\(\int (b+3 d x^2) (b x+d x^3)^7 \, dx\) [196]

Optimal result

Integrand size = 20, antiderivative size = 15 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^7 \, dx=\frac {1}{8} \left (b x+d x^3\right )^8 \]

[Out]

1/8*(d*x^3+b*x)^8

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {1602} \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^7 \, dx=\frac {1}{8} \left (b x+d x^3\right )^8 \]

[In]

Int[(b + 3*d*x^2)*(b*x + d*x^3)^7,x]

[Out]

(b*x + d*x^3)^8/8

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \left (b x+d x^3\right )^8 \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(98\) vs. \(2(15)=30\).

Time = 0.00 (sec) , antiderivative size = 98, normalized size of antiderivative = 6.53 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^7 \, dx=\frac {b^8 x^8}{8}+b^7 d x^{10}+\frac {7}{2} b^6 d^2 x^{12}+7 b^5 d^3 x^{14}+\frac {35}{4} b^4 d^4 x^{16}+7 b^3 d^5 x^{18}+\frac {7}{2} b^2 d^6 x^{20}+b d^7 x^{22}+\frac {d^8 x^{24}}{8} \]

[In]

Integrate[(b + 3*d*x^2)*(b*x + d*x^3)^7,x]

[Out]

(b^8*x^8)/8 + b^7*d*x^10 + (7*b^6*d^2*x^12)/2 + 7*b^5*d^3*x^14 + (35*b^4*d^4*x^16)/4 + 7*b^3*d^5*x^18 + (7*b^2
*d^6*x^20)/2 + b*d^7*x^22 + (d^8*x^24)/8

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\left (x^{3} d +b x \right )^{8}}{8}\)	\(14\)
gosper	\(\frac {x^{8} \left (d \,x^{2}+b \right )^{8}}{8}\)	\(15\)
norman	\(\frac {1}{8} d^{8} x^{24}+x^{22} b \,d^{7}+7 x^{18} b^{3} d^{5}+\frac {7}{2} x^{20} b^{2} d^{6}+7 x^{14} b^{5} d^{3}+\frac {35}{4} x^{16} b^{4} d^{4}+\frac {1}{8} x^{8} b^{8}+x^{10} b^{7} d +\frac {7}{2} x^{12} b^{6} d^{2}\)	\(89\)
risch	\(\frac {1}{8} d^{8} x^{24}+x^{22} b \,d^{7}+7 x^{18} b^{3} d^{5}+\frac {7}{2} x^{20} b^{2} d^{6}+7 x^{14} b^{5} d^{3}+\frac {35}{4} x^{16} b^{4} d^{4}+\frac {1}{8} x^{8} b^{8}+x^{10} b^{7} d +\frac {7}{2} x^{12} b^{6} d^{2}\)	\(89\)
parallelrisch	\(\frac {1}{8} d^{8} x^{24}+x^{22} b \,d^{7}+7 x^{18} b^{3} d^{5}+\frac {7}{2} x^{20} b^{2} d^{6}+7 x^{14} b^{5} d^{3}+\frac {35}{4} x^{16} b^{4} d^{4}+\frac {1}{8} x^{8} b^{8}+x^{10} b^{7} d +\frac {7}{2} x^{12} b^{6} d^{2}\)	\(89\)

[In]

int((3*d*x^2+b)*(d*x^3+b*x)^7,x,method=_RETURNVERBOSE)

[Out]

1/8*(d*x^3+b*x)^8

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (13) = 26\).

Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 5.87 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^7 \, dx=\frac {1}{8} \, d^{8} x^{24} + b d^{7} x^{22} + \frac {7}{2} \, b^{2} d^{6} x^{20} + 7 \, b^{3} d^{5} x^{18} + \frac {35}{4} \, b^{4} d^{4} x^{16} + 7 \, b^{5} d^{3} x^{14} + \frac {7}{2} \, b^{6} d^{2} x^{12} + b^{7} d x^{10} + \frac {1}{8} \, b^{8} x^{8} \]

[In]

integrate((3*d*x^2+b)*(d*x^3+b*x)^7,x, algorithm="fricas")

[Out]

1/8*d^8*x^24 + b*d^7*x^22 + 7/2*b^2*d^6*x^20 + 7*b^3*d^5*x^18 + 35/4*b^4*d^4*x^16 + 7*b^5*d^3*x^14 + 7/2*b^6*d
^2*x^12 + b^7*d*x^10 + 1/8*b^8*x^8

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (10) = 20\).

Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 6.47 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^7 \, dx=\frac {b^{8} x^{8}}{8} + b^{7} d x^{10} + \frac {7 b^{6} d^{2} x^{12}}{2} + 7 b^{5} d^{3} x^{14} + \frac {35 b^{4} d^{4} x^{16}}{4} + 7 b^{3} d^{5} x^{18} + \frac {7 b^{2} d^{6} x^{20}}{2} + b d^{7} x^{22} + \frac {d^{8} x^{24}}{8} \]

[In]

integrate((3*d*x**2+b)*(d*x**3+b*x)**7,x)

[Out]

b**8*x**8/8 + b**7*d*x**10 + 7*b**6*d**2*x**12/2 + 7*b**5*d**3*x**14 + 35*b**4*d**4*x**16/4 + 7*b**3*d**5*x**1
8 + 7*b**2*d**6*x**20/2 + b*d**7*x**22 + d**8*x**24/8

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^7 \, dx=\frac {1}{8} \, {\left (d x^{3} + b x\right )}^{8} \]

[In]

integrate((3*d*x^2+b)*(d*x^3+b*x)^7,x, algorithm="maxima")

[Out]

1/8*(d*x^3 + b*x)^8

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^7 \, dx=\frac {1}{8} \, {\left (d x^{3} + b x\right )}^{8} \]

[In]

integrate((3*d*x^2+b)*(d*x^3+b*x)^7,x, algorithm="giac")

[Out]

1/8*(d*x^3 + b*x)^8

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 5.87 \[ \int \left (b+3 d x^2\right ) \left (b x+d x^3\right )^7 \, dx=\frac {b^8\,x^8}{8}+b^7\,d\,x^{10}+\frac {7\,b^6\,d^2\,x^{12}}{2}+7\,b^5\,d^3\,x^{14}+\frac {35\,b^4\,d^4\,x^{16}}{4}+7\,b^3\,d^5\,x^{18}+\frac {7\,b^2\,d^6\,x^{20}}{2}+b\,d^7\,x^{22}+\frac {d^8\,x^{24}}{8} \]

[In]

int((b*x + d*x^3)^7*(b + 3*d*x^2),x)

[Out]

(b^8*x^8)/8 + (d^8*x^24)/8 + b^7*d*x^10 + b*d^7*x^22 + (7*b^6*d^2*x^12)/2 + 7*b^5*d^3*x^14 + (35*b^4*d^4*x^16)
/4 + 7*b^3*d^5*x^18 + (7*b^2*d^6*x^20)/2