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

Optimal. Leaf size=170 \[ \frac {32 b^3 \left (b x+c x^2\right )^{7/2} (8 b B-15 A c)}{45045 c^5 x^{7/2}}-\frac {16 b^2 \left (b x+c x^2\right )^{7/2} (8 b B-15 A c)}{6435 c^4 x^{5/2}}+\frac {4 b \left (b x+c x^2\right )^{7/2} (8 b B-15 A c)}{715 c^3 x^{3/2}}-\frac {2 \left (b x+c x^2\right )^{7/2} (8 b B-15 A c)}{195 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{7/2}}{15 c} \]

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Rubi [A]  time = 0.14, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 5, 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} (8 b B-15 A c)}{6435 c^4 x^{5/2}}+\frac {32 b^3 \left (b x+c x^2\right )^{7/2} (8 b B-15 A c)}{45045 c^5 x^{7/2}}-\frac {2 \left (b x+c x^2\right )^{7/2} (8 b B-15 A c)}{195 c^2 \sqrt {x}}+\frac {4 b \left (b x+c x^2\right )^{7/2} (8 b B-15 A c)}{715 c^3 x^{3/2}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{7/2}}{15 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^3*(8*b*B - 15*A*c)*(b*x + c*x^2)^(7/2))/(45045*c^5*x^(7/2)) - (16*b^2*(8*b*B - 15*A*c)*(b*x + c*x^2)^(7/
2))/(6435*c^4*x^(5/2)) + (4*b*(8*b*B - 15*A*c)*(b*x + c*x^2)^(7/2))/(715*c^3*x^(3/2)) - (2*(8*b*B - 15*A*c)*(b
*x + c*x^2)^(7/2))/(195*c^2*Sqrt[x]) + (2*B*Sqrt[x]*(b*x + c*x^2)^(7/2))/(15*c)

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 \sqrt {x} (A+B x) \left (b x+c x^2\right )^{5/2} \, dx &=\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{7/2}}{15 c}+\frac {\left (2 \left (\frac {1}{2} (-b B+A c)+\frac {7}{2} (-b B+2 A c)\right )\right ) \int \sqrt {x} \left (b x+c x^2\right )^{5/2} \, dx}{15 c}\\ &=-\frac {2 (8 b B-15 A c) \left (b x+c x^2\right )^{7/2}}{195 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{7/2}}{15 c}+\frac {(2 b (8 b B-15 A c)) \int \frac {\left (b x+c x^2\right )^{5/2}}{\sqrt {x}} \, dx}{65 c^2}\\ &=\frac {4 b (8 b B-15 A c) \left (b x+c x^2\right )^{7/2}}{715 c^3 x^{3/2}}-\frac {2 (8 b B-15 A c) \left (b x+c x^2\right )^{7/2}}{195 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{7/2}}{15 c}-\frac {\left (8 b^2 (8 b B-15 A c)\right ) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^{3/2}} \, dx}{715 c^3}\\ &=-\frac {16 b^2 (8 b B-15 A c) \left (b x+c x^2\right )^{7/2}}{6435 c^4 x^{5/2}}+\frac {4 b (8 b B-15 A c) \left (b x+c x^2\right )^{7/2}}{715 c^3 x^{3/2}}-\frac {2 (8 b B-15 A c) \left (b x+c x^2\right )^{7/2}}{195 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{7/2}}{15 c}+\frac {\left (16 b^3 (8 b B-15 A c)\right ) \int \frac {\left (b x+c x^2\right )^{5/2}}{x^{5/2}} \, dx}{6435 c^4}\\ &=\frac {32 b^3 (8 b B-15 A c) \left (b x+c x^2\right )^{7/2}}{45045 c^5 x^{7/2}}-\frac {16 b^2 (8 b B-15 A c) \left (b x+c x^2\right )^{7/2}}{6435 c^4 x^{5/2}}+\frac {4 b (8 b B-15 A c) \left (b x+c x^2\right )^{7/2}}{715 c^3 x^{3/2}}-\frac {2 (8 b B-15 A c) \left (b x+c x^2\right )^{7/2}}{195 c^2 \sqrt {x}}+\frac {2 B \sqrt {x} \left (b x+c x^2\right )^{7/2}}{15 c}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 101, normalized size = 0.59 \begin {gather*} \frac {2 (b+c x)^3 \sqrt {x (b+c x)} \left (-16 b^3 c (15 A+28 B x)+168 b^2 c^2 x (5 A+6 B x)-42 b c^3 x^2 (45 A+44 B x)+231 c^4 x^3 (15 A+13 B x)+128 b^4 B\right )}{45045 c^5 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b + c*x)^3*Sqrt[x*(b + c*x)]*(128*b^4*B + 168*b^2*c^2*x*(5*A + 6*B*x) + 231*c^4*x^3*(15*A + 13*B*x) - 16*b
^3*c*(15*A + 28*B*x) - 42*b*c^3*x^2*(45*A + 44*B*x)))/(45045*c^5*Sqrt[x])

