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

Optimal. Leaf size=219 \[ \frac {(6 b c-a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^4 \sqrt {b}}+\frac {c \sqrt {c+d x} (6 b c-7 a d)}{4 a^2 x (a+b x)}-\frac {\sqrt {c} \left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^4}+\frac {\sqrt {c+d x} \left (4 a^2 d^2-17 a b c d+12 b^2 c^2\right )}{4 a^3 (a+b x)}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)} \]

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Rubi [A]  time = 0.24, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {98, 149, 151, 156, 63, 208} \begin {gather*} \frac {\sqrt {c+d x} \left (4 a^2 d^2-17 a b c d+12 b^2 c^2\right )}{4 a^3 (a+b x)}-\frac {\sqrt {c} \left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^4}+\frac {c \sqrt {c+d x} (6 b c-7 a d)}{4 a^2 x (a+b x)}+\frac {(6 b c-a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^4 \sqrt {b}}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((12*b^2*c^2 - 17*a*b*c*d + 4*a^2*d^2)*Sqrt[c + d*x])/(4*a^3*(a + b*x)) + (c*(6*b*c - 7*a*d)*Sqrt[c + d*x])/(4
*a^2*x*(a + b*x)) - (c*(c + d*x)^(3/2))/(2*a*x^2*(a + b*x)) - (Sqrt[c]*(24*b^2*c^2 - 40*a*b*c*d + 15*a^2*d^2)*
ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(4*a^4) + ((b*c - a*d)^(3/2)*(6*b*c - a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqr
t[b*c - a*d]])/(a^4*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 151

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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && 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)^2} \, dx &=-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac {\int \frac {\sqrt {c+d x} \left (\frac {1}{2} c (6 b c-7 a d)+\frac {1}{2} d (3 b c-4 a d) x\right )}{x^2 (a+b x)^2} \, dx}{2 a}\\ &=\frac {c (6 b c-7 a d) \sqrt {c+d x}}{4 a^2 x (a+b x)}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac {\int \frac {-\frac {1}{4} c \left (24 b^2 c^2-40 a b c d+15 a^2 d^2\right )-\frac {1}{4} d \left (18 b^2 c^2-27 a b c d+8 a^2 d^2\right ) x}{x (a+b x)^2 \sqrt {c+d x}} \, dx}{2 a^2}\\ &=\frac {\left (12 b^2 c^2-17 a b c d+4 a^2 d^2\right ) \sqrt {c+d x}}{4 a^3 (a+b x)}+\frac {c (6 b c-7 a d) \sqrt {c+d x}}{4 a^2 x (a+b x)}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac {\int \frac {-\frac {1}{4} c (b c-a d) \left (24 b^2 c^2-40 a b c d+15 a^2 d^2\right )-\frac {1}{4} d (b c-a d) \left (12 b^2 c^2-17 a b c d+4 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}} \, dx}{2 a^3 (b c-a d)}\\ &=\frac {\left (12 b^2 c^2-17 a b c d+4 a^2 d^2\right ) \sqrt {c+d x}}{4 a^3 (a+b x)}+\frac {c (6 b c-7 a d) \sqrt {c+d x}}{4 a^2 x (a+b x)}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac {\left ((b c-a d)^2 (6 b c-a d)\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{2 a^4}+\frac {\left (c \left (24 b^2 c^2-40 a b c d+15 a^2 d^2\right )\right ) \int \frac {1}{x \sqrt {c+d x}} \, dx}{8 a^4}\\ &=\frac {\left (12 b^2 c^2-17 a b c d+4 a^2 d^2\right ) \sqrt {c+d x}}{4 a^3 (a+b x)}+\frac {c (6 b c-7 a d) \sqrt {c+d x}}{4 a^2 x (a+b x)}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac {\left ((b c-a d)^2 (6 b c-a d)\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^4 d}+\frac {\left (c \left (24 b^2 c^2-40 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^4 d}\\ &=\frac {\left (12 b^2 c^2-17 a b c d+4 a^2 d^2\right ) \sqrt {c+d x}}{4 a^3 (a+b x)}+\frac {c (6 b c-7 a d) \sqrt {c+d x}}{4 a^2 x (a+b x)}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac {\sqrt {c} \left (24 b^2 c^2-40 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^4}+\frac {(b c-a d)^{3/2} (6 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^4 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 192, normalized size = 0.88 \begin {gather*} \frac {\frac {a \sqrt {c+d x} \left (a^2 \left (-2 c^2-9 c d x+4 d^2 x^2\right )+a b c x (6 c-17 d x)+12 b^2 c^2 x^2\right )}{x^2 (a+b x)}-\sqrt {c} \left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+\frac {4 \sqrt {b c-a d} \left (a^2 d^2-7 a b c d+6 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{\sqrt {b}}}{4 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a*Sqrt[c + d*x]*(12*b^2*c^2*x^2 + a*b*c*x*(6*c - 17*d*x) + a^2*(-2*c^2 - 9*c*d*x + 4*d^2*x^2)))/(x^2*(a + b*
x)) - Sqrt[c]*(24*b^2*c^2 - 40*a*b*c*d + 15*a^2*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]] + (4*Sqrt[b*c - a*d]*(6*b^
2*c^2 - 7*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/Sqrt[b])/(4*a^4)

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IntegrateAlgebraic [A]  time = 0.85, size = 318, normalized size = 1.45 \begin {gather*} \frac {\left (-15 a^2 \sqrt {c} d^2+40 a b c^{3/2} d-24 b^2 c^{5/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^4}+\frac {\sqrt {c+d x} \left (11 a^2 c^2 d^2-17 a^2 c d^2 (c+d x)+4 a^2 d^2 (c+d x)^2-23 a b c^3 d+40 a b c^2 d (c+d x)-17 a b c d (c+d x)^2+12 b^2 c^4-24 b^2 c^3 (c+d x)+12 b^2 c^2 (c+d x)^2\right )}{4 a^3 d x^2 (a d+b (c+d x)-b c)}+\frac {\left (-a^4 d^4+9 a^3 b c d^3-21 a^2 b^2 c^2 d^2+19 a b^3 c^3 d-6 b^4 c^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{a^4 \sqrt {b} (a d-b c)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(12*b^2*c^4 - 23*a*b*c^3*d + 11*a^2*c^2*d^2 - 24*b^2*c^3*(c + d*x) + 40*a*b*c^2*d*(c + d*x) - 1
7*a^2*c*d^2*(c + d*x) + 12*b^2*c^2*(c + d*x)^2 - 17*a*b*c*d*(c + d*x)^2 + 4*a^2*d^2*(c + d*x)^2))/(4*a^3*d*x^2
*(-(b*c) + a*d + b*(c + d*x))) + ((-6*b^4*c^4 + 19*a*b^3*c^3*d - 21*a^2*b^2*c^2*d^2 + 9*a^3*b*c*d^3 - a^4*d^4)
*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x])/(b*c - a*d)])/(a^4*Sqrt[b]*(-(b*c) + a*d)^(3/2)) + ((-24*b^
2*c^(5/2) + 40*a*b*c^(3/2)*d - 15*a^2*Sqrt[c]*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(4*a^4)

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fricas [A]  time = 1.91, size = 1173, normalized size = 5.36

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(4*((6*b^3*c^2 - 7*a*b^2*c*d + a^2*b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a^2*b*c*d + a^3*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)) + ((24*b^3*c^2 - 40*a*b^2*c
*d + 15*a^2*b*d^2)*x^3 + (24*a*b^2*c^2 - 40*a^2*b*c*d + 15*a^3*d^2)*x^2)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sq
rt(c) + 2*c)/x) - 2*(2*a^3*c^2 - (12*a*b^2*c^2 - 17*a^2*b*c*d + 4*a^3*d^2)*x^2 - 3*(2*a^2*b*c^2 - 3*a^3*c*d)*x
)*sqrt(d*x + c))/(a^4*b*x^3 + a^5*x^2), 1/8*(8*((6*b^3*c^2 - 7*a*b^2*c*d + a^2*b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a
^2*b*c*d + a^3*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)) + ((24
*b^3*c^2 - 40*a*b^2*c*d + 15*a^2*b*d^2)*x^3 + (24*a*b^2*c^2 - 40*a^2*b*c*d + 15*a^3*d^2)*x^2)*sqrt(c)*log((d*x
- 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(2*a^3*c^2 - (12*a*b^2*c^2 - 17*a^2*b*c*d + 4*a^3*d^2)*x^2 - 3*(2*a^2
*b*c^2 - 3*a^3*c*d)*x)*sqrt(d*x + c))/(a^4*b*x^3 + a^5*x^2), 1/4*(((24*b^3*c^2 - 40*a*b^2*c*d + 15*a^2*b*d^2)*
x^3 + (24*a*b^2*c^2 - 40*a^2*b*c*d + 15*a^3*d^2)*x^2)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) + 2*((6*b^3*c^
2 - 7*a*b^2*c*d + a^2*b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a^2*b*c*d + a^3*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^3*c^2 - (12*a*b^2*c^2 - 17*a^2*b*c*d +
4*a^3*d^2)*x^2 - 3*(2*a^2*b*c^2 - 3*a^3*c*d)*x)*sqrt(d*x + c))/(a^4*b*x^3 + a^5*x^2), 1/4*(4*((6*b^3*c^2 - 7*
a*b^2*c*d + a^2*b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a^2*b*c*d + a^3*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)) + ((24*b^3*c^2 - 40*a*b^2*c*d + 15*a^2*b*d^2)*x^3 + (24*a*b^2*c^2 - 4
0*a^2*b*c*d + 15*a^3*d^2)*x^2)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) - (2*a^3*c^2 - (12*a*b^2*c^2 - 17*a^2
*b*c*d + 4*a^3*d^2)*x^2 - 3*(2*a^2*b*c^2 - 3*a^3*c*d)*x)*sqrt(d*x + c))/(a^4*b*x^3 + a^5*x^2)]

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giac [A]  time = 1.36, size = 265, normalized size = 1.21 \begin {gather*} -\frac {{\left (6 \, b^{3} c^{3} - 13 \, a b^{2} c^{2} d + 8 \, 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^{4}} + \frac {{\left (24 \, b^{2} c^{3} - 40 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{4 \, a^{4} \sqrt {-c}} + \frac {\sqrt {d x + c} b^{2} c^{2} d - 2 \, \sqrt {d x + c} a b c d^{2} + \sqrt {d x + c} a^{2} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} a^{3}} + \frac {8 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{2} d - 8 \, \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^{3} 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)^2,x, algorithm="giac")

[Out]

-(6*b^3*c^3 - 13*a*b^2*c^2*d + 8*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^4) + 1/4*(24*b^2*c^3 - 40*a*b*c^2*d + 15*a^2*c*d^2)*arctan(sqrt(d*x + c)/sqrt(-c))/(a^4*sqrt(-c
)) + (sqrt(d*x + c)*b^2*c^2*d - 2*sqrt(d*x + c)*a*b*c*d^2 + sqrt(d*x + c)*a^2*d^3)/(((d*x + c)*b - b*c + a*d)*
a^3) + 1/4*(8*(d*x + c)^(3/2)*b*c^2*d - 8*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^3*d^2*x^2)

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maple [B]  time = 0.02, size = 403, normalized size = 1.84 \begin {gather*} \frac {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}\, a}-\frac {8 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^{2}}+\frac {13 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^{3}}-\frac {6 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^{4}}+\frac {\sqrt {d x +c}\, d^{3}}{\left (b d x +a d \right ) a}-\frac {2 \sqrt {d x +c}\, b c \,d^{2}}{\left (b d x +a d \right ) a^{2}}+\frac {\sqrt {d x +c}\, b^{2} c^{2} d}{\left (b d x +a d \right ) a^{3}}-\frac {15 \sqrt {c}\, d^{2} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{4 a^{2}}+\frac {10 b \,c^{\frac {3}{2}} d \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{3}}-\frac {6 b^{2} c^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{4}}+\frac {7 \sqrt {d x +c}\, c^{2}}{4 a^{2} x^{2}}-\frac {2 \sqrt {d x +c}\, b \,c^{3}}{a^{3} d \,x^{2}}-\frac {9 \left (d x +c \right )^{\frac {3}{2}} c}{4 a^{2} x^{2}}+\frac {2 \left (d x +c \right )^{\frac {3}{2}} b \,c^{2}}{a^{3} 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)^2,x)

[Out]

d^3/a*(d*x+c)^(1/2)/(b*d*x+a*d)-2*d^2/a^2*(d*x+c)^(1/2)/(b*d*x+a*d)*b*c+d/a^3*(d*x+c)^(1/2)/(b*d*x+a*d)*b^2*c^
2+d^3/a/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)-8*d^2/a^2/((a*d-b*c)*b)^(1/2)*arctan((
d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*b*c+13*d/a^3/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*
b)*b^2*c^2-6/a^4/((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/a^2/x^2*(d*x+c)
^(3/2)+2/d*c^2/a^3/x^2*(d*x+c)^(3/2)*b+7/4*c^2/a^2/x^2*(d*x+c)^(1/2)-2/d*c^3/a^3/x^2*(d*x+c)^(1/2)*b-15/4*d^2*
c^(1/2)/a^2*arctanh((d*x+c)^(1/2)/c^(1/2))+10*d*c^(3/2)/a^3*arctanh((d*x+c)^(1/2)/c^(1/2))*b-6*c^(5/2)/a^4*arc
tanh((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)^2,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 = 0.95, size = 670, normalized size = 3.06 \begin {gather*} \frac {\frac {\sqrt {c+d\,x}\,\left (11\,a^2\,c^2\,d^3-23\,a\,b\,c^3\,d^2+12\,b^2\,c^4\,d\right )}{4\,a^3}-\frac {{\left (c+d\,x\right )}^{3/2}\,\left (17\,a^2\,c\,d^3-40\,a\,b\,c^2\,d^2+24\,b^2\,c^3\,d\right )}{4\,a^3}+\frac {d\,{\left (c+d\,x\right )}^{5/2}\,\left (4\,a^2\,d^2-17\,a\,b\,c\,d+12\,b^2\,c^2\right )}{4\,a^3}}{b\,{\left (c+d\,x\right )}^3+\left (a\,d-3\,b\,c\right )\,{\left (c+d\,x\right )}^2-b\,c^3+\left (3\,b\,c^2-2\,a\,c\,d\right )\,\left (c+d\,x\right )+a\,c^2\,d}+\frac {\sqrt {c}\,\ln \left (\sqrt {c+d\,x}-\sqrt {c}\right )\,\left (\frac {15\,a^2\,d^2}{8}-5\,a\,b\,c\,d+3\,b^2\,c^2\right )}{a^4}-\frac {\sqrt {c}\,\ln \left (\sqrt {c+d\,x}+\sqrt {c}\right )\,\left (15\,a^2\,d^2-40\,a\,b\,c\,d+24\,b^2\,c^2\right )}{8\,a^4}+\frac {\mathrm {atan}\left (-\frac {b^2\,c^2\,d^7\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b\,d^3+3\,a^2\,b^2\,c\,d^2-3\,a\,b^3\,c^2\,d+b^4\,c^3}\,111{}\mathrm {i}}{8\,\left (41\,a\,b^3\,c^3\,d^8-\frac {291\,b^4\,c^4\,d^7}{8}-\frac {143\,a^2\,b^2\,c^2\,d^9}{8}+\frac {45\,b^5\,c^5\,d^6}{4\,a}+2\,a^3\,b\,c\,d^{10}\right )}+\frac {b^3\,c^3\,d^6\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b\,d^3+3\,a^2\,b^2\,c\,d^2-3\,a\,b^3\,c^2\,d+b^4\,c^3}\,45{}\mathrm {i}}{4\,\left (2\,a^4\,b\,c\,d^{10}-\frac {143\,a^3\,b^2\,c^2\,d^9}{8}+41\,a^2\,b^3\,c^3\,d^8-\frac {291\,a\,b^4\,c^4\,d^7}{8}+\frac {45\,b^5\,c^5\,d^6}{4}\right )}+\frac {b\,c\,d^8\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b\,d^3+3\,a^2\,b^2\,c\,d^2-3\,a\,b^3\,c^2\,d+b^4\,c^3}\,2{}\mathrm {i}}{41\,b^3\,c^3\,d^8-\frac {143\,a\,b^2\,c^2\,d^9}{8}-\frac {291\,b^4\,c^4\,d^7}{8\,a}+\frac {45\,b^5\,c^5\,d^6}{4\,a^2}+2\,a^2\,b\,c\,d^{10}}\right )\,\left (a\,d-6\,b\,c\right )\,\sqrt {-b\,{\left (a\,d-b\,c\right )}^3}\,1{}\mathrm {i}}{a^4\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((c + d*x)^(1/2)*(12*b^2*c^4*d + 11*a^2*c^2*d^3 - 23*a*b*c^3*d^2))/(4*a^3) - ((c + d*x)^(3/2)*(17*a^2*c*d^3 +
24*b^2*c^3*d - 40*a*b*c^2*d^2))/(4*a^3) + (d*(c + d*x)^(5/2)*(4*a^2*d^2 + 12*b^2*c^2 - 17*a*b*c*d))/(4*a^3))/
(b*(c + d*x)^3 + (a*d - 3*b*c)*(c + d*x)^2 - b*c^3 + (3*b*c^2 - 2*a*c*d)*(c + d*x) + a*c^2*d) + