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

Optimal. Leaf size=133 \[ -\frac {16 b^2 \left (b x+c x^2\right )^{7/2} (6 b B-13 A c)}{9009 c^4 x^{7/2}}+\frac {8 b \left (b x+c x^2\right )^{7/2} (6 b B-13 A c)}{1287 c^3 x^{5/2}}-\frac {2 \left (b x+c x^2\right )^{7/2} (6 b B-13 A c)}{143 c^2 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{13 c \sqrt {x}} \]

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Rubi [A]  time = 0.11, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {16 b^2 \left (b x+c x^2\right )^{7/2} (6 b B-13 A c)}{9009 c^4 x^{7/2}}-\frac {2 \left (b x+c x^2\right )^{7/2} (6 b B-13 A c)}{143 c^2 x^{3/2}}+\frac {8 b \left (b x+c x^2\right )^{7/2} (6 b B-13 A c)}{1287 c^3 x^{5/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{13 c \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*b^2*(6*b*B - 13*A*c)*(b*x + c*x^2)^(7/2))/(9009*c^4*x^(7/2)) + (8*b*(6*b*B - 13*A*c)*(b*x + c*x^2)^(7/2))
/(1287*c^3*x^(5/2)) - (2*(6*b*B - 13*A*c)*(b*x + c*x^2)^(7/2))/(143*c^2*x^(3/2)) + (2*B*(b*x + c*x^2)^(7/2))/(
13*c*Sqrt[x])

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}}{\sqrt {x}} \, dx &=\frac {2 B \left (b x+c x^2\right )^{7/2}}{13 c \sqrt {x}}+\frac {\left (2 \left (\frac {1}{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}}{\sqrt {x}} \, dx}{13 c}\\ &=-\frac {2 (6 b B-13 A c) \left (b x+c x^2\right )^{7/2}}{143 c^2 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{13 c \sqrt {x}}+\frac {(4 b (6 b B-13 A c)) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^{3/2}} \, dx}{143 c^2}\\ &=\frac {8 b (6 b B-13 A c) \left (b x+c x^2\right )^{7/2}}{1287 c^3 x^{5/2}}-\frac {2 (6 b B-13 A c) \left (b x+c x^2\right )^{7/2}}{143 c^2 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{13 c \sqrt {x}}-\frac {\left (8 b^2 (6 b B-13 A c)\right ) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^{5/2}} \, dx}{1287 c^3}\\ &=-\frac {16 b^2 (6 b B-13 A c) \left (b x+c x^2\right )^{7/2}}{9009 c^4 x^{7/2}}+\frac {8 b (6 b B-13 A c) \left (b x+c x^2\right )^{7/2}}{1287 c^3 x^{5/2}}-\frac {2 (6 b B-13 A c) \left (b x+c x^2\right )^{7/2}}{143 c^2 x^{3/2}}+\frac {2 B \left (b x+c x^2\right )^{7/2}}{13 c \sqrt {x}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 82, normalized size = 0.62 \begin {gather*} \frac {2 (b+c x)^3 \sqrt {x (b+c x)} \left (8 b^2 c (13 A+21 B x)-14 b c^2 x (26 A+27 B x)+63 c^3 x^2 (13 A+11 B x)-48 b^3 B\right )}{9009 c^4 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b + c*x)^3*Sqrt[x*(b + c*x)]*(-48*b^3*B + 63*c^3*x^2*(13*A + 11*B*x) + 8*b^2*c*(13*A + 21*B*x) - 14*b*c^2*
x*(26*A + 27*B*x)))/(9009*c^4*Sqrt[x])

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IntegrateAlgebraic [A]  time = 0.53, size = 83, normalized size = 0.62 \begin {gather*} \frac {2 \left (b x+c x^2\right )^{7/2} \left (104 A b^2 c-364 A b c^2 x+819 A c^3 x^2-48 b^3 B+168 b^2 B c x-378 b B c^2 x^2+693 B c^3 x^3\right )}{9009 c^4 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*x + c*x^2)^(7/2)*(-48*b^3*B + 104*A*b^2*c + 168*b^2*B*c*x - 364*A*b*c^2*x - 378*b*B*c^2*x^2 + 819*A*c^3*
x^2 + 693*B*c^3*x^3))/(9009*c^4*x^(7/2))

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fricas [A]  time = 0.40, size = 150, normalized size = 1.13 \begin {gather*} \frac {2 \, {\left (693 \, B c^{6} x^{6} - 48 \, B b^{6} + 104 \, A b^{5} c + 63 \, {\left (27 \, B b c^{5} + 13 \, A c^{6}\right )} x^{5} + 7 \, {\left (159 \, B b^{2} c^{4} + 299 \, A b c^{5}\right )} x^{4} + {\left (15 \, B b^{3} c^{3} + 1469 \, A b^{2} c^{4}\right )} x^{3} - 3 \, {\left (6 \, B b^{4} c^{2} - 13 \, A b^{3} c^{3}\right )} x^{2} + 4 \, {\left (6 \, B b^{5} c - 13 \, A b^{4} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{9009 \, c^{4} \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^(1/2),x, algorithm="fricas")

[Out]

