\(\int (b d+2 c d x)^3 (a+b x+c x^2)^{5/2} \, dx\) [1219]

Optimal result

Integrand size = 26, antiderivative size = 59 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {4}{63} \left (b^2-4 a c\right ) d^3 \left (a+b x+c x^2\right )^{7/2}+\frac {2}{9} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{7/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {706, 643} \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {4}{63} d^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{7/2}+\frac {2}{9} d^3 (b+2 c x)^2 \left (a+b x+c x^2\right )^{7/2} \]

[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 643

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 706

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 2.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2} \, dx=-\frac {2}{63} d^3 (a+x (b+c x))^{7/2} \left (-9 b^2-28 b c x+4 c \left (2 a-7 c x^2\right )\right ) \]

[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

Maple [A] (verified)

Time = 2.60 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69

method	result	size
gosper	\(-\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} \left (-28 c^{2} x^{2}-28 b c x +8 a c -9 b^{2}\right ) d^{3}}{63}\)	\(41\)
pseudoelliptic	\(-\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} \left (-28 c^{2} x^{2}-28 b c x +8 a c -9 b^{2}\right ) d^{3}}{63}\)	\(41\)
trager	\(d^{3} \left (\frac {8}{9} c^{5} x^{8}+\frac {32}{9} b \,c^{4} x^{7}+\frac {152}{63} a \,c^{4} x^{6}+\frac {118}{21} b^{2} c^{3} x^{6}+\frac {152}{21} a b \,c^{3} x^{5}+\frac {278}{63} c^{2} x^{5} b^{3}+\frac {40}{21} a^{2} c^{3} x^{4}+\frac {170}{21} a \,b^{2} c^{2} x^{4}+\frac {110}{63} b^{4} c \,x^{4}+\frac {80}{21} a^{2} b \,c^{2} x^{3}+\frac {260}{63} a \,b^{3} c \,x^{3}+\frac {2}{7} x^{3} b^{5}+\frac {8}{63} a^{3} c^{2} x^{2}+\frac {58}{21} c \,x^{2} a^{2} b^{2}+\frac {6}{7} b^{4} x^{2} a +\frac {8}{63} a^{3} b c x +\frac {6}{7} a^{2} b^{3} x -\frac {16}{63} a^{4} c +\frac {2}{7} a^{3} b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\)	\(201\)
risch	\(-\frac {2 d^{3} \left (-28 c^{5} x^{8}-112 b \,c^{4} x^{7}-76 a \,c^{4} x^{6}-177 b^{2} c^{3} x^{6}-228 a b \,c^{3} x^{5}-139 c^{2} x^{5} b^{3}-60 a^{2} c^{3} x^{4}-255 a \,b^{2} c^{2} x^{4}-55 b^{4} c \,x^{4}-120 a^{2} b \,c^{2} x^{3}-130 a \,b^{3} c \,x^{3}-9 x^{3} b^{5}-4 a^{3} c^{2} x^{2}-87 c \,x^{2} a^{2} b^{2}-27 b^{4} x^{2} a -4 a^{3} b c x -27 a^{2} b^{3} x +8 a^{4} c -9 a^{3} b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{63}\)	\(202\)
default	\(\text {Expression too large to display}\)	\(1212\)

[In]

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

[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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (51) = 102\).

Time = 0.37 (sec) , antiderivative size = 215, normalized size of antiderivative = 3.64 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {2}{63} \, {\left (28 \, c^{5} d^{3} x^{8} + 112 \, b c^{4} d^{3} x^{7} + {\left (177 \, b^{2} c^{3} + 76 \, a c^{4}\right )} d^{3} x^{6} + {\left (139 \, b^{3} c^{2} + 228 \, a b c^{3}\right )} d^{3} x^{5} + 5 \, {\left (11 \, b^{4} c + 51 \, a b^{2} c^{2} + 12 \, a^{2} c^{3}\right )} d^{3} x^{4} + {\left (9 \, b^{5} + 130 \, a b^{3} c + 120 \, a^{2} b c^{2}\right )} d^{3} x^{3} + {\left (27 \, a b^{4} + 87 \, a^{2} b^{2} c + 4 \, a^{3} c^{2}\right )} d^{3} x^{2} + {\left (27 \, a^{2} b^{3} + 4 \, a^{3} b c\right )} d^{3} x + {\left (9 \, a^{3} b^{2} - 8 \, a^{4} c\right )} d^{3}\right )} \sqrt {c x^{2} + b x + a} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 559 vs. \(2 (58) = 116\).

