3.3.49 \(\int \frac {A+B x}{x^{3/2} (b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=214 \[ -\frac {35 c^2 (2 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{11/2}}+\frac {35 c^2 \sqrt {x} (2 b B-3 A c)}{8 b^5 \sqrt {b x+c x^2}}+\frac {35 c (2 b B-3 A c)}{24 b^4 \sqrt {x} \sqrt {b x+c x^2}}-\frac {7 c \sqrt {x} (2 b B-3 A c)}{12 b^3 \left (b x+c x^2\right )^{3/2}}-\frac {2 b B-3 A c}{4 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.18, antiderivative size = 214, 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^2 \sqrt {x} (2 b B-3 A c)}{8 b^5 \sqrt {b x+c x^2}}-\frac {35 c^2 (2 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{11/2}}+\frac {35 c (2 b B-3 A c)}{24 b^4 \sqrt {x} \sqrt {b x+c x^2}}-\frac {7 c \sqrt {x} (2 b B-3 A c)}{12 b^3 \left (b x+c x^2\right )^{3/2}}-\frac {2 b B-3 A c}{4 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(3/2)*(b*x + c*x^2)^(5/2)),x]

[Out]

-A/(3*b*x^(3/2)*(b*x + c*x^2)^(3/2)) - (2*b*B - 3*A*c)/(4*b^2*Sqrt[x]*(b*x + c*x^2)^(3/2)) - (7*c*(2*b*B - 3*A
*c)*Sqrt[x])/(12*b^3*(b*x + c*x^2)^(3/2)) + (35*c*(2*b*B - 3*A*c))/(24*b^4*Sqrt[x]*Sqrt[b*x + c*x^2]) + (35*c^
2*(2*b*B - 3*A*c)*Sqrt[x])/(8*b^5*Sqrt[b*x + c*x^2]) - (35*c^2*(2*b*B - 3*A*c)*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt
[b]*Sqrt[x])])/(8*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^{3/2} \left (b x+c x^2\right )^{5/2}} \, dx &=-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}--\frac {\left (-\frac {3}{2} (-b B+A c)-\frac {3}{2} (-b B+2 A c)\right ) \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{5/2}} \, dx}{3 b}\\ &=-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}-\frac {2 b B-3 A c}{4 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}-\frac {(7 c (2 b B-3 A c)) \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{5/2}} \, dx}{8 b^2}\\ &=-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}-\frac {2 b B-3 A c}{4 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}-\frac {7 c (2 b B-3 A c) \sqrt {x}}{12 b^3 \left (b x+c x^2\right )^{3/2}}-\frac {(35 c (2 b B-3 A c)) \int \frac {1}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx}{24 b^3}\\ &=-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}-\frac {2 b B-3 A c}{4 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}-\frac {7 c (2 b B-3 A c) \sqrt {x}}{12 b^3 \left (b x+c x^2\right )^{3/2}}+\frac {35 c (2 b B-3 A c)}{24 b^4 \sqrt {x} \sqrt {b x+c x^2}}+\frac {\left (35 c^2 (2 b B-3 A c)\right ) \int \frac {\sqrt {x}}{\left (b x+c x^2\right )^{3/2}} \, dx}{16 b^4}\\ &=-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}-\frac {2 b B-3 A c}{4 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}-\frac {7 c (2 b B-3 A c) \sqrt {x}}{12 b^3 \left (b x+c x^2\right )^{3/2}}+\frac {35 c (2 b B-3 A c)}{24 b^4 \sqrt {x} \sqrt {b x+c x^2}}+\frac {35 c^2 (2 b B-3 A c) \sqrt {x}}{8 b^5 \sqrt {b x+c x^2}}+\frac {\left (35 c^2 (2 b B-3 A c)\right ) \int \frac {1}{\sqrt {x} \sqrt {b x+c x^2}} \, dx}{16 b^5}\\ &=-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}-\frac {2 b B-3 A c}{4 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}-\frac {7 c (2 b B-3 A c) \sqrt {x}}{12 b^3 \left (b x+c x^2\right )^{3/2}}+\frac {35 c (2 b B-3 A c)}{24 b^4 \sqrt {x} \sqrt {b x+c x^2}}+\frac {35 c^2 (2 b B-3 A c) \sqrt {x}}{8 b^5 \sqrt {b x+c x^2}}+\frac {\left (35 c^2 (2 b B-3 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{-b+x^2} \, dx,x,\frac {\sqrt {b x+c x^2}}{\sqrt {x}}\right )}{8 b^5}\\ &=-\frac {A}{3 b x^{3/2} \left (b x+c x^2\right )^{3/2}}-\frac {2 b B-3 A c}{4 b^2 \sqrt {x} \left (b x+c x^2\right )^{3/2}}-\frac {7 c (2 b B-3 A c) \sqrt {x}}{12 b^3 \left (b x+c x^2\right )^{3/2}}+\frac {35 c (2 b B-3 A c)}{24 b^4 \sqrt {x} \sqrt {b x+c x^2}}+\frac {35 c^2 (2 b B-3 A c) \sqrt {x}}{8 b^5 \sqrt {b x+c x^2}}-\frac {35 c^2 (2 b B-3 A c) \tanh ^{-1}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{11/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 62, normalized size = 0.29 \begin {gather*} \frac {c^2 x^3 (2 b B-3 A c) \, _2F_1\left (-\frac {3}{2},3;-\frac {1}{2};\frac {c x}{b}+1\right )-A b^3}{3 b^4 x^{3/2} (x (b+c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(3/2)*(b*x + c*x^2)^(5/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 3.88, size = 166, normalized size = 0.78 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-8 A b^4+18 A b^3 c x-63 A b^2 c^2 x^2-420 A b c^3 x^3-315 A c^4 x^4-12 b^4 B x+42 b^3 B c x^2+280 b^2 B c^2 x^3+210 b B c^3 x^4\right )}{24 b^5 x^{7/2} (b+c x)^2}-\frac {35 \left (2 b B c^2-3 A c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {b x+c x^2}}\right )}{8 b^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^(3/2)*(b*x + c*x^2)^(5/2)),x]

