3.1.92 \(\int \frac {(b x+c x^2)^{3/2}}{x^{13/2}} \, dx\)

Optimal. Leaf size=139 \[ -\frac {3 c^4 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{5/2}}+\frac {3 c^3 \sqrt {b x+c x^2}}{64 b^2 x^{3/2}}-\frac {c^2 \sqrt {b x+c x^2}}{32 b x^{5/2}}-\frac {c \sqrt {b x+c x^2}}{8 x^{7/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {662, 672, 660, 207} \begin {gather*} \frac {3 c^3 \sqrt {b x+c x^2}}{64 b^2 x^{3/2}}-\frac {3 c^4 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{5/2}}-\frac {c^2 \sqrt {b x+c x^2}}{32 b x^{5/2}}-\frac {c \sqrt {b x+c x^2}}{8 x^{7/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^(13/2),x]

[Out]

-(c*Sqrt[b*x + c*x^2])/(8*x^(7/2)) - (c^2*Sqrt[b*x + c*x^2])/(32*b*x^(5/2)) + (3*c^3*Sqrt[b*x + c*x^2])/(64*b^
2*x^(3/2)) - (b*x + c*x^2)^(3/2)/(4*x^(11/2)) - (3*c^4*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(64*b^(5/
2))

Rule 207

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

Rule 660

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 662

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

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(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
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{13/2}} \, dx &=-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}+\frac {1}{8} (3 c) \int \frac {\sqrt {b x+c x^2}}{x^{9/2}} \, dx\\ &=-\frac {c \sqrt {b x+c x^2}}{8 x^{7/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}+\frac {1}{16} c^2 \int \frac {1}{x^{5/2} \sqrt {b x+c x^2}} \, dx\\ &=-\frac {c \sqrt {b x+c x^2}}{8 x^{7/2}}-\frac {c^2 \sqrt {b x+c x^2}}{32 b x^{5/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}-\frac {\left (3 c^3\right ) \int \frac {1}{x^{3/2} \sqrt {b x+c x^2}} \, dx}{64 b}\\ &=-\frac {c \sqrt {b x+c x^2}}{8 x^{7/2}}-\frac {c^2 \sqrt {b x+c x^2}}{32 b x^{5/2}}+\frac {3 c^3 \sqrt {b x+c x^2}}{64 b^2 x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}+\frac {\left (3 c^4\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{128 b^2}\\ &=-\frac {c \sqrt {b x+c x^2}}{8 x^{7/2}}-\frac {c^2 \sqrt {b x+c x^2}}{32 b x^{5/2}}+\frac {3 c^3 \sqrt {b x+c x^2}}{64 b^2 x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}+\frac {\left (3 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{64 b^2}\\ &=-\frac {c \sqrt {b x+c x^2}}{8 x^{7/2}}-\frac {c^2 \sqrt {b x+c x^2}}{32 b x^{5/2}}+\frac {3 c^3 \sqrt {b x+c x^2}}{64 b^2 x^{3/2}}-\frac {\left (b x+c x^2\right )^{3/2}}{4 x^{11/2}}-\frac {3 c^4 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 42, normalized size = 0.30 \begin {gather*} -\frac {2 c^4 (x (b+c x))^{5/2} \, _2F_1\left (\frac {5}{2},5;\frac {7}{2};\frac {c x}{b}+1\right )}{5 b^5 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^(13/2),x]

[Out]

(-2*c^4*(x*(b + c*x))^(5/2)*Hypergeometric2F1[5/2, 5, 7/2, 1 + (c*x)/b])/(5*b^5*x^(5/2))

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IntegrateAlgebraic [A]  time = 0.71, size = 93, normalized size = 0.67 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-16 b^3-24 b^2 c x-2 b c^2 x^2+3 c^3 x^3\right )}{64 b^2 x^{9/2}}-\frac {3 c^4 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{64 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*x + c*x^2)^(3/2)/x^(13/2),x]

[Out]

(Sqrt[b*x + c*x^2]*(-16*b^3 - 24*b^2*c*x - 2*b*c^2*x^2 + 3*c^3*x^3))/(64*b^2*x^(9/2)) - (3*c^4*ArcTanh[(Sqrt[b
]*Sqrt[x])/Sqrt[b*x + c*x^2]])/(64*b^(5/2))

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fricas [A]  time = 0.43, size = 196, normalized size = 1.41 \begin {gather*} \left [\frac {3 \, \sqrt {b} c^{4} x^{5} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (3 \, b c^{3} x^{3} - 2 \, b^{2} c^{2} x^{2} - 24 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{128 \, b^{3} x^{5}}, \frac {3 \, \sqrt {-b} c^{4} x^{5} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (3 \, b c^{3} x^{3} - 2 \, b^{2} c^{2} x^{2} - 24 \, b^{3} c x - 16 \, b^{4}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{64 \, b^{3} x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(13/2),x, algorithm="fricas")

[Out]

[1/128*(3*sqrt(b)*c^4*x^5*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(3*b*c^3*x^3 - 2
*b^2*c^2*x^2 - 24*b^3*c*x - 16*b^4)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^5), 1/64*(3*sqrt(-b)*c^4*x^5*arctan(sqrt
(-b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (3*b*c^3*x^3 - 2*b^2*c^2*x^2 - 24*b^3*c*x - 16*b^4)*sqrt(c*x^2 + b*x)*sqrt(x
))/(b^3*x^5)]

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giac [A]  time = 0.29, size = 99, normalized size = 0.71 \begin {gather*} \frac {\frac {3 \, c^{5} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {3 \, {\left (c x + b\right )}^{\frac {7}{2}} c^{5} - 11 \, {\left (c x + b\right )}^{\frac {5}{2}} b c^{5} - 11 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{2} c^{5} + 3 \, \sqrt {c x + b} b^{3} c^{5}}{b^{2} c^{4} x^{4}}}{64 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(13/2),x, algorithm="giac")

[Out]

1/64*(3*c^5*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (3*(c*x + b)^(7/2)*c^5 - 11*(c*x + b)^(5/2)*b*c^5
- 11*(c*x + b)^(3/2)*b^2*c^5 + 3*sqrt(c*x + b)*b^3*c^5)/(b^2*c^4*x^4))/c

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maple [A]  time = 0.06, size = 108, normalized size = 0.78 \begin {gather*} -\frac {\sqrt {\left (c x +b \right ) x}\, \left (3 c^{4} x^{4} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-3 \sqrt {c x +b}\, \sqrt {b}\, c^{3} x^{3}+2 \sqrt {c x +b}\, b^{\frac {3}{2}} c^{2} x^{2}+24 \sqrt {c x +b}\, b^{\frac {5}{2}} c x +16 \sqrt {c x +b}\, b^{\frac {7}{2}}\right )}{64 \sqrt {c x +b}\, b^{\frac {5}{2}} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/x^(13/2),x)

[Out]

-1/64*((c*x+b)*x)^(1/2)/b^(5/2)*(3*arctanh((c*x+b)^(1/2)/b^(1/2))*c^4*x^4-3*(c*x+b)^(1/2)*b^(1/2)*c^3*x^3+2*(c
*x+b)^(1/2)*b^(3/2)*c^2*x^2+24*(c*x+b)^(1/2)*b^(5/2)*c*x+16*(c*x+b)^(1/2)*b^(7/2))/x^(9/2)/(c*x+b)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}}}{x^{\frac {13}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^(13/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(3/2)/x^(13/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}}{x^{13/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(3/2)/x^(13/2),x)

[Out]

int((b*x + c*x^2)^(3/2)/x^(13/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**(13/2),x)

[Out]

Timed out

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