3.1 \(\int (b x+c x^2)^{7/2} \, dx\)

Optimal. Leaf size=147 \[ \frac {35 b^8 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{16384 c^{9/2}}-\frac {35 b^6 (b+2 c x) \sqrt {b x+c x^2}}{16384 c^4}+\frac {35 b^4 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{6144 c^3}-\frac {7 b^2 (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{7/2}}{16 c} \]

[Out]

35/6144*b^4*(2*c*x+b)*(c*x^2+b*x)^(3/2)/c^3-7/384*b^2*(2*c*x+b)*(c*x^2+b*x)^(5/2)/c^2+1/16*(2*c*x+b)*(c*x^2+b*
x)^(7/2)/c+35/16384*b^8*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(9/2)-35/16384*b^6*(2*c*x+b)*(c*x^2+b*x)^(1/2)/
c^4

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Rubi [A]  time = 0.05, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {612, 620, 206} \[ -\frac {35 b^6 (b+2 c x) \sqrt {b x+c x^2}}{16384 c^4}+\frac {35 b^4 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{6144 c^3}-\frac {7 b^2 (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^2}+\frac {35 b^8 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{16384 c^{9/2}}+\frac {(b+2 c x) \left (b x+c x^2\right )^{7/2}}{16 c} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(7/2),x]

[Out]

(-35*b^6*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(16384*c^4) + (35*b^4*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(6144*c^3) - (7
*b^2*(b + 2*c*x)*(b*x + c*x^2)^(5/2))/(384*c^2) + ((b + 2*c*x)*(b*x + c*x^2)^(7/2))/(16*c) + (35*b^8*ArcTanh[(
Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(16384*c^(9/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rubi steps

\begin {align*} \int \left (b x+c x^2\right )^{7/2} \, dx &=\frac {(b+2 c x) \left (b x+c x^2\right )^{7/2}}{16 c}-\frac {\left (7 b^2\right ) \int \left (b x+c x^2\right )^{5/2} \, dx}{32 c}\\ &=-\frac {7 b^2 (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{7/2}}{16 c}+\frac {\left (35 b^4\right ) \int \left (b x+c x^2\right )^{3/2} \, dx}{768 c^2}\\ &=\frac {35 b^4 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{6144 c^3}-\frac {7 b^2 (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{7/2}}{16 c}-\frac {\left (35 b^6\right ) \int \sqrt {b x+c x^2} \, dx}{4096 c^3}\\ &=-\frac {35 b^6 (b+2 c x) \sqrt {b x+c x^2}}{16384 c^4}+\frac {35 b^4 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{6144 c^3}-\frac {7 b^2 (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{7/2}}{16 c}+\frac {\left (35 b^8\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{32768 c^4}\\ &=-\frac {35 b^6 (b+2 c x) \sqrt {b x+c x^2}}{16384 c^4}+\frac {35 b^4 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{6144 c^3}-\frac {7 b^2 (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{7/2}}{16 c}+\frac {\left (35 b^8\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{16384 c^4}\\ &=-\frac {35 b^6 (b+2 c x) \sqrt {b x+c x^2}}{16384 c^4}+\frac {35 b^4 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{6144 c^3}-\frac {7 b^2 (b+2 c x) \left (b x+c x^2\right )^{5/2}}{384 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{7/2}}{16 c}+\frac {35 b^8 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{16384 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 142, normalized size = 0.97 \[ \frac {\sqrt {x (b+c x)} \left (\frac {105 b^{15/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}+\sqrt {c} \left (-105 b^7+70 b^6 c x-56 b^5 c^2 x^2+48 b^4 c^3 x^3+10880 b^3 c^4 x^4+25856 b^2 c^5 x^5+21504 b c^6 x^6+6144 c^7 x^7\right )\right )}{49152 c^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(7/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^7 + 70*b^6*c*x - 56*b^5*c^2*x^2 + 48*b^4*c^3*x^3 + 10880*b^3*c^4*x^4 + 258
56*b^2*c^5*x^5 + 21504*b*c^6*x^6 + 6144*c^7*x^7) + (105*b^(15/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*
Sqrt[1 + (c*x)/b])))/(49152*c^(9/2))

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fricas [A]  time = 0.70, size = 258, normalized size = 1.76 \[ \left [\frac {105 \, b^{8} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (6144 \, c^{8} x^{7} + 21504 \, b c^{7} x^{6} + 25856 \, b^{2} c^{6} x^{5} + 10880 \, b^{3} c^{5} x^{4} + 48 \, b^{4} c^{4} x^{3} - 56 \, b^{5} c^{3} x^{2} + 70 \, b^{6} c^{2} x - 105 \, b^{7} c\right )} \sqrt {c x^{2} + b x}}{98304 \, c^{5}}, -\frac {105 \, b^{8} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (6144 \, c^{8} x^{7} + 21504 \, b c^{7} x^{6} + 25856 \, b^{2} c^{6} x^{5} + 10880 \, b^{3} c^{5} x^{4} + 48 \, b^{4} c^{4} x^{3} - 56 \, b^{5} c^{3} x^{2} + 70 \, b^{6} c^{2} x - 105 \, b^{7} c\right )} \sqrt {c x^{2} + b x}}{49152 \, c^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(7/2),x, algorithm="fricas")

[Out]

[1/98304*(105*b^8*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(6144*c^8*x^7 + 21504*b*c^7*x^6 + 2
5856*b^2*c^6*x^5 + 10880*b^3*c^5*x^4 + 48*b^4*c^4*x^3 - 56*b^5*c^3*x^2 + 70*b^6*c^2*x - 105*b^7*c)*sqrt(c*x^2
+ b*x))/c^5, -1/49152*(105*b^8*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (6144*c^8*x^7 + 21504*b*c^7
*x^6 + 25856*b^2*c^6*x^5 + 10880*b^3*c^5*x^4 + 48*b^4*c^4*x^3 - 56*b^5*c^3*x^2 + 70*b^6*c^2*x - 105*b^7*c)*sqr
t(c*x^2 + b*x))/c^5]

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giac [A]  time = 0.63, size = 132, normalized size = 0.90 \[ -\frac {35 \, b^{8} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{32768 \, c^{\frac {9}{2}}} - \frac {1}{49152} \, {\left (\frac {105 \, b^{7}}{c^{4}} - 2 \, {\left (\frac {35 \, b^{6}}{c^{3}} - 4 \, {\left (\frac {7 \, b^{5}}{c^{2}} - 2 \, {\left (\frac {3 \, b^{4}}{c} + 8 \, {\left (85 \, b^{3} + 2 \, {\left (101 \, b^{2} c + 12 \, {\left (2 \, c^{3} x + 7 \, b c^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {c x^{2} + b x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(7/2),x, algorithm="giac")

[Out]

-35/32768*b^8*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2) - 1/49152*(105*b^7/c^4 - 2*(35*
b^6/c^3 - 4*(7*b^5/c^2 - 2*(3*b^4/c + 8*(85*b^3 + 2*(101*b^2*c + 12*(2*c^3*x + 7*b*c^2)*x)*x)*x)*x)*x)*x)*sqrt
(c*x^2 + b*x)

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maple [A]  time = 0.05, size = 173, normalized size = 1.18 \[ \frac {35 b^{8} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{32768 c^{\frac {9}{2}}}-\frac {35 \sqrt {c \,x^{2}+b x}\, b^{6} x}{8192 c^{3}}-\frac {35 \sqrt {c \,x^{2}+b x}\, b^{7}}{16384 c^{4}}+\frac {35 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{4} x}{3072 c^{2}}+\frac {35 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{5}}{6144 c^{3}}-\frac {7 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} b^{2} x}{192 c}-\frac {7 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} b^{3}}{384 c^{2}}+\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {7}{2}}}{16 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(7/2),x)

[Out]

1/16*(2*c*x+b)*(c*x^2+b*x)^(7/2)/c-7/192*b^2/c*(c*x^2+b*x)^(5/2)*x-7/384*b^3/c^2*(c*x^2+b*x)^(5/2)+35/3072*b^4
/c^2*(c*x^2+b*x)^(3/2)*x+35/6144*b^5/c^3*(c*x^2+b*x)^(3/2)-35/8192*b^6/c^3*(c*x^2+b*x)^(1/2)*x-35/16384*b^7/c^
4*(c*x^2+b*x)^(1/2)+35/32768*b^8/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))

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maxima [A]  time = 1.34, size = 180, normalized size = 1.22 \[ \frac {1}{8} \, {\left (c x^{2} + b x\right )}^{\frac {7}{2}} x - \frac {35 \, \sqrt {c x^{2} + b x} b^{6} x}{8192 \, c^{3}} + \frac {35 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{4} x}{3072 \, c^{2}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} b^{2} x}{192 \, c} + \frac {35 \, b^{8} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{32768 \, c^{\frac {9}{2}}} - \frac {35 \, \sqrt {c x^{2} + b x} b^{7}}{16384 \, c^{4}} + \frac {35 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{5}}{6144 \, c^{3}} - \frac {7 \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} b^{3}}{384 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {7}{2}} b}{16 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(7/2),x, algorithm="maxima")

[Out]

1/8*(c*x^2 + b*x)^(7/2)*x - 35/8192*sqrt(c*x^2 + b*x)*b^6*x/c^3 + 35/3072*(c*x^2 + b*x)^(3/2)*b^4*x/c^2 - 7/19
2*(c*x^2 + b*x)^(5/2)*b^2*x/c + 35/32768*b^8*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(9/2) - 35/16384*s
qrt(c*x^2 + b*x)*b^7/c^4 + 35/6144*(c*x^2 + b*x)^(3/2)*b^5/c^3 - 7/384*(c*x^2 + b*x)^(5/2)*b^3/c^2 + 1/16*(c*x
^2 + b*x)^(7/2)*b/c

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mupad [B]  time = 0.74, size = 151, normalized size = 1.03 \[ \frac {{\left (c\,x^2+b\,x\right )}^{7/2}\,\left (\frac {b}{2}+c\,x\right )}{8\,c}-\frac {7\,b^2\,\left (\frac {{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (\frac {b}{2}+c\,x\right )}{6\,c}-\frac {5\,b^2\,\left (\frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (\frac {b}{2}+c\,x\right )}{4\,c}-\frac {3\,b^2\,\left (\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {b^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}\right )}{16\,c}\right )}{24\,c}\right )}{32\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(7/2),x)

[Out]

((b*x + c*x^2)^(7/2)*(b/2 + c*x))/(8*c) - (7*b^2*(((b*x + c*x^2)^(5/2)*(b/2 + c*x))/(6*c) - (5*b^2*(((b*x + c*
x^2)^(3/2)*(b/2 + c*x))/(4*c) - (3*b^2*((b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) - (b^2*log((b/2 + c*x)/c^(1/2) + (
b*x + c*x^2)^(1/2)))/(8*c^(3/2))))/(16*c)))/(24*c)))/(32*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b x + c x^{2}\right )^{\frac {7}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(7/2),x)

[Out]

Integral((b*x + c*x**2)**(7/2), x)

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