3.11.31 \(\int (b d+2 c d x)^3 (a+b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=59 \[ \frac {4}{63} d^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{7/2}+\frac {2}{9} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{7/2} \]

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Rubi [A]  time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {692, 629} \begin {gather*} \frac {4}{63} d^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{7/2}+\frac {2}{9} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{7/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(b^2 - 4*a*c)*d^3*(a + b*x + c*x^2)^(7/2))/63 + (2*d^3*(b + 2*c*x)^2*(a + b*x + c*x^2)^(7/2))/9

Rule 629

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

Rule 692

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*d*(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1))/(b*(m + 2*p + 1)), x] + Dist[(d^2*(m - 1)*(b^2 - 4*a*c))/(b^2*(m + 2*p + 1)), In
t[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[
2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

Rubi steps

\begin {align*} \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2} \, dx &=\frac {2}{9} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{7/2}+\frac {1}{9} \left (2 \left (b^2-4 a c\right ) d^2\right ) \int (b d+2 c d x) \left (a+b x+c x^2\right )^{5/2} \, dx\\ &=\frac {4}{63} \left (b^2-4 a c\right ) d^3 \left (a+b x+c x^2\right )^{7/2}+\frac {2}{9} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{7/2}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 44, normalized size = 0.75 \begin {gather*} \frac {2}{63} d^3 (a+x (b+c x))^{7/2} \left (4 c \left (7 c x^2-2 a\right )+9 b^2+28 b c x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^3*(a + x*(b + c*x))^(7/2)*(9*b^2 + 28*b*c*x + 4*c*(-2*a + 7*c*x^2)))/63

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IntegrateAlgebraic [B]  time = 1.00, size = 259, normalized size = 4.39 \begin {gather*} -\frac {2}{63} \sqrt {a+b x+c x^2} \left (8 a^4 c d^3-9 a^3 b^2 d^3-4 a^3 b c d^3 x-4 a^3 c^2 d^3 x^2-27 a^2 b^3 d^3 x-87 a^2 b^2 c d^3 x^2-120 a^2 b c^2 d^3 x^3-60 a^2 c^3 d^3 x^4-27 a b^4 d^3 x^2-130 a b^3 c d^3 x^3-255 a b^2 c^2 d^3 x^4-228 a b c^3 d^3 x^5-76 a c^4 d^3 x^6-9 b^5 d^3 x^3-55 b^4 c d^3 x^4-139 b^3 c^2 d^3 x^5-177 b^2 c^3 d^3 x^6-112 b c^4 d^3 x^7-28 c^5 d^3 x^8\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[a + b*x + c*x^2]*(-9*a^3*b^2*d^3 + 8*a^4*c*d^3 - 27*a^2*b^3*d^3*x - 4*a^3*b*c*d^3*x - 27*a*b^4*d^3*x^
2 - 87*a^2*b^2*c*d^3*x^2 - 4*a^3*c^2*d^3*x^2 - 9*b^5*d^3*x^3 - 130*a*b^3*c*d^3*x^3 - 120*a^2*b*c^2*d^3*x^3 - 5
5*b^4*c*d^3*x^4 - 255*a*b^2*c^2*d^3*x^4 - 60*a^2*c^3*d^3*x^4 - 139*b^3*c^2*d^3*x^5 - 228*a*b*c^3*d^3*x^5 - 177
*b^2*c^3*d^3*x^6 - 76*a*c^4*d^3*x^6 - 112*b*c^4*d^3*x^7 - 28*c^5*d^3*x^8))/63

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fricas [B]  time = 0.45, size = 215, normalized size = 3.64 \begin {gather*} \frac {2}{63} \, {\left (28 \, c^{5} d^{3} x^{8} + 112 \, b c^{4} d^{3} x^{7} + {\left (177 \, b^{2} c^{3} + 76 \, a c^{4}\right )} d^{3} x^{6} + {\left (139 \, b^{3} c^{2} + 228 \, a b c^{3}\right )} d^{3} x^{5} + 5 \, {\left (11 \, b^{4} c + 51 \, a b^{2} c^{2} + 12 \, a^{2} c^{3}\right )} d^{3} x^{4} + {\left (9 \, b^{5} + 130 \, a b^{3} c + 120 \, a^{2} b c^{2}\right )} d^{3} x^{3} + {\left (27 \, a b^{4} + 87 \, a^{2} b^{2} c + 4 \, a^{3} c^{2}\right )} d^{3} x^{2} + {\left (27 \, a^{2} b^{3} + 4 \, a^{3} b c\right )} d^{3} x + {\left (9 \, a^{3} b^{2} - 8 \, a^{4} c\right )} d^{3}\right )} \sqrt {c x^{2} + b x + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/63*(28*c^5*d^3*x^8 + 112*b*c^4*d^3*x^7 + (177*b^2*c^3 + 76*a*c^4)*d^3*x^6 + (139*b^3*c^2 + 228*a*b*c^3)*d^3*
x^5 + 5*(11*b^4*c + 51*a*b^2*c^2 + 12*a^2*c^3)*d^3*x^4 + (9*b^5 + 130*a*b^3*c + 120*a^2*b*c^2)*d^3*x^3 + (27*a
*b^4 + 87*a^2*b^2*c + 4*a^3*c^2)*d^3*x^2 + (27*a^2*b^3 + 4*a^3*b*c)*d^3*x + (9*a^3*b^2 - 8*a^4*c)*d^3)*sqrt(c*
x^2 + b*x + a)

