3.5.36 \(\int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx\)

Optimal. Leaf size=151 \[ \frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 \sqrt {b}}+\frac {c \sqrt {c+d x} (4 b c-7 a d)}{4 a^2 x}-\frac {\sqrt {c} \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^3}-\frac {c (c+d x)^{3/2}}{2 a x^2} \]

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Rubi [A]  time = 0.15, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {98, 149, 156, 63, 208} \begin {gather*} -\frac {\sqrt {c} \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^3}+\frac {c \sqrt {c+d x} (4 b c-7 a d)}{4 a^2 x}+\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 \sqrt {b}}-\frac {c (c+d x)^{3/2}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(4*b*c - 7*a*d)*Sqrt[c + d*x])/(4*a^2*x) - (c*(c + d*x)^(3/2))/(2*a*x^2) - (Sqrt[c]*(8*b^2*c^2 - 20*a*b*c*d
+ 15*a^2*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(4*a^3) + (2*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/
Sqrt[b*c - a*d]])/(a^3*Sqrt[b])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 98

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

Rule 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 208

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

Rubi steps

\begin {align*} \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)} \, dx &=-\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {\int \frac {\sqrt {c+d x} \left (\frac {1}{2} c (4 b c-7 a d)+\frac {1}{2} d (b c-4 a d) x\right )}{x^2 (a+b x)} \, dx}{2 a}\\ &=\frac {c (4 b c-7 a d) \sqrt {c+d x}}{4 a^2 x}-\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {\int \frac {-\frac {1}{4} c \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right )-\frac {1}{4} d \left (4 b^2 c^2-9 a b c d+8 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}} \, dx}{2 a^2}\\ &=\frac {c (4 b c-7 a d) \sqrt {c+d x}}{4 a^2 x}-\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {(b c-a d)^3 \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{a^3}+\frac {\left (c \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {c+d x}} \, dx}{8 a^3}\\ &=\frac {c (4 b c-7 a d) \sqrt {c+d x}}{4 a^2 x}-\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {\left (2 (b c-a d)^3\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{a^3 d}+\frac {\left (c \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{4 a^3 d}\\ &=\frac {c (4 b c-7 a d) \sqrt {c+d x}}{4 a^2 x}-\frac {c (c+d x)^{3/2}}{2 a x^2}-\frac {\sqrt {c} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^3}+\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^3 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 131, normalized size = 0.87 \begin {gather*} \frac {-\sqrt {c} \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+\frac {a c \sqrt {c+d x} (-2 a c-9 a d x+4 b c x)}{x^2}+\frac {8 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b}}}{4 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a*c*Sqrt[c + d*x]*(-2*a*c + 4*b*c*x - 9*a*d*x))/x^2 - Sqrt[c]*(8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*ArcTanh[
Sqrt[c + d*x]/Sqrt[c]] + (8*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/Sqrt[b])/(4*a^
3)

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IntegrateAlgebraic [A]  time = 0.43, size = 171, normalized size = 1.13 \begin {gather*} -\frac {2 (a d-b c)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{a^3 \sqrt {b}}+\frac {c \sqrt {c+d x} \left (-9 a d (c+d x)+7 a c d-4 b c^2+4 b c (c+d x)\right )}{4 a^2 d x^2}+\frac {\left (-15 a^2 \sqrt {c} d^2+20 a b c^{3/2} d-8 b^2 c^{5/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c + d*x]*(-4*b*c^2 + 7*a*c*d + 4*b*c*(c + d*x) - 9*a*d*(c + d*x)))/(4*a^2*d*x^2) - (2*(-(b*c) + a*d)^(
5/2)*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x])/(b*c - a*d)])/(a^3*Sqrt[b]) + ((-8*b^2*c^(5/2) + 20*a*b
*c^(3/2)*d - 15*a^2*Sqrt[c]*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(4*a^3)