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IntegrateAlgebraic [A]  time = 0.60, size = 107, normalized size = 0.63 \begin {gather*} \frac {2 \left (b x+c x^2\right )^{7/2} \left (-240 A b^3 c+840 A b^2 c^2 x-1890 A b c^3 x^2+3465 A c^4 x^3+128 b^4 B-448 b^3 B c x+1008 b^2 B c^2 x^2-1848 b B c^3 x^3+3003 B c^4 x^4\right )}{45045 c^5 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*x + c*x^2)^(7/2)*(128*b^4*B - 240*A*b^3*c - 448*b^3*B*c*x + 840*A*b^2*c^2*x + 1008*b^2*B*c^2*x^2 - 1890*
A*b*c^3*x^2 - 1848*b*B*c^3*x^3 + 3465*A*c^4*x^3 + 3003*B*c^4*x^4))/(45045*c^5*x^(7/2))

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fricas [A]  time = 0.40, size = 174, normalized size = 1.02 \begin {gather*} \frac {2 \, {\left (3003 \, B c^{7} x^{7} + 128 \, B b^{7} - 240 \, A b^{6} c + 231 \, {\left (31 \, B b c^{6} + 15 \, A c^{7}\right )} x^{6} + 63 \, {\left (71 \, B b^{2} c^{5} + 135 \, A b c^{6}\right )} x^{5} + 35 \, {\left (B b^{3} c^{4} + 159 \, A b^{2} c^{5}\right )} x^{4} - 5 \, {\left (8 \, B b^{4} c^{3} - 15 \, A b^{3} c^{4}\right )} x^{3} + 6 \, {\left (8 \, B b^{5} c^{2} - 15 \, A b^{4} c^{3}\right )} x^{2} - 8 \, {\left (8 \, B b^{6} c - 15 \, A b^{5} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{45045 \, c^{5} \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]

2/45045*(3003*B*c^7*x^7 + 128*B*b^7 - 240*A*b^6*c + 231*(31*B*b*c^6 + 15*A*c^7)*x^6 + 63*(71*B*b^2*c^5 + 135*A
*b*c^6)*x^5 + 35*(B*b^3*c^4 + 159*A*b^2*c^5)*x^4 - 5*(8*B*b^4*c^3 - 15*A*b^3*c^4)*x^3 + 6*(8*B*b^5*c^2 - 15*A*
b^4*c^3)*x^2 - 8*(8*B*b^6*c - 15*A*b^5*c^2)*x)*sqrt(c*x^2 + b*x)/(c^5*sqrt(x))

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giac [B]  time = 0.28, size = 488, normalized size = 2.87 \begin {gather*} -\frac {2}{45045} \, B c^{2} {\left (\frac {1024 \, b^{\frac {15}{2}}}{c^{7}} - \frac {3003 \, {\left (c x + b\right )}^{\frac {15}{2}} - 20790 \, {\left (c x + b\right )}^{\frac {13}{2}} b + 61425 \, {\left (c x + b\right )}^{\frac {11}{2}} b^{2} - 100100 \, {\left (c x + b\right )}^{\frac {9}{2}} b^{3} + 96525 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{4} - 54054 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{5} + 15015 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{6}}{c^{7}}\right )} + \frac {4}{9009} \, B b c {\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 {2}{9009} \, A 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 {2}{3465} \, B b^{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}{3465} \, A 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}{315} \, A 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 )} \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]

