3.3.41 \(\int \frac {A+B x}{x^{7/2} (b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=216 \[ \frac {35 c^3 (8 b B-9 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{11/2}}-\frac {35 c^3 \sqrt {x} (8 b B-9 A c)}{64 b^5 \sqrt {b x+c x^2}}-\frac {35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt {x} \sqrt {b x+c x^2}}+\frac {7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt {b x+c x^2}}-\frac {8 b B-9 A c}{24 b^2 x^{5/2} \sqrt {b x+c x^2}}-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}} \]

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Rubi [A]  time = 0.19, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {792, 672, 666, 660, 207} \begin {gather*} -\frac {35 c^3 \sqrt {x} (8 b B-9 A c)}{64 b^5 \sqrt {b x+c x^2}}-\frac {35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt {x} \sqrt {b x+c x^2}}+\frac {35 c^3 (8 b B-9 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{11/2}}+\frac {7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt {b x+c x^2}}-\frac {8 b B-9 A c}{24 b^2 x^{5/2} \sqrt {b x+c x^2}}-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(7/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

-A/(4*b*x^(7/2)*Sqrt[b*x + c*x^2]) - (8*b*B - 9*A*c)/(24*b^2*x^(5/2)*Sqrt[b*x + c*x^2]) + (7*c*(8*b*B - 9*A*c)
)/(96*b^3*x^(3/2)*Sqrt[b*x + c*x^2]) - (35*c^2*(8*b*B - 9*A*c))/(192*b^4*Sqrt[x]*Sqrt[b*x + c*x^2]) - (35*c^3*
(8*b*B - 9*A*c)*Sqrt[x])/(64*b^5*Sqrt[b*x + c*x^2]) + (35*c^3*(8*b*B - 9*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[
b]*Sqrt[x])])/(64*b^(11/2))

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 666

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

Rule 792

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

Rubi steps

\begin {align*} \int \frac {A+B x}{x^{7/2} \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}+\frac {\left (\frac {1}{2} (b B-2 A c)-\frac {7}{2} (-b B+A c)\right ) \int \frac {1}{x^{5/2} \left (b x+c x^2\right )^{3/2}} \, dx}{4 b}\\ &=-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}-\frac {8 b B-9 A c}{24 b^2 x^{5/2} \sqrt {b x+c x^2}}-\frac {(7 c (8 b B-9 A c)) \int \frac {1}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx}{48 b^2}\\ &=-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}-\frac {8 b B-9 A c}{24 b^2 x^{5/2} \sqrt {b x+c x^2}}+\frac {7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt {b x+c x^2}}+\frac {\left (35 c^2 (8 b B-9 A c)\right ) \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx}{192 b^3}\\ &=-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}-\frac {8 b B-9 A c}{24 b^2 x^{5/2} \sqrt {b x+c x^2}}+\frac {7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt {b x+c x^2}}-\frac {35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt {x} \sqrt {b x+c x^2}}-\frac {\left (35 c^3 (8 b B-9 A c)\right ) \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{128 b^4}\\ &=-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}-\frac {8 b B-9 A c}{24 b^2 x^{5/2} \sqrt {b x+c x^2}}+\frac {7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt {b x+c x^2}}-\frac {35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt {x} \sqrt {b x+c x^2}}-\frac {35 c^3 (8 b B-9 A c) \sqrt {x}}{64 b^5 \sqrt {b x+c x^2}}-\frac {\left (35 c^3 (8 b B-9 A c)\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{128 b^5}\\ &=-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}-\frac {8 b B-9 A c}{24 b^2 x^{5/2} \sqrt {b x+c x^2}}+\frac {7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt {b x+c x^2}}-\frac {35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt {x} \sqrt {b x+c x^2}}-\frac {35 c^3 (8 b B-9 A c) \sqrt {x}}{64 b^5 \sqrt {b x+c x^2}}-\frac {\left (35 c^3 (8 b B-9 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{64 b^5}\\ &=-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}-\frac {8 b B-9 A c}{24 b^2 x^{5/2} \sqrt {b x+c x^2}}+\frac {7 c (8 b B-9 A c)}{96 b^3 x^{3/2} \sqrt {b x+c x^2}}-\frac {35 c^2 (8 b B-9 A c)}{192 b^4 \sqrt {x} \sqrt {b x+c x^2}}-\frac {35 c^3 (8 b B-9 A c) \sqrt {x}}{64 b^5 \sqrt {b x+c x^2}}+\frac {35 c^3 (8 b B-9 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 62, normalized size = 0.29 \begin {gather*} \frac {c^3 x^4 (9 A c-8 b B) \, _2F_1\left (-\frac {1}{2},4;\frac {1}{2};\frac {c x}{b}+1\right )-A b^4}{4 b^5 x^{7/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(7/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