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fricas [A]  time = 1.99, size = 713, normalized size = 4.72 \begin {gather*} \left [\frac {8 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) + {\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {c} x^{2} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{8 \, a^{3} x^{2}}, \frac {16 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {c} x^{2} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{8 \, a^{3} x^{2}}, \frac {{\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) + 4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d + 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) - {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{4 \, a^{3} x^{2}}, \frac {8 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - {\left (2 \, a^{2} c^{2} - {\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {d x + c}}{4 \, a^{3} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(8*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d + 2*sqrt(d*x + c)*b*s
qrt((b*c - a*d)/b))/(b*x + a)) + (8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(c)*x^2*log((d*x - 2*sqrt(d*x + c)*
sqrt(c) + 2*c)/x) - 2*(2*a^2*c^2 - (4*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c))/(a^3*x^2), 1/8*(16*(b^2*c^2 - 2*a
*b*c*d + a^2*d^2)*x^2*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + (8*b^2*
c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(c)*x^2*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(2*a^2*c^2 - (4*a*
b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c))/(a^3*x^2), 1/4*((8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(-c)*x^2*arctan
(sqrt(d*x + c)*sqrt(-c)/c) + 4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*
d + 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) - (2*a^2*c^2 - (4*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c))
/(a^3*x^2), 1/4*(8*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c
- a*d)/b)/(b*c - a*d)) + (8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(-c)*x^2*arctan(sqrt(d*x + c)*sqrt(-c)/c)
- (2*a^2*c^2 - (4*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c))/(a^3*x^2)]

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giac [A]  time = 1.41, size = 198, normalized size = 1.31 \begin {gather*} -\frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{3}} + \frac {{\left (8 \, b^{2} c^{3} - 20 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{4 \, a^{3} \sqrt {-c}} + \frac {4 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{2} d - 4 \, \sqrt {d x + c} b c^{3} d - 9 \, {\left (d x + c\right )}^{\frac {3}{2}} a c d^{2} + 7 \, \sqrt {d x + c} a c^{2} d^{2}}{4 \, a^{2} d^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2
*c + a*b*d)*a^3) + 1/4*(8*b^2*c^3 - 20*a*b*c^2*d + 15*a^2*c*d^2)*arctan(sqrt(d*x + c)/sqrt(-c))/(a^3*sqrt(-c))
+ 1/4*(4*(d*x + c)^(3/2)*b*c^2*d - 4*sqrt(d*x + c)*b*c^3*d - 9*(d*x + c)^(3/2)*a*c*d^2 + 7*sqrt(d*x + c)*a*c^
2*d^2)/(a^2*d^2*x^2)

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maple [B]  time = 0.02, size = 321, normalized size = 2.13 \begin {gather*} -\frac {6 b c \,d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a}+\frac {6 b^{2} c^{2} d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a^{2}}-\frac {2 b^{3} c^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a^{3}}+\frac {2 d^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\frac {15 \sqrt {c}\, d^{2} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{4 a}+\frac {5 b \,c^{\frac {3}{2}} d \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{2}}-\frac {2 b^{2} c^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{3}}+\frac {7 \sqrt {d x +c}\, c^{2}}{4 a \,x^{2}}-\frac {\sqrt {d x +c}\, b \,c^{3}}{a^{2} d \,x^{2}}-\frac {9 \left (d x +c \right )^{\frac {3}{2}} c}{4 a \,x^{2}}+\frac {\left (d x +c \right )^{\frac {3}{2}} b \,c^{2}}{a^{2} d \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)/x^3/(b*x+a),x)

[Out]

