3.3.17 \(\int \frac {(A+B x) (b x+c x^2)^{5/2}}{x^{3/2}} \, dx\)

Optimal. Leaf size=96 \[ \frac {4 b \left (b x+c x^2\right )^{7/2} (4 b B-11 A c)}{693 c^3 x^{7/2}}-\frac {2 \left (b x+c x^2\right )^{7/2} (4 b B-11 A c)}{99 c^2 x^{5/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{11 c x^{3/2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {794, 656, 648} \begin {gather*} -\frac {2 \left (b x+c x^2\right )^{7/2} (4 b B-11 A c)}{99 c^2 x^{5/2}}+\frac {4 b \left (b x+c x^2\right )^{7/2} (4 b B-11 A c)}{693 c^3 x^{7/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{11 c x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*(4*b*B - 11*A*c)*(b*x + c*x^2)^(7/2))/(693*c^3*x^(7/2)) - (2*(4*b*B - 11*A*c)*(b*x + c*x^2)^(7/2))/(99*c^
2*x^(5/2)) + (2*B*(b*x + c*x^2)^(7/2))/(11*c*x^(3/2))

Rule 648

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rule 656

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (b x+c x^2\right )^{5/2}}{x^{3/2}} \, dx &=\frac {2 B \left (b x+c x^2\right )^{7/2}}{11 c x^{3/2}}+\frac {\left (2 \left (-\frac {3}{2} (-b B+A c)+\frac {7}{2} (-b B+2 A c)\right )\right ) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^{3/2}} \, dx}{11 c}\\ &=-\frac {2 (4 b B-11 A c) \left (b x+c x^2\right )^{7/2}}{99 c^2 x^{5/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{11 c x^{3/2}}+\frac {(2 b (4 b B-11 A c)) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^{5/2}} \, dx}{99 c^2}\\ &=\frac {4 b (4 b B-11 A c) \left (b x+c x^2\right )^{7/2}}{693 c^3 x^{7/2}}-\frac {2 (4 b B-11 A c) \left (b x+c x^2\right )^{7/2}}{99 c^2 x^{5/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{11 c x^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 63, normalized size = 0.66 \begin {gather*} \frac {2 (b+c x)^3 \sqrt {x (b+c x)} \left (-2 b c (11 A+14 B x)+7 c^2 x (11 A+9 B x)+8 b^2 B\right )}{693 c^3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b + c*x)^3*Sqrt[x*(b + c*x)]*(8*b^2*B + 7*c^2*x*(11*A + 9*B*x) - 2*b*c*(11*A + 14*B*x)))/(693*c^3*Sqrt[x])

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IntegrateAlgebraic [A]  time = 0.56, size = 59, normalized size = 0.61 \begin {gather*} \frac {2 \left (b x+c x^2\right )^{7/2} \left (-22 A b c+77 A c^2 x+8 b^2 B-28 b B c x+63 B c^2 x^2\right )}{693 c^3 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^(5/2))/x^(3/2),x]

[Out]

(2*(b*x + c*x^2)^(7/2)*(8*b^2*B - 22*A*b*c - 28*b*B*c*x + 77*A*c^2*x + 63*B*c^2*x^2))/(693*c^3*x^(7/2))

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fricas [A]  time = 0.41, size = 125, normalized size = 1.30 \begin {gather*} \frac {2 \, {\left (63 \, B c^{5} x^{5} + 8 \, B b^{5} - 22 \, A b^{4} c + 7 \, {\left (23 \, B b c^{4} + 11 \, A c^{5}\right )} x^{4} + {\left (113 \, B b^{2} c^{3} + 209 \, A b c^{4}\right )} x^{3} + 3 \, {\left (B b^{3} c^{2} + 55 \, A b^{2} c^{3}\right )} x^{2} - {\left (4 \, B b^{4} c - 11 \, A b^{3} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{693 \, c^{3} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/693*(63*B*c^5*x^5 + 8*B*b^5 - 22*A*b^4*c + 7*(23*B*b*c^4 + 11*A*c^5)*x^4 + (113*B*b^2*c^3 + 209*A*b*c^4)*x^3
+ 3*(B*b^3*c^2 + 55*A*b^2*c^3)*x^2 - (4*B*b^4*c - 11*A*b^3*c^2)*x)*sqrt(c*x^2 + b*x)/(c^3*sqrt(x))