(-(A*b^4) + c^3*(-8*b*B + 9*A*c)*x^4*Hypergeometric2F1[-1/2, 4, 1/2, 1 + (c*x)/b])/(4*b^5*x^(7/2)*Sqrt[x*(b +
c*x)])

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IntegrateAlgebraic [A]  time = 2.68, size = 166, normalized size = 0.77 \begin {gather*} \frac {35 \left (8 b B c^3-9 A c^4\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{64 b^{11/2}}+\frac {\sqrt {b x+c x^2} \left (-48 A b^4+72 A b^3 c x-126 A b^2 c^2 x^2+315 A b c^3 x^3+945 A c^4 x^4-64 b^4 B x+112 b^3 B c x^2-280 b^2 B c^2 x^3-840 b B c^3 x^4\right )}{192 b^5 x^{9/2} (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(7/2)*(b*x + c*x^2)^(3/2)),x]

[Out]

(Sqrt[b*x + c*x^2]*(-48*A*b^4 - 64*b^4*B*x + 72*A*b^3*c*x + 112*b^3*B*c*x^2 - 126*A*b^2*c^2*x^2 - 280*b^2*B*c^
2*x^3 + 315*A*b*c^3*x^3 - 840*b*B*c^3*x^4 + 945*A*c^4*x^4))/(192*b^5*x^(9/2)*(b + c*x)) + (35*(8*b*B*c^3 - 9*A
*c^4)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]])/(64*b^(11/2))

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fricas [A]  time = 0.44, size = 406, normalized size = 1.88 \begin {gather*} \left [-\frac {105 \, {\left ({\left (8 \, B b c^{4} - 9 \, A c^{5}\right )} x^{6} + {\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{5}\right )} \sqrt {b} \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 (48 \, A b^{5} + 105 \, {\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{4} + 35 \, {\left (8 \, B b^{3} c^{2} - 9 \, A b^{2} c^{3}\right )} x^{3} - 14 \, {\left (8 \, B b^{4} c - 9 \, A b^{3} c^{2}\right )} x^{2} + 8 \, {\left (8 \, B b^{5} - 9 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{384 \, {\left (b^{6} c x^{6} + b^{7} x^{5}\right )}}, -\frac {105 \, {\left ({\left (8 \, B b c^{4} - 9 \, A c^{5}\right )} x^{6} + {\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{5}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) + {\left (48 \, A b^{5} + 105 \, {\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{4} + 35 \, {\left (8 \, B b^{3} c^{2} - 9 \, A b^{2} c^{3}\right )} x^{3} - 14 \, {\left (8 \, B b^{4} c - 9 \, A b^{3} c^{2}\right )} x^{2} + 8 \, {\left (8 \, B b^{5} - 9 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{192 \, {\left (b^{6} c x^{6} + b^{7} x^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

[-1/384*(105*((8*B*b*c^4 - 9*A*c^5)*x^6 + (8*B*b^2*c^3 - 9*A*b*c^4)*x^5)*sqrt(b)*log(-(c*x^2 + 2*b*x - 2*sqrt(
c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(48*A*b^5 + 105*(8*B*b^2*c^3 - 9*A*b*c^4)*x^4 + 35*(8*B*b^3*c^2 - 9*A*b
^2*c^3)*x^3 - 14*(8*B*b^4*c - 9*A*b^3*c^2)*x^2 + 8*(8*B*b^5 - 9*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^6*c*
x^6 + b^7*x^5), -1/192*(105*((8*B*b*c^4 - 9*A*c^5)*x^6 + (8*B*b^2*c^3 - 9*A*b*c^4)*x^5)*sqrt(-b)*arctan(sqrt(-
b)*sqrt(x)/sqrt(c*x^2 + b*x)) + (48*A*b^5 + 105*(8*B*b^2*c^3 - 9*A*b*c^4)*x^4 + 35*(8*B*b^3*c^2 - 9*A*b^2*c^3)
*x^3 - 14*(8*B*b^4*c - 9*A*b^3*c^2)*x^2 + 8*(8*B*b^5 - 9*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^6*c*x^6 + b
^7*x^5)]

