}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (2 A c^{2} e + 4 B b c e - 10 B c^{2} d\right )}{5 e^{6}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (4 A b c e^{2} - 8 A c^{2} d e + 2 B b^{2} e^{2} - 16 B b c d e + 20 B c^{2} d^{2}\right )}{3 e^{6}} + \frac {\sqrt {d + e x} \left (2 A b^{2} e^{3} - 12 A b c d e^{2} + 12 A c^{2} d^{2} e - 6 B b^{2} d e^{2} + 24 B b c d^{2} e - 20 B c^{2} d^{3}\right )}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B*c**2*(d + e*x)**(7/2)/(7*e**6) + 2*d**2*(-A*e + B*d)*(b*e - c*d)**2/(3*e**6*(d + e*x)**(3/2)) - 2*d*(b*e -
c*d)*(-2*A*b*e**2 + 4*A*c*d*e + 3*B*b*d*e - 5*B*c*d**2)/(e**6*sqrt(d + e*x)) + (d + e*x)**(5/2)*(2*A*c**2*e +
4*B*b*c*e - 10*B*c**2*d)/(5*e**6) + (d + e*x)**(3/2)*(4*A*b*c*e**2 - 8*A*c**2*d*e + 2*B*b**2*e**2 - 16*B*b*c*
d*e + 20*B*c**2*d**2)/(3*e**6) + sqrt(d + e*x)*(2*A*b**2*e**3 - 12*A*b*c*d*e**2 + 12*A*c**2*d**2*e - 6*B*b**2*
d*e**2 + 24*B*b*c*d**2*e - 20*B*c**2*d**3)/e**6

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