Time = 0.32 (sec) , antiderivative size = 559, normalized size of antiderivative = 9.47 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2} \, dx=- \frac {16 a^{4} c d^{3} \sqrt {a + b x + c x^{2}}}{63} + \frac {2 a^{3} b^{2} d^{3} \sqrt {a + b x + c x^{2}}}{7} + \frac {8 a^{3} b c d^{3} x \sqrt {a + b x + c x^{2}}}{63} + \frac {8 a^{3} c^{2} d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{63} + \frac {6 a^{2} b^{3} d^{3} x \sqrt {a + b x + c x^{2}}}{7} + \frac {58 a^{2} b^{2} c d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{21} + \frac {80 a^{2} b c^{2} d^{3} x^{3} \sqrt {a + b x + c x^{2}}}{21} + \frac {40 a^{2} c^{3} d^{3} x^{4} \sqrt {a + b x + c x^{2}}}{21} + \frac {6 a b^{4} d^{3} x^{2} \sqrt {a + b x + c x^{2}}}{7} + \frac {260 a b^{3} c d^{3} x^{3} \sqrt {a + b x + c x^{2}}}{63} + \frac {170 a b^{2} c^{2} d^{3} x^{4} \sqrt {a + b x + c x^{2}}}{21} + \frac {152 a b c^{3} d^{3} x^{5} \sqrt {a + b x + c x^{2}}}{21} + \frac {152 a c^{4} d^{3} x^{6} \sqrt {a + b x + c x^{2}}}{63} + \frac {2 b^{5} d^{3} x^{3} \sqrt {a + b x + c x^{2}}}{7} + \frac {110 b^{4} c d^{3} x^{4} \sqrt {a + b x + c x^{2}}}{63} + \frac {278 b^{3} c^{2} d^{3} x^{5} \sqrt {a + b x + c x^{2}}}{63} + \frac {118 b^{2} c^{3} d^{3} x^{6} \sqrt {a + b x + c x^{2}}}{21} + \frac {32 b c^{4} d^{3} x^{7} \sqrt {a + b x + c x^{2}}}{9} + \frac {8 c^{5} d^{3} x^{8} \sqrt {a + b x + c x^{2}}}{9} \]

[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

Maxima [F(-2)]

Exception generated. \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2} \, dx=\text {Exception raised: ValueError} \]

[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 deta

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.98 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {2}{7} \, {\left (c x^{2} + b x + a\right )}^{\frac {7}{2}} b^{2} d^{3} + \frac {8}{9} \, {\left (c x^{2} + b x + a\right )}^{\frac {9}{2}} c d^{3} - \frac {8}{7} \, {\left (c x^{2} + b x + a\right )}^{\frac {7}{2}} a c d^{3} \]

[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

Mupad [B] (verification not implemented)

Time = 10.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.98 \[ \int (b d+2 c d x)^3 \left (a+b x+c x^2\right )^{5/2} \, dx=\frac {8\,c\,d^3\,{\left (c\,x^2+b\,x+a\right )}^{9/2}}{9}+\frac {2\,b^2\,d^3\,{\left (c\,x^2+b\,x+a\right )}^{7/2}}{7}-\frac {8\,a\,c\,d^3\,{\left (c\,x^2+b\,x+a\right )}^{7/2}}{7} \]

[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