2/9009*(693*B*c^6*x^6 - 48*B*b^6 + 104*A*b^5*c + 63*(27*B*b*c^5 + 13*A*c^6)*x^5 + 7*(159*B*b^2*c^4 + 299*A*b*c
^5)*x^4 + (15*B*b^3*c^3 + 1469*A*b^2*c^4)*x^3 - 3*(6*B*b^4*c^2 - 13*A*b^3*c^3)*x^2 + 4*(6*B*b^5*c - 13*A*b^4*c
^2)*x)*sqrt(c*x^2 + b*x)/(c^4*sqrt(x))

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giac [B]  time = 0.27, size = 416, normalized size = 3.13 \begin {gather*} \frac {2}{9009} \, B c^{2} {\left (\frac {256 \, b^{\frac {13}{2}}}{c^{6}} + \frac {693 \, {\left (c x + b\right )}^{\frac {13}{2}} - 4095 \, {\left (c x + b\right )}^{\frac {11}{2}} b + 10010 \, {\left (c x + b\right )}^{\frac {9}{2}} b^{2} - 12870 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{3} + 9009 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{4} - 3003 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{5}}{c^{6}}\right )} - \frac {4}{3465} \, B b c {\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 {2}{3465} \, A 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 {2}{315} \, B b^{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}{315} \, A 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}{105} \, A 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 )} \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^(1/2),x, algorithm="giac")

[Out]

2/9009*B*c^2*(256*b^(13/2)/c^6 + (693*(c*x + b)^(13/2) - 4095*(c*x + b)^(11/2)*b + 10010*(c*x + b)^(9/2)*b^2 -
12870*(c*x + b)^(7/2)*b^3 + 9009*(c*x + b)^(5/2)*b^4 - 3003*(c*x + b)^(3/2)*b^5)/c^6) - 4/3465*B*b*c*(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) - 2/3465*A*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) + 2/315*B*b^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) + 4/315*A*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) - 2/105*A*b^2*(8*b^(7/2)/c^3 - (15*(c*x + b)^(7/2) - 42*(c*x + b)^(5/2)*b + 3
5*(c*x + b)^(3/2)*b^2)/c^3)

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maple [A]  time = 0.05, size = 83, normalized size = 0.62 \begin {gather*} \frac {2 \left (c x +b \right ) \left (693 B \,c^{3} x^{3}+819 A \,c^{3} x^{2}-378 B b \,c^{2} x^{2}-364 A b \,c^{2} x +168 B \,b^{2} c x +104 A \,b^{2} c -48 b^{3} B \right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{9009 c^{4} 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^(1/2),x)

[Out]

2/9009*(c*x+b)*(693*B*c^3*x^3+819*A*c^3*x^2-378*B*b*c^2*x^2-364*A*b*c^2*x+168*B*b^2*c*x+104*A*b^2*c-48*B*b^3)*
(c*x^2+b*x)^(5/2)/c^4/x^(5/2)

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maxima [B]  time = 0.71, size = 375, normalized size = 2.82 \begin {gather*} \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} A}{3465 \, c^{3} x^{4}} + \frac {2 \, {\left (5 \, {\left (693 \, c^{6} x^{6} + 63 \, b c^{5} x^{5} - 70 \, b^{2} c^{4} x^{4} + 80 \, b^{3} c^{3} x^{3} - 96 \, b^{4} c^{2} x^{2} + 128 \, b^{5} c x - 256 \, b^{6}\right )} x^{5} + 26 \, {\left (315 \, b c^{5} x^{6} + 35 \, b^{2} c^{4} x^{5} - 40 \, b^{3} c^{3} x^{4} + 48 \, b^{4} c^{2} x^{3} - 64 \, b^{5} c x^{2} + 128 \, b^{6} x\right )} x^{4} + 143 \, {\left (35 \, b^{2} c^{4} x^{6} + 5 \, b^{3} c^{3} x^{5} - 6 \, b^{4} c^{2} x^{4} + 8 \, b^{5} c x^{3} - 16 \, b^{6} x^{2}\right )} x^{3}\right )} \sqrt {c x + b} B}{45045 \, c^{4} x^{5}} \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^(1/2),x, algorithm="maxima")

[Out]

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)*A/(c^3*x^4) + 2/45045*(5*(693*c^6*x^6 + 63*b*c^5*x^5 - 70*b^2*c^4*x^
4 + 80*b^3*c^3*x^3 - 96*b^4*c^2*x^2 + 128*b^5*c*x - 256*b^6)*x^5 + 26*(315*b*c^5*x^6 + 35*b^2*c^4*x^5 - 40*b^3
*c^3*x^4 + 48*b^4*c^2*x^3 - 64*b^5*c*x^2 + 128*b^6*x)*x^4 + 143*(35*b^2*c^4*x^6 + 5*b^3*c^3*x^5 - 6*b^4*c^2*x^
4 + 8*b^5*c*x^3 - 16*b^6*x^2)*x^3)*sqrt(c*x + b)*B/(c^4*x^5)

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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 )}{\sqrt {x}} \,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^(1/2),x)

[Out]

int(((b*x + c*x^2)^(5/2)*(A + B*x))/x^(1/2), 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 )}{\sqrt {x}}\, 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**(1/2),x)

[Out]

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

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