[Out]

(Sqrt[b*x + c*x^2]*(-8*A*b^4 - 12*b^4*B*x + 18*A*b^3*c*x + 42*b^3*B*c*x^2 - 63*A*b^2*c^2*x^2 + 280*b^2*B*c^2*x
^3 - 420*A*b*c^3*x^3 + 210*b*B*c^3*x^4 - 315*A*c^4*x^4))/(24*b^5*x^(7/2)*(b + c*x)^2) - (35*(2*b*B*c^2 - 3*A*c
^3)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[b*x + c*x^2]])/(8*b^(11/2))

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fricas [A]  time = 0.45, size = 477, normalized size = 2.23 \begin {gather*} \left [-\frac {105 \, {\left ({\left (2 \, B b c^{4} - 3 \, A c^{5}\right )} x^{6} + 2 \, {\left (2 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{5} + {\left (2 \, B b^{3} c^{2} - 3 \, A b^{2} c^{3}\right )} x^{4}\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 (8 \, A b^{5} - 105 \, {\left (2 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} - 140 \, {\left (2 \, B b^{3} c^{2} - 3 \, A b^{2} c^{3}\right )} x^{3} - 21 \, {\left (2 \, B b^{4} c - 3 \, A b^{3} c^{2}\right )} x^{2} + 6 \, {\left (2 \, B b^{5} - 3 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, {\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}}, \frac {105 \, {\left ({\left (2 \, B b c^{4} - 3 \, A c^{5}\right )} x^{6} + 2 \, {\left (2 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{5} + {\left (2 \, B b^{3} c^{2} - 3 \, A b^{2} c^{3}\right )} x^{4}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {x}}{\sqrt {c x^{2} + b x}}\right ) - {\left (8 \, A b^{5} - 105 \, {\left (2 \, B b^{2} c^{3} - 3 \, A b c^{4}\right )} x^{4} - 140 \, {\left (2 \, B b^{3} c^{2} - 3 \, A b^{2} c^{3}\right )} x^{3} - 21 \, {\left (2 \, B b^{4} c - 3 \, A b^{3} c^{2}\right )} x^{2} + 6 \, {\left (2 \, B b^{5} - 3 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, {\left (b^{6} c^{2} x^{6} + 2 \, b^{7} c x^{5} + b^{8} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")

[Out]

[-1/48*(105*((2*B*b*c^4 - 3*A*c^5)*x^6 + 2*(2*B*b^2*c^3 - 3*A*b*c^4)*x^5 + (2*B*b^3*c^2 - 3*A*b^2*c^3)*x^4)*sq
rt(b)*log(-(c*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(8*A*b^5 - 105*(2*B*b^2*c^3 - 3*A*b*
c^4)*x^4 - 140*(2*B*b^3*c^2 - 3*A*b^2*c^3)*x^3 - 21*(2*B*b^4*c - 3*A*b^3*c^2)*x^2 + 6*(2*B*b^5 - 3*A*b^4*c)*x)
*sqrt(c*x^2 + b*x)*sqrt(x))/(b^6*c^2*x^6 + 2*b^7*c*x^5 + b^8*x^4), 1/24*(105*((2*B*b*c^4 - 3*A*c^5)*x^6 + 2*(2
*B*b^2*c^3 - 3*A*b*c^4)*x^5 + (2*B*b^3*c^2 - 3*A*b^2*c^3)*x^4)*sqrt(-b)*arctan(sqrt(-b)*sqrt(x)/sqrt(c*x^2 + b
*x)) - (8*A*b^5 - 105*(2*B*b^2*c^3 - 3*A*b*c^4)*x^4 - 140*(2*B*b^3*c^2 - 3*A*b^2*c^3)*x^3 - 21*(2*B*b^4*c - 3*
A*b^3*c^2)*x^2 + 6*(2*B*b^5 - 3*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^6*c^2*x^6 + 2*b^7*c*x^5 + b^8*x^4)]