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giac [A]  time = 0.34, size = 197, normalized size = 0.91 \begin {gather*} -\frac {35 \, {\left (8 \, B b c^{3} - 9 \, A c^{4}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{64 \, \sqrt {-b} b^{5}} - \frac {2 \, {\left (B b c^{3} - A c^{4}\right )}}{\sqrt {c x + b} b^{5}} - \frac {456 \, {\left (c x + b\right )}^{\frac {7}{2}} B b c^{3} - 1544 \, {\left (c x + b\right )}^{\frac {5}{2}} B b^{2} c^{3} + 1784 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{3} c^{3} - 696 \, \sqrt {c x + b} B b^{4} c^{3} - 561 \, {\left (c x + b\right )}^{\frac {7}{2}} A c^{4} + 1929 \, {\left (c x + b\right )}^{\frac {5}{2}} A b c^{4} - 2295 \, {\left (c x + b\right )}^{\frac {3}{2}} A b^{2} c^{4} + 975 \, \sqrt {c x + b} A b^{3} c^{4}}{192 \, b^{5} c^{4} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

-35/64*(8*B*b*c^3 - 9*A*c^4)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^5) - 2*(B*b*c^3 - A*c^4)/(sqrt(c*x + b
)*b^5) - 1/192*(456*(c*x + b)^(7/2)*B*b*c^3 - 1544*(c*x + b)^(5/2)*B*b^2*c^3 + 1784*(c*x + b)^(3/2)*B*b^3*c^3
- 696*sqrt(c*x + b)*B*b^4*c^3 - 561*(c*x + b)^(7/2)*A*c^4 + 1929*(c*x + b)^(5/2)*A*b*c^4 - 2295*(c*x + b)^(3/2
)*A*b^2*c^4 + 975*sqrt(c*x + b)*A*b^3*c^4)/(b^5*c^4*x^4)

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maple [A]  time = 0.09, size = 174, normalized size = 0.81 \begin {gather*} -\frac {\sqrt {\left (c x +b \right ) x}\, \left (945 \sqrt {c x +b}\, A \,c^{4} x^{4} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-840 \sqrt {c x +b}\, B b \,c^{3} x^{4} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-945 A \sqrt {b}\, c^{4} x^{4}+840 B \,b^{\frac {3}{2}} c^{3} x^{4}-315 A \,b^{\frac {3}{2}} c^{3} x^{3}+280 B \,b^{\frac {5}{2}} c^{2} x^{3}+126 A \,b^{\frac {5}{2}} c^{2} x^{2}-112 B \,b^{\frac {7}{2}} c \,x^{2}-72 A \,b^{\frac {7}{2}} c x +64 B \,b^{\frac {9}{2}} x +48 A \,b^{\frac {9}{2}}\right )}{192 \left (c x +b \right ) b^{\frac {11}{2}} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x)

[Out]

-1/192/x^(9/2)*((c*x+b)*x)^(1/2)*(945*A*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^4*c^4+64*B*b^(9/2)*x-11
2*B*b^(7/2)*x^2*c+280*B*b^(5/2)*x^3*c^2+840*B*b^(3/2)*x^4*c^3-840*B*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/
2)*x^4*b*c^3+48*A*b^(9/2)-72*A*b^(7/2)*x*c+126*A*b^(5/2)*x^2*c^2-315*A*b^(3/2)*x^3*c^3-945*A*b^(1/2)*x^4*c^4)/
(c*x+b)/b^(11/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{\frac {7}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*x^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+B\,x}{x^{7/2}\,{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(7/2)*(b*x + c*x^2)^(3/2)),x)

[Out]

int((A + B*x)/(x^(7/2)*(b*x + c*x^2)^(3/2)), 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((B*x+A)/x**(7/2)/(c*x**2+b*x)**(3/2),x)

[Out]

Timed out

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