2*d^3/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)-6*d^2/a/((a*d-b*c)*b)^(1/2)*arctan((d*x+
c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*b*c+6*d/a^2/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*b^
2*c^2-2/a^3/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*b^3*c^3-9/4*c*(d*x+c)^(3/2)/a/x^2+
1/d*c^2/a^2/x^2*(d*x+c)^(3/2)*b+7/4*c^2/a/x^2*(d*x+c)^(1/2)-1/d*c^3/a^2/x^2*(d*x+c)^(1/2)*b-15/4*d^2*c^(1/2)/a
*arctanh((d*x+c)^(1/2)/c^(1/2))+5*d*c^(3/2)/a^2*arctanh((d*x+c)^(1/2)/c^(1/2))*b-2*c^(5/2)/a^3*arctanh((d*x+c)
^(1/2)/c^(1/2))*b^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

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mupad [B]  time = 0.77, size = 1204, normalized size = 7.97 \begin {gather*} \frac {\frac {\left (7\,a\,c^2\,d^2-4\,b\,c^3\,d\right )\,\sqrt {c+d\,x}}{4\,a^2}-\frac {\left (9\,a\,c\,d^2-4\,b\,c^2\,d\right )\,{\left (c+d\,x\right )}^{3/2}}{4\,a^2}}{{\left (c+d\,x\right )}^2-2\,c\,\left (c+d\,x\right )+c^2}+\frac {2\,\mathrm {atanh}\left (\frac {95\,b^2\,c^2\,d^6\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b\,d^5+5\,a^4\,b^2\,c\,d^4-10\,a^3\,b^3\,c^2\,d^3+10\,a^2\,b^4\,c^3\,d^2-5\,a\,b^5\,c^4\,d+b^6\,c^5}}{4\,\left (\frac {215\,b^5\,c^5\,d^6}{4}-\frac {469\,a\,b^4\,c^4\,d^7}{4}+\frac {517\,a^2\,b^3\,c^3\,d^8}{4}-\frac {287\,a^3\,b^2\,c^2\,d^9}{4}-\frac {10\,b^6\,c^6\,d^5}{a}+16\,a^4\,b\,c\,d^{10}\right )}+\frac {10\,b^3\,c^3\,d^5\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b\,d^5+5\,a^4\,b^2\,c\,d^4-10\,a^3\,b^3\,c^2\,d^3+10\,a^2\,b^4\,c^3\,d^2-5\,a\,b^5\,c^4\,d+b^6\,c^5}}{-16\,a^5\,b\,c\,d^{10}+\frac {287\,a^4\,b^2\,c^2\,d^9}{4}-\frac {517\,a^3\,b^3\,c^3\,d^8}{4}+\frac {469\,a^2\,b^4\,c^4\,d^7}{4}-\frac {215\,a\,b^5\,c^5\,d^6}{4}+10\,b^6\,c^6\,d^5}+\frac {16\,b\,c\,d^7\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b\,d^5+5\,a^4\,b^2\,c\,d^4-10\,a^3\,b^3\,c^2\,d^3+10\,a^2\,b^4\,c^3\,d^2-5\,a\,b^5\,c^4\,d+b^6\,c^5}}{\frac {469\,b^4\,c^4\,d^7}{4}-\frac {517\,a\,b^3\,c^3\,d^8}{4}+\frac {287\,a^2\,b^2\,c^2\,d^9}{4}-\frac {215\,b^5\,c^5\,d^6}{4\,a}+\frac {10\,b^6\,c^6\,d^5}{a^2}-16\,a^3\,b\,c\,d^{10}}\right )\,\sqrt {-b\,{\left (a\,d-b\,c\right )}^5}}{a^3\,b}-\frac {\sqrt {c}\,\mathrm {atanh}\left (\frac {3665\,b^2\,c^{3/2}\,d^9\,\sqrt {c+d\,x}}{32\,\left (\frac {3665\,b^2\,c^2\,d^9}{32}-30\,a\,b\,c\,d^{10}-\frac {5717\,b^3\,c^3\,d^8}{32\,a}+\frac {1143\,b^4\,c^4\,d^7}{8\,a^2}-\frac {235\,b^5\,c^5\,d^6}{4\,a^3}+\frac {10\,b^6\,c^6\,d^5}{a^4}\right )}+\frac {5717\,b^3\,c^{5/2}\,d^8\,\sqrt {c+d\,x}}{32\,\left (\frac {5717\,b^3\,c^3\,d^8}{32}-\frac {3665\,a\,b^2\,c^2\,d^9}{32}-\frac {1143\,b^4\,c^4\,d^7}{8\,a}+\frac {235\,b^5\,c^5\,d^6}{4\,a^2}-\frac {10\,b^6\,c^6\,d^5}{a^3}+30\,a^2\,b\,c\,d^{10}\right )}+\frac {1143\,b^4\,c^{7/2}\,d^7\,\sqrt {c+d\,x}}{8\,\left (\frac {1143\,b^4\,c^4\,d^7}{8}-\frac {5717\,a\,b^3\,c^3\,d^8}{32}+\frac {3665\,a^2\,b^2\,c^2\,d^9}{32}-\frac {235\,b^5\,c^5\,d^6}{4\,a}+\frac {10\,b^6\,c^6\,d^5}{a^2}-30\,a^3\,b\,c\,d^{10}\right )}+\frac {235\,b^5\,c^{9/2}\,d^6\,\sqrt {c+d\,x}}{4\,\left (\frac {235\,b^5\,c^5\,d^6}{4}-\frac {1143\,a\,b^4\,c^4\,d^7}{8}+\frac {5717\,a^2\,b^3\,c^3\,d^8}{32}-\frac {3665\,a^3\,b^2\,c^2\,d^9}{32}-\frac {10\,b^6\,c^6\,d^5}{a}+30\,a^4\,b\,c\,d^{10}\right )}+\frac {10\,b^6\,c^{11/2}\,d^5\,\sqrt {c+d\,x}}{-30\,a^5\,b\,c\,d^{10}+\frac {3665\,a^4\,b^2\,c^2\,d^9}{32}-\frac {5717\,a^3\,b^3\,c^3\,d^8}{32}+\frac {1143\,a^2\,b^4\,c^4\,d^7}{8}-\frac {235\,a\,b^5\,c^5\,d^6}{4}+10\,b^6\,c^6\,d^5}-\frac {30\,a\,b\,\sqrt {c}\,d^{10}\,\sqrt {c+d\,x}}{\frac {3665\,b^2\,c^2\,d^9}{32}-30\,a\,b\,c\,d^{10}-\frac {5717\,b^3\,c^3\,d^8}{32\,a}+\frac {1143\,b^4\,c^4\,d^7}{8\,a^2}-\frac {235\,b^5\,c^5\,d^6}{4\,a^3}+\frac {10\,b^6\,c^6\,d^5}{a^4}}\right )\,\left (15\,a^2\,d^2-20\,a\,b\,c\,d+8\,b^2\,c^2\right )}{4\,a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(x^3*(a + b*x)),x)