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giac [B]  time = 0.28, size = 344, normalized size = 3.58 \begin {gather*} -\frac {2}{3465} \, B c^{2} {\left (\frac {128 \, b^{\frac {11}{2}}}{c^{5}} - \frac {315 \, {\left (c x + b\right )}^{\frac {11}{2}} - 1540 \, {\left (c x + b\right )}^{\frac {9}{2}} b + 2970 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{2} - 2772 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}}{c^{5}}\right )} + \frac {4}{315} \, B b c {\left (\frac {16 \, b^{\frac {9}{2}}}{c^{4}} + \frac {35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} + \frac {2}{315} \, A c^{2} {\left (\frac {16 \, b^{\frac {9}{2}}}{c^{4}} + \frac {35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} - \frac {2}{105} \, B b^{2} {\left (\frac {8 \, b^{\frac {7}{2}}}{c^{3}} - \frac {15 \, {\left (c x + b\right )}^{\frac {7}{2}} - 42 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}}{c^{3}}\right )} - \frac {4}{105} \, A b c {\left (\frac {8 \, b^{\frac {7}{2}}}{c^{3}} - \frac {15 \, {\left (c x + b\right )}^{\frac {7}{2}} - 42 \, {\left (c x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2}}{c^{3}}\right )} + \frac {2}{15} \, A b^{2} {\left (\frac {2 \, b^{\frac {5}{2}}}{c^{2}} + \frac {3 \, {\left (c x + b\right )}^{\frac {5}{2}} - 5 \, {\left (c x + b\right )}^{\frac {3}{2}} b}{c^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*B*c^2*(128*b^(11/2)/c^5 - (315*(c*x + b)^(11/2) - 1540*(c*x + b)^(9/2)*b + 2970*(c*x + b)^(7/2)*b^2 -
2772*(c*x + b)^(5/2)*b^3 + 1155*(c*x + b)^(3/2)*b^4)/c^5) + 4/315*B*b*c*(16*b^(9/2)/c^4 + (35*(c*x + b)^(9/2)
- 135*(c*x + b)^(7/2)*b + 189*(c*x + b)^(5/2)*b^2 - 105*(c*x + b)^(3/2)*b^3)/c^4) + 2/315*A*c^2*(16*b^(9/2)/c^
4 + (35*(c*x + b)^(9/2) - 135*(c*x + b)^(7/2)*b + 189*(c*x + b)^(5/2)*b^2 - 105*(c*x + b)^(3/2)*b^3)/c^4) - 2/
105*B*b^2*(8*b^(7/2)/c^3 - (15*(c*x + b)^(7/2) - 42*(c*x + b)^(5/2)*b + 35*(c*x + b)^(3/2)*b^2)/c^3) - 4/105*A
*b*c*(8*b^(7/2)/c^3 - (15*(c*x + b)^(7/2) - 42*(c*x + b)^(5/2)*b + 35*(c*x + b)^(3/2)*b^2)/c^3) + 2/15*A*b^2*(
2*b^(5/2)/c^2 + (3*(c*x + b)^(5/2) - 5*(c*x + b)^(3/2)*b)/c^2)

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maple [A]  time = 0.05, size = 59, normalized size = 0.61 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-63 B \,c^{2} x^{2}-77 A \,c^{2} x +28 B b c x +22 A b c -8 b^{2} B \right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{693 c^{3} x^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^(3/2),x)

[Out]

-2/693*(c*x+b)*(-63*B*c^2*x^2-77*A*c^2*x+28*B*b*c*x+22*A*b*c-8*B*b^2)*(c*x^2+b*x)^(5/2)/c^3/x^(5/2)

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maxima [B]  time = 0.86, size = 305, normalized size = 3.18 \begin {gather*} \frac {2 \, {\left ({\left (35 \, c^{4} x^{4} + 5 \, b c^{3} x^{3} - 6 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 16 \, b^{4}\right )} x^{3} + 6 \, {\left (15 \, b c^{3} x^{4} + 3 \, b^{2} c^{2} x^{3} - 4 \, b^{3} c x^{2} + 8 \, b^{4} x\right )} x^{2} + 21 \, {\left (3 \, b^{2} c^{2} x^{4} + b^{3} c x^{3} - 2 \, b^{4} x^{2}\right )} x\right )} \sqrt {c x + b} A}{315 \, c^{2} x^{3}} + \frac {2 \, {\left ({\left (315 \, c^{5} x^{5} + 35 \, b c^{4} x^{4} - 40 \, b^{2} c^{3} x^{3} + 48 \, b^{3} c^{2} x^{2} - 64 \, b^{4} c x + 128 \, b^{5}\right )} x^{4} + 22 \, {\left (35 \, b c^{4} x^{5} + 5 \, b^{2} c^{3} x^{4} - 6 \, b^{3} c^{2} x^{3} + 8 \, b^{4} c x^{2} - 16 \, b^{5} x\right )} x^{3} + 33 \, {\left (15 \, b^{2} c^{3} x^{5} + 3 \, b^{3} c^{2} x^{4} - 4 \, b^{4} c x^{3} + 8 \, b^{5} x^{2}\right )} x^{2}\right )} \sqrt {c x + b} B}{3465 \, c^{3} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*((35*c^4*x^4 + 5*b*c^3*x^3 - 6*b^2*c^2*x^2 + 8*b^3*c*x - 16*b^4)*x^3 + 6*(15*b*c^3*x^4 + 3*b^2*c^2*x^3 -
4*b^3*c*x^2 + 8*b^4*x)*x^2 + 21*(3*b^2*c^2*x^4 + b^3*c*x^3 - 2*b^4*x^2)*x)*sqrt(c*x + b)*A/(c^2*x^3) + 2/3465
*((315*c^5*x^5 + 35*b*c^4*x^4 - 40*b^2*c^3*x^3 + 48*b^3*c^2*x^2 - 64*b^4*c*x + 128*b^5)*x^4 + 22*(35*b*c^4*x^5
+ 5*b^2*c^3*x^4 - 6*b^3*c^2*x^3 + 8*b^4*c*x^2 - 16*b^5*x)*x^3 + 33*(15*b^2*c^3*x^5 + 3*b^3*c^2*x^4 - 4*b^4*c*
x^3 + 8*b^5*x^2)*x^2)*sqrt(c*x + b)*B/(c^3*x^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(3/2),x)

[Out]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{x^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(5/2)/x**(3/2),x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**(3/2), x)

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