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

Optimal. Leaf size=61 \[ \frac {2 B \left (b x+c x^2\right )^{7/2}}{9 c x^{5/2}}-\frac {2 \left (b x+c x^2\right )^{7/2} (2 b B-9 A c)}{63 c^2 x^{7/2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {794, 648} \begin {gather*} \frac {2 B \left (b x+c x^2\right )^{7/2}}{9 c x^{5/2}}-\frac {2 \left (b x+c x^2\right )^{7/2} (2 b B-9 A c)}{63 c^2 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*b*B - 9*A*c)*(b*x + c*x^2)^(7/2))/(63*c^2*x^(7/2)) + (2*B*(b*x + c*x^2)^(7/2))/(9*c*x^(5/2))

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 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^{5/2}} \, dx &=\frac {2 B \left (b x+c x^2\right )^{7/2}}{9 c x^{5/2}}+\frac {\left (2 \left (-\frac {5}{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^{5/2}} \, dx}{9 c}\\ &=-\frac {2 (2 b B-9 A c) \left (b x+c x^2\right )^{7/2}}{63 c^2 x^{7/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{9 c x^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 44, normalized size = 0.72 \begin {gather*} \frac {2 (b+c x)^3 \sqrt {x (b+c x)} (9 A c-2 b B+7 B c x)}{63 c^2 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b + c*x)^3*Sqrt[x*(b + c*x)]*(-2*b*B + 9*A*c + 7*B*c*x))/(63*c^2*Sqrt[x])

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IntegrateAlgebraic [A]  time = 0.58, size = 39, normalized size = 0.64 \begin {gather*} \frac {2 \left (b x+c x^2\right )^{7/2} (9 A c-2 b B+7 B c x)}{63 c^2 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-2*b*B + 9*A*c + 7*B*c*x)*(b*x + c*x^2)^(7/2))/(63*c^2*x^(7/2))

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fricas [B]  time = 0.41, size = 100, normalized size = 1.64 \begin {gather*} \frac {2 \, {\left (7 \, B c^{4} x^{4} - 2 \, B b^{4} + 9 \, A b^{3} c + {\left (19 \, B b c^{3} + 9 \, A c^{4}\right )} x^{3} + 3 \, {\left (5 \, B b^{2} c^{2} + 9 \, A b c^{3}\right )} x^{2} + {\left (B b^{3} c + 27 \, A b^{2} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{63 \, c^{2} \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^(5/2),x, algorithm="fricas")

[Out]

2/63*(7*B*c^4*x^4 - 2*B*b^4 + 9*A*b^3*c + (19*B*b*c^3 + 9*A*c^4)*x^3 + 3*(5*B*b^2*c^2 + 9*A*b*c^3)*x^2 + (B*b^
3*c + 27*A*b^2*c^2)*x)*sqrt(c*x^2 + b*x)/(c^2*sqrt(x))

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giac [B]  time = 0.24, size = 270, normalized size = 4.43 \begin {gather*} \frac {2}{315} \, B 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 {4}{105} \, B 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}{105} \, A c^{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 {2}{15} \, B 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 )} + \frac {4}{15} \, A b c {\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 )} + \frac {2}{3} \, A b^{2} {\left (\frac {{\left (c x + b\right )}^{\frac {3}{2}}}{c} - \frac {b^{\frac {3}{2}}}{c}\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^(5/2),x, algorithm="giac")

[Out]

2/315*B*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) - 4/105*B*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) - 2/105*A*c^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) + 2/15*B*b^2*(2*b^(5/2)/c^2 + (3*(c*x + b)^(5/2) - 5*(c*x + b)^(3/2)*b)/c^2) + 4/15*A*b*c*(2*b^(5/
2)/c^2 + (3*(c*x + b)^(5/2) - 5*(c*x + b)^(3/2)*b)/c^2) + 2/3*A*b^2*((c*x + b)^(3/2)/c - b^(3/2)/c)

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maple [A]  time = 0.04, size = 39, normalized size = 0.64 \begin {gather*} \frac {2 \left (c x +b \right ) \left (7 B c x +9 A c -2 b B \right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{63 c^{2} 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^(5/2),x)

[Out]

2/63*(c*x+b)*(7*B*c*x+9*A*c-2*B*b)*(c*x^2+b*x)^(5/2)/c^2/x^(5/2)

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maxima [B]  time = 0.61, size = 230, normalized size = 3.77 \begin {gather*} \frac {2 \, {\left (35 \, b^{2} c x^{3} + 35 \, b^{3} x^{2} + {\left (15 \, c^{3} x^{3} + 3 \, b c^{2} x^{2} - 4 \, b^{2} c x + 8 \, b^{3}\right )} x^{2} + 14 \, {\left (3 \, b c^{2} x^{3} + b^{2} c x^{2} - 2 \, b^{3} x\right )} x\right )} \sqrt {c x + b} A}{105 \, c x^{2}} + \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} B}{315 \, c^{2} x^{3}} \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^(5/2),x, algorithm="maxima")

[Out]

2/105*(35*b^2*c*x^3 + 35*b^3*x^2 + (15*c^3*x^3 + 3*b*c^2*x^2 - 4*b^2*c*x + 8*b^3)*x^2 + 14*(3*b*c^2*x^3 + b^2*
c*x^2 - 2*b^3*x)*x)*sqrt(c*x + b)*A/(c*x^2) + 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 - 1
6*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)*B/(c^2*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (A+B\,x\right )}{x^{5/2}} \,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^(5/2),x)

[Out]

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

[Out]

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

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