-2/45045*B*c^2*(1024*b^(15/2)/c^7 - (3003*(c*x + b)^(15/2) - 20790*(c*x + b)^(13/2)*b + 61425*(c*x + b)^(11/2)
*b^2 - 100100*(c*x + b)^(9/2)*b^3 + 96525*(c*x + b)^(7/2)*b^4 - 54054*(c*x + b)^(5/2)*b^5 + 15015*(c*x + b)^(3
/2)*b^6)/c^7) + 4/9009*B*b*c*(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) + 2/900
9*A*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 - 1287
0*(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) - 2/3465*B*b^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) - 4/3465*A*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) + 2/315*A*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)

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maple [A]  time = 0.05, size = 107, normalized size = 0.63 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-3003 B \,x^{4} c^{4}-3465 A \,c^{4} x^{3}+1848 B b \,c^{3} x^{3}+1890 A b \,c^{3} x^{2}-1008 B \,b^{2} c^{2} x^{2}-840 A \,b^{2} c^{2} x +448 B \,b^{3} c x +240 A \,b^{3} c -128 b^{4} B \right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{45045 c^{5} 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]

-2/45045*(c*x+b)*(-3003*B*c^4*x^4-3465*A*c^4*x^3+1848*B*b*c^3*x^3+1890*A*b*c^3*x^2-1008*B*b^2*c^2*x^2-840*A*b^
2*c^2*x+448*B*b^3*c*x+240*A*b^3*c-128*B*b^4)*(c*x^2+b*x)^(5/2)/c^5/x^(5/2)

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maxima [B]  time = 0.82, size = 441, normalized size = 2.59 \begin {gather*} \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} A}{45045 \, c^{4} x^{5}} + \frac {2 \, {\left ({\left (3003 \, c^{7} x^{7} + 231 \, b c^{6} x^{6} - 252 \, b^{2} c^{5} x^{5} + 280 \, b^{3} c^{4} x^{4} - 320 \, b^{4} c^{3} x^{3} + 384 \, b^{5} c^{2} x^{2} - 512 \, b^{6} c x + 1024 \, b^{7}\right )} x^{6} + 10 \, {\left (693 \, b c^{6} x^{7} + 63 \, b^{2} c^{5} x^{6} - 70 \, b^{3} c^{4} x^{5} + 80 \, b^{4} c^{3} x^{4} - 96 \, b^{5} c^{2} x^{3} + 128 \, b^{6} c x^{2} - 256 \, b^{7} x\right )} x^{5} + 13 \, {\left (315 \, b^{2} c^{5} x^{7} + 35 \, b^{3} c^{4} x^{6} - 40 \, b^{4} c^{3} x^{5} + 48 \, b^{5} c^{2} x^{4} - 64 \, b^{6} c x^{3} + 128 \, b^{7} x^{2}\right )} x^{4}\right )} \sqrt {c x + b} B}{45045 \, c^{5} x^{6}} \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]

2/45045*(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)*A/(c^4*
x^5) + 2/45045*((3003*c^7*x^7 + 231*b*c^6*x^6 - 252*b^2*c^5*x^5 + 280*b^3*c^4*x^4 - 320*b^4*c^3*x^3 + 384*b^5*
c^2*x^2 - 512*b^6*c*x + 1024*b^7)*x^6 + 10*(693*b*c^6*x^7 + 63*b^2*c^5*x^6 - 70*b^3*c^4*x^5 + 80*b^4*c^3*x^4 -
96*b^5*c^2*x^3 + 128*b^6*c*x^2 - 256*b^7*x)*x^5 + 13*(315*b^2*c^5*x^7 + 35*b^3*c^4*x^6 - 40*b^4*c^3*x^5 + 48*
b^5*c^2*x^4 - 64*b^6*c*x^3 + 128*b^7*x^2)*x^4)*sqrt(c*x + b)*B/(c^5*x^6)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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