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giac [A]  time = 0.20, size = 58, normalized size = 0.98 \begin {gather*} \frac {2}{7} \, {\left (c x^{2} + b x + a\right )}^{\frac {7}{2}} b^{2} d^{3} + \frac {8}{9} \, {\left (c x^{2} + b x + a\right )}^{\frac {9}{2}} c d^{3} - \frac {8}{7} \, {\left (c x^{2} + b x + a\right )}^{\frac {7}{2}} a c d^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(c*x^2 + b*x + a)^(7/2)*b^2*d^3 + 8/9*(c*x^2 + b*x + a)^(9/2)*c*d^3 - 8/7*(c*x^2 + b*x + a)^(7/2)*a*c*d^3

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maple [A]  time = 0.05, size = 41, normalized size = 0.69 \begin {gather*} -\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} \left (-28 c^{2} x^{2}-28 b c x +8 a c -9 b^{2}\right ) d^{3}}{63} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^3*(c*x^2+b*x+a)^(5/2),x)

[Out]

-2/63*(c*x^2+b*x+a)^(7/2)*(-28*c^2*x^2-28*b*c*x+8*a*c-9*b^2)*d^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 1.01, size = 58, normalized size = 0.98 \begin {gather*} \frac {8\,c\,d^3\,{\left (c\,x^2+b\,x+a\right )}^{9/2}}{9}+\frac {2\,b^2\,d^3\,{\left (c\,x^2+b\,x+a\right )}^{7/2}}{7}-\frac {8\,a\,c\,d^3\,{\left (c\,x^2+b\,x+a\right )}^{7/2}}{7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(5/2),x)

[Out]

(8*c*d^3*(a + b*x + c*x^2)^(9/2))/9 + (2*b^2*d^3*(a + b*x + c*x^2)^(7/2))/7 - (8*a*c*d^3*(a + b*x + c*x^2)^(7/
2))/7

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sympy [B]  time = 6.00, size = 559, normalized size = 9.47 \begin {gather*} - \frac {16 a^{4} c d^{3} \sqrt {a + b x + c x^{2}}}{63} + \frac {2 a^{3} b^{2} d^{3} \sqrt {a + b x + c x^{2}}}{7} + \frac {8 a^{3} b c d^{3} x \sqrt {a + b x + c x^{2}}}{63} + \frac {8 a^{3} c^{2} d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{63} + \frac {6 a^{2} b^{3} d^{3} x \sqrt {a + b x + c x^{2}}}{7} + \frac {58 a^{2} b^{2} c d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{21} + \frac {80 a^{2} b c^{2} d^{3} x^{3} \sqrt {a + b x + c x^{2}}}{21} + \frac {40 a^{2} c^{3} d^{3} x^{4} \sqrt {a + b x + c x^{2}}}{21} + \frac {6 a b^{4} d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{7} + \frac {260 a b^{3} c d^{3} x^{3} \sqrt {a + b x + c x^{2}}}{63} + \frac {170 a b^{2} c^{2} d^{3} x^{4} \sqrt {a + b x + c x^{2}}}{21} + \frac {152 a b c^{3} d^{3} x^{5} \sqrt {a + b x + c x^{2}}}{21} + \frac {152 a c^{4} d^{3} x^{6} \sqrt {a + b x + c x^{2}}}{63} + \frac {2 b^{5} d^{3} x^{3} \sqrt {a + b x + c x^{2}}}{7} + \frac {110 b^{4} c d^{3} x^{4} \sqrt {a + b x + c x^{2}}}{63} + \frac {278 b^{3} c^{2} d^{3} x^{5} \sqrt {a + b x + c x^{2}}}{63} + \frac {118 b^{2} c^{3} d^{3} x^{6} \sqrt {a + b x + c x^{2}}}{21} + \frac {32 b c^{4} d^{3} x^{7} \sqrt {a + b x + c x^{2}}}{9} + \frac {8 c^{5} d^{3} x^{8} \sqrt {a + b x + c x^{2}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**3*(c*x**2+b*x+a)**(5/2),x)

[Out]

-16*a**4*c*d**3*sqrt(a + b*x + c*x**2)/63 + 2*a**3*b**2*d**3*sqrt(a + b*x + c*x**2)/7 + 8*a**3*b*c*d**3*x*sqrt
(a + b*x + c*x**2)/63 + 8*a**3*c**2*d**3*x**2*sqrt(a + b*x + c*x**2)/63 + 6*a**2*b**3*d**3*x*sqrt(a + b*x + c*
x**2)/7 + 58*a**2*b**2*c*d**3*x**2*sqrt(a + b*x + c*x**2)/21 + 80*a**2*b*c**2*d**3*x**3*sqrt(a + b*x + c*x**2)
/21 + 40*a**2*c**3*d**3*x**4*sqrt(a + b*x + c*x**2)/21 + 6*a*b**4*d**3*x**2*sqrt(a + b*x + c*x**2)/7 + 260*a*b
**3*c*d**3*x**3*sqrt(a + b*x + c*x**2)/63 + 170*a*b**2*c**2*d**3*x**4*sqrt(a + b*x + c*x**2)/21 + 152*a*b*c**3
*d**3*x**5*sqrt(a + b*x + c*x**2)/21 + 152*a*c**4*d**3*x**6*sqrt(a + b*x + c*x**2)/63 + 2*b**5*d**3*x**3*sqrt(
a + b*x + c*x**2)/7 + 110*b**4*c*d**3*x**4*sqrt(a + b*x + c*x**2)/63 + 278*b**3*c**2*d**3*x**5*sqrt(a + b*x +
c*x**2)/63 + 118*b**2*c**3*d**3*x**6*sqrt(a + b*x + c*x**2)/21 + 32*b*c**4*d**3*x**7*sqrt(a + b*x + c*x**2)/9
+ 8*c**5*d**3*x**8*sqrt(a + b*x + c*x**2)/9

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