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giac [A]  time = 0.31, size = 200, normalized size = 0.93 \begin {gather*} \frac {35 \, {\left (2 \, B b c^{2} - 3 \, A c^{3}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{8 \, \sqrt {-b} b^{5}} + \frac {210 \, {\left (c x + b\right )}^{4} B b c^{2} - 560 \, {\left (c x + b\right )}^{3} B b^{2} c^{2} + 462 \, {\left (c x + b\right )}^{2} B b^{3} c^{2} - 96 \, {\left (c x + b\right )} B b^{4} c^{2} - 16 \, B b^{5} c^{2} - 315 \, {\left (c x + b\right )}^{4} A c^{3} + 840 \, {\left (c x + b\right )}^{3} A b c^{3} - 693 \, {\left (c x + b\right )}^{2} A b^{2} c^{3} + 144 \, {\left (c x + b\right )} A b^{3} c^{3} + 16 \, A b^{4} c^{3}}{24 \, {\left ({\left (c x + b\right )}^{\frac {3}{2}} - \sqrt {c x + b} b\right )}^{3} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x)^(5/2),x, algorithm="giac")

[Out]

35/8*(2*B*b*c^2 - 3*A*c^3)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^5) + 1/24*(210*(c*x + b)^4*B*b*c^2 - 560
*(c*x + b)^3*B*b^2*c^2 + 462*(c*x + b)^2*B*b^3*c^2 - 96*(c*x + b)*B*b^4*c^2 - 16*B*b^5*c^2 - 315*(c*x + b)^4*A
*c^3 + 840*(c*x + b)^3*A*b*c^3 - 693*(c*x + b)^2*A*b^2*c^3 + 144*(c*x + b)*A*b^3*c^3 + 16*A*b^4*c^3)/(((c*x +
b)^(3/2) - sqrt(c*x + b)*b)^3*b^5)

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maple [A]  time = 0.07, size = 234, normalized size = 1.09 \begin {gather*} \frac {\sqrt {\left (c x +b \right ) x}\, \left (315 \sqrt {c x +b}\, A \,c^{4} x^{4} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-210 \sqrt {c x +b}\, B b \,c^{3} x^{4} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-315 A \sqrt {b}\, c^{4} x^{4}+210 B \,b^{\frac {3}{2}} c^{3} x^{4}+315 \sqrt {c x +b}\, A b \,c^{3} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-210 \sqrt {c x +b}\, B \,b^{2} c^{2} x^{3} \arctanh \left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )-420 A \,b^{\frac {3}{2}} c^{3} x^{3}+280 B \,b^{\frac {5}{2}} c^{2} x^{3}-63 A \,b^{\frac {5}{2}} c^{2} x^{2}+42 B \,b^{\frac {7}{2}} c \,x^{2}+18 A \,b^{\frac {7}{2}} c x -12 B \,b^{\frac {9}{2}} x -8 A \,b^{\frac {9}{2}}\right )}{24 \left (c x +b \right )^{2} b^{\frac {11}{2}} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^(3/2)/(c*x^2+b*x)^(5/2),x)

[Out]

1/24*((c*x+b)*x)^(1/2)*(315*A*arctanh((c*x+b)^(1/2)/b^(1/2))*(c*x+b)^(1/2)*x^4*c^4-210*B*arctanh((c*x+b)^(1/2)
/b^(1/2))*(c*x+b)^(1/2)*x^4*b*c^3+315*A*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*b*c^3*(c*x+b)^(1/2)-315*A*b^(1/2)*c
^4*x^4-210*B*arctanh((c*x+b)^(1/2)/b^(1/2))*x^3*b^2*c^2*(c*x+b)^(1/2)+210*B*b^(3/2)*c^3*x^4-420*A*b^(3/2)*c^3*
x^3+280*B*b^(5/2)*c^2*x^3-63*A*b^(5/2)*c^2*x^2+42*B*b^(7/2)*c*x^2+18*A*b^(7/2)*c*x-12*B*b^(9/2)*x-8*A*b^(9/2))
/x^(7/2)/(c*x+b)^2/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 {5}{2}} x^{\frac {3}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(c*x^2+b*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/((c*x^2 + b*x)^(5/2)*x^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+B\,x}{x^{3/2}\,{\left (c\,x^2+b\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(3/2)*(b*x + c*x^2)^(5/2)),x)

[Out]

int((A + B*x)/(x^(3/2)*(b*x + c*x^2)^(5/2)), 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**(3/2)/(c*x**2+b*x)**(5/2),x)

[Out]

Timed out

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