[Out]

(((7*a*c^2*d^2 - 4*b*c^3*d)*(c + d*x)^(1/2))/(4*a^2) - ((9*a*c*d^2 - 4*b*c^2*d)*(c + d*x)^(3/2))/(4*a^2))/((c
+ d*x)^2 - 2*c*(c + d*x) + c^2) + (2*atanh((95*b^2*c^2*d^6*(c + d*x)^(1/2)*(b^6*c^5 - a^5*b*d^5 + 5*a^4*b^2*c*
d^4 + 10*a^2*b^4*c^3*d^2 - 10*a^3*b^3*c^2*d^3 - 5*a*b^5*c^4*d)^(1/2))/(4*((215*b^5*c^5*d^6)/4 - (469*a*b^4*c^4
*d^7)/4 + (517*a^2*b^3*c^3*d^8)/4 - (287*a^3*b^2*c^2*d^9)/4 - (10*b^6*c^6*d^5)/a + 16*a^4*b*c*d^10)) + (10*b^3
*c^3*d^5*(c + d*x)^(1/2)*(b^6*c^5 - a^5*b*d^5 + 5*a^4*b^2*c*d^4 + 10*a^2*b^4*c^3*d^2 - 10*a^3*b^3*c^2*d^3 - 5*
a*b^5*c^4*d)^(1/2))/(10*b^6*c^6*d^5 - (215*a*b^5*c^5*d^6)/4 + (469*a^2*b^4*c^4*d^7)/4 - (517*a^3*b^3*c^3*d^8)/
4 + (287*a^4*b^2*c^2*d^9)/4 - 16*a^5*b*c*d^10) + (16*b*c*d^7*(c + d*x)^(1/2)*(b^6*c^5 - a^5*b*d^5 + 5*a^4*b^2*
c*d^4 + 10*a^2*b^4*c^3*d^2 - 10*a^3*b^3*c^2*d^3 - 5*a*b^5*c^4*d)^(1/2))/((469*b^4*c^4*d^7)/4 - (517*a*b^3*c^3*
d^8)/4 + (287*a^2*b^2*c^2*d^9)/4 - (215*b^5*c^5*d^6)/(4*a) + (10*b^6*c^6*d^5)/a^2 - 16*a^3*b*c*d^10))*(-b*(a*d
- b*c)^5)^(1/2))/(a^3*b) - (c^(1/2)*atanh((3665*b^2*c^(3/2)*d^9*(c + d*x)^(1/2))/(32*((3665*b^2*c^2*d^9)/32 -
30*a*b*c*d^10 - (5717*b^3*c^3*d^8)/(32*a) + (1143*b^4*c^4*d^7)/(8*a^2) - (235*b^5*c^5*d^6)/(4*a^3) + (10*b^6*
c^6*d^5)/a^4)) + (5717*b^3*c^(5/2)*d^8*(c + d*x)^(1/2))/(32*((5717*b^3*c^3*d^8)/32 - (3665*a*b^2*c^2*d^9)/32 -
(1143*b^4*c^4*d^7)/(8*a) + (235*b^5*c^5*d^6)/(4*a^2) - (10*b^6*c^6*d^5)/a^3 + 30*a^2*b*c*d^10)) + (1143*b^4*c
^(7/2)*d^7*(c + d*x)^(1/2))/(8*((1143*b^4*c^4*d^7)/8 - (5717*a*b^3*c^3*d^8)/32 + (3665*a^2*b^2*c^2*d^9)/32 - (
235*b^5*c^5*d^6)/(4*a) + (10*b^6*c^6*d^5)/a^2 - 30*a^3*b*c*d^10)) + (235*b^5*c^(9/2)*d^6*(c + d*x)^(1/2))/(4*(
(235*b^5*c^5*d^6)/4 - (1143*a*b^4*c^4*d^7)/8 + (5717*a^2*b^3*c^3*d^8)/32 - (3665*a^3*b^2*c^2*d^9)/32 - (10*b^6
*c^6*d^5)/a + 30*a^4*b*c*d^10)) + (10*b^6*c^(11/2)*d^5*(c + d*x)^(1/2))/(10*b^6*c^6*d^5 - (235*a*b^5*c^5*d^6)/
4 + (1143*a^2*b^4*c^4*d^7)/8 - (5717*a^3*b^3*c^3*d^8)/32 + (3665*a^4*b^2*c^2*d^9)/32 - 30*a^5*b*c*d^10) - (30*
a*b*c^(1/2)*d^10*(c + d*x)^(1/2))/((3665*b^2*c^2*d^9)/32 - 30*a*b*c*d^10 - (5717*b^3*c^3*d^8)/(32*a) + (1143*b
^4*c^4*d^7)/(8*a^2) - (235*b^5*c^5*d^6)/(4*a^3) + (10*b^6*c^6*d^5)/a^4))*(15*a^2*d^2 + 8*b^2*c^2 - 20*a*b*c*d)
)/(4*a^3)

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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((d*x+c)**(5/2)/x**3/(b*x+a),x)

[Out]

Timed out

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