3.3.83 \(\int (b x+c x^2)^{5/2} \, dx\)

Optimal. Leaf size=118 \[ -\frac {5 b^6 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{7/2}}+\frac {5 b^4 (b+2 c x) \sqrt {b x+c x^2}}{512 c^3}-\frac {5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c} \]

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Rubi [A]  time = 0.04, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {612, 620, 206} \begin {gather*} \frac {5 b^4 (b+2 c x) \sqrt {b x+c x^2}}{512 c^3}-\frac {5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}-\frac {5 b^6 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{7/2}}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(5/2),x]

[Out]

(5*b^4*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(512*c^3) - (5*b^2*(b + 2*c*x)*(b*x + c*x^2)^(3/2))/(192*c^2) + ((b + 2*
c*x)*(b*x + c*x^2)^(5/2))/(12*c) - (5*b^6*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(512*c^(7/2))

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 )^{5/2} \, dx &=\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {\left (5 b^2\right ) \int \left (b x+c x^2\right )^{3/2} \, dx}{24 c}\\ &=-\frac {5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}+\frac {\left (5 b^4\right ) \int \sqrt {b x+c x^2} \, dx}{128 c^2}\\ &=\frac {5 b^4 (b+2 c x) \sqrt {b x+c x^2}}{512 c^3}-\frac {5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {\left (5 b^6\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{1024 c^3}\\ &=\frac {5 b^4 (b+2 c x) \sqrt {b x+c x^2}}{512 c^3}-\frac {5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {\left (5 b^6\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{512 c^3}\\ &=\frac {5 b^4 (b+2 c x) \sqrt {b x+c x^2}}{512 c^3}-\frac {5 b^2 (b+2 c x) \left (b x+c x^2\right )^{3/2}}{192 c^2}+\frac {(b+2 c x) \left (b x+c x^2\right )^{5/2}}{12 c}-\frac {5 b^6 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{512 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 120, normalized size = 1.02 \begin {gather*} \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (15 b^5-10 b^4 c x+8 b^3 c^2 x^2+432 b^2 c^3 x^3+640 b c^4 x^4+256 c^5 x^5\right )-\frac {15 b^{11/2} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{1536 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(5/2),x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^5 - 10*b^4*c*x + 8*b^3*c^2*x^2 + 432*b^2*c^3*x^3 + 640*b*c^4*x^4 + 256*c^5*x
^5) - (15*b^(11/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(1536*c^(7/2))

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IntegrateAlgebraic [A]  time = 0.41, size = 112, normalized size = 0.95 \begin {gather*} \frac {5 b^6 \log \left (-2 \sqrt {c} \sqrt {b x+c x^2}+b+2 c x\right )}{1024 c^{7/2}}+\frac {\sqrt {b x+c x^2} \left (15 b^5-10 b^4 c x+8 b^3 c^2 x^2+432 b^2 c^3 x^3+640 b c^4 x^4+256 c^5 x^5\right )}{1536 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(15*b^5 - 10*b^4*c*x + 8*b^3*c^2*x^2 + 432*b^2*c^3*x^3 + 640*b*c^4*x^4 + 256*c^5*x^5))/(153
6*c^3) + (5*b^6*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[b*x + c*x^2]])/(1024*c^(7/2))

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fricas [A]  time = 0.42, size = 213, normalized size = 1.81 \begin {gather*} \left [\frac {15 \, b^{6} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (256 \, c^{6} x^{5} + 640 \, b c^{5} x^{4} + 432 \, b^{2} c^{4} x^{3} + 8 \, b^{3} c^{3} x^{2} - 10 \, b^{4} c^{2} x + 15 \, b^{5} c\right )} \sqrt {c x^{2} + b x}}{3072 \, c^{4}}, \frac {15 \, b^{6} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (256 \, c^{6} x^{5} + 640 \, b c^{5} x^{4} + 432 \, b^{2} c^{4} x^{3} + 8 \, b^{3} c^{3} x^{2} - 10 \, b^{4} c^{2} x + 15 \, b^{5} c\right )} \sqrt {c x^{2} + b x}}{1536 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3072*(15*b^6*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(256*c^6*x^5 + 640*b*c^5*x^4 + 432*b^
2*c^4*x^3 + 8*b^3*c^3*x^2 - 10*b^4*c^2*x + 15*b^5*c)*sqrt(c*x^2 + b*x))/c^4, 1/1536*(15*b^6*sqrt(-c)*arctan(sq
rt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (256*c^6*x^5 + 640*b*c^5*x^4 + 432*b^2*c^4*x^3 + 8*b^3*c^3*x^2 - 10*b^4*c^2*
x + 15*b^5*c)*sqrt(c*x^2 + b*x))/c^4]

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giac [A]  time = 0.24, size = 107, normalized size = 0.91 \begin {gather*} \frac {5 \, b^{6} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{1024 \, c^{\frac {7}{2}}} + \frac {1}{1536} \, \sqrt {c x^{2} + b x} {\left (\frac {15 \, b^{5}}{c^{3}} - 2 \, {\left (\frac {5 \, b^{4}}{c^{2}} - 4 \, {\left (\frac {b^{3}}{c} + 2 \, {\left (27 \, b^{2} + 8 \, {\left (2 \, c^{2} x + 5 \, b c\right )} x\right )} x\right )} x\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/1024*b^6*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2) + 1/1536*sqrt(c*x^2 + b*x)*(15*b^5
/c^3 - 2*(5*b^4/c^2 - 4*(b^3/c + 2*(27*b^2 + 8*(2*c^2*x + 5*b*c)*x)*x)*x)*x)

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maple [A]  time = 0.04, size = 134, normalized size = 1.14 \begin {gather*} -\frac {5 b^{6} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{1024 c^{\frac {7}{2}}}+\frac {5 \sqrt {c \,x^{2}+b x}\, b^{4} x}{256 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x}\, b^{5}}{512 c^{3}}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{2} x}{96 c}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{3}}{192 c^{2}}+\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{12 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*c*x+b)*(c*x^2+b*x)^(5/2)/c-5/96*b^2/c*(c*x^2+b*x)^(3/2)*x-5/192*b^3/c^2*(c*x^2+b*x)^(3/2)+5/256*b^4/c^
2*(c*x^2+b*x)^(1/2)*x+5/512*b^5/c^3*(c*x^2+b*x)^(1/2)-5/1024*b^6/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1
/2))

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maxima [A]  time = 1.38, size = 141, normalized size = 1.19 \begin {gather*} \frac {1}{6} \, {\left (c x^{2} + b x\right )}^{\frac {5}{2}} x + \frac {5 \, \sqrt {c x^{2} + b x} b^{4} x}{256 \, c^{2}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2} x}{96 \, c} - \frac {5 \, b^{6} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{1024 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} b^{5}}{512 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{3}}{192 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {5}{2}} b}{12 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(c*x^2 + b*x)^(5/2)*x + 5/256*sqrt(c*x^2 + b*x)*b^4*x/c^2 - 5/96*(c*x^2 + b*x)^(3/2)*b^2*x/c - 5/1024*b^6*
log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 5/512*sqrt(c*x^2 + b*x)*b^5/c^3 - 5/192*(c*x^2 + b*x)^(
3/2)*b^3/c^2 + 1/12*(c*x^2 + b*x)^(5/2)*b/c

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mupad [B]  time = 0.38, size = 119, normalized size = 1.01 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(5/2),x)

[Out]

((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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*x + c*x**2)**(5/2), x)

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