3.2.1 \(\int \frac {1}{x^{7/2} \sqrt {b x+c x^2}} \, dx\)

Optimal. Leaf size=117 \[ \frac {5 c^3 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{7/2}}-\frac {5 c^2 \sqrt {b x+c x^2}}{8 b^3 x^{3/2}}+\frac {5 c \sqrt {b x+c x^2}}{12 b^2 x^{5/2}}-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {672, 660, 207} \begin {gather*} -\frac {5 c^2 \sqrt {b x+c x^2}}{8 b^3 x^{3/2}}+\frac {5 c^3 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{7/2}}+\frac {5 c \sqrt {b x+c x^2}}{12 b^2 x^{5/2}}-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*Sqrt[b*x + c*x^2]),x]

[Out]

-Sqrt[b*x + c*x^2]/(3*b*x^(7/2)) + (5*c*Sqrt[b*x + c*x^2])/(12*b^2*x^(5/2)) - (5*c^2*Sqrt[b*x + c*x^2])/(8*b^3
*x^(3/2)) + (5*c^3*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/(8*b^(7/2))

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 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 {1}{x^{7/2} \sqrt {b x+c x^2}} \, dx &=-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}}-\frac {(5 c) \int \frac {1}{x^{5/2} \sqrt {b x+c x^2}} \, dx}{6 b}\\ &=-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}}+\frac {5 c \sqrt {b x+c x^2}}{12 b^2 x^{5/2}}+\frac {\left (5 c^2\right ) \int \frac {1}{x^{3/2} \sqrt {b x+c x^2}} \, dx}{8 b^2}\\ &=-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}}+\frac {5 c \sqrt {b x+c x^2}}{12 b^2 x^{5/2}}-\frac {5 c^2 \sqrt {b x+c x^2}}{8 b^3 x^{3/2}}-\frac {\left (5 c^3\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{16 b^3}\\ &=-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}}+\frac {5 c \sqrt {b x+c x^2}}{12 b^2 x^{5/2}}-\frac {5 c^2 \sqrt {b x+c x^2}}{8 b^3 x^{3/2}}-\frac {\left (5 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{8 b^3}\\ &=-\frac {\sqrt {b x+c x^2}}{3 b x^{7/2}}+\frac {5 c \sqrt {b x+c x^2}}{12 b^2 x^{5/2}}-\frac {5 c^2 \sqrt {b x+c x^2}}{8 b^3 x^{3/2}}+\frac {5 c^3 \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 40, normalized size = 0.34 \begin {gather*} \frac {2 c^3 \sqrt {x (b+c x)} \, _2F_1\left (\frac {1}{2},4;\frac {3}{2};\frac {c x}{b}+1\right )}{b^4 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(7/2)*Sqrt[b*x + c*x^2]),x]

[Out]

(2*c^3*Sqrt[x*(b + c*x)]*Hypergeometric2F1[1/2, 4, 3/2, 1 + (c*x)/b])/(b^4*Sqrt[x])

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IntegrateAlgebraic [A]  time = 0.22, size = 82, normalized size = 0.70 \begin {gather*} \frac {5 c^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{8 b^{7/2}}+\frac {\sqrt {b x+c x^2} \left (-8 b^2+10 b c x-15 c^2 x^2\right )}{24 b^3 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^(7/2)*Sqrt[b*x + c*x^2]),x]

[Out]

(Sqrt[b*x + c*x^2]*(-8*b^2 + 10*b*c*x - 15*c^2*x^2))/(24*b^3*x^(7/2)) + (5*c^3*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[
b*x + c*x^2]])/(8*b^(7/2))

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fricas [A]  time = 0.43, size = 174, normalized size = 1.49 \begin {gather*} \left [\frac {15 \, \sqrt {b} c^{3} x^{4} \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 (15 \, b c^{2} x^{2} - 10 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, b^{4} x^{4}}, -\frac {15 \, \sqrt {-b} c^{3} x^{4} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (15 \, b c^{2} x^{2} - 10 \, b^{2} c x + 8 \, b^{3}\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, b^{4} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(15*sqrt(b)*c^3*x^4*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) - 2*(15*b*c^2*x^2 -
10*b^2*c*x + 8*b^3)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^4), -1/24*(15*sqrt(-b)*c^3*x^4*arctan(sqrt(-b)*sqrt(x)/s
qrt(c*x^2 + b*x)) + (15*b*c^2*x^2 - 10*b^2*c*x + 8*b^3)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*x^4)]

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giac [A]  time = 0.32, size = 84, normalized size = 0.72 \begin {gather*} -\frac {\frac {15 \, c^{4} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{3}} + \frac {15 \, {\left (c x + b\right )}^{\frac {5}{2}} c^{4} - 40 \, {\left (c x + b\right )}^{\frac {3}{2}} b c^{4} + 33 \, \sqrt {c x + b} b^{2} c^{4}}{b^{3} c^{3} x^{3}}}{24 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

-1/24*(15*c^4*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) + (15*(c*x + b)^(5/2)*c^4 - 40*(c*x + b)^(3/2)*b*c
^4 + 33*sqrt(c*x + b)*b^2*c^4)/(b^3*c^3*x^3))/c

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maple [A]  time = 0.06, size = 90, normalized size = 0.77 \begin {gather*} \frac {\sqrt {\left (c x +b \right ) x}\, \left (15 c^{3} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-15 \sqrt {c x +b}\, \sqrt {b}\, c^{2} x^{2}+10 \sqrt {c x +b}\, b^{\frac {3}{2}} c x -8 \sqrt {c x +b}\, b^{\frac {5}{2}}\right )}{24 \sqrt {c x +b}\, b^{\frac {7}{2}} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(7/2)/(c*x^2+b*x)^(1/2),x)

[Out]

1/24*((c*x+b)*x)^(1/2)/b^(7/2)*(15*arctanh((c*x+b)^(1/2)/b^(1/2))*c^3*x^3-15*(c*x+b)^(1/2)*b^(1/2)*c^2*x^2+10*
(c*x+b)^(1/2)*b^(3/2)*c*x-8*(c*x+b)^(1/2)*b^(5/2))/x^(7/2)/(c*x+b)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {c x^{2} + b x} x^{\frac {7}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(7/2)/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(c*x^2 + b*x)*x^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^{7/2}\,\sqrt {c\,x^2+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*(b*x + c*x^2)^(1/2)),x)

[Out]

int(1/(x^(7/2)*(b*x + c*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{\frac {7}{2}} \sqrt {x \left (b + c x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(7/2)/(c*x**2+b*x)**(1/2),x)

[Out]

Integral(1/(x**(7/2)*sqrt(x*(b + c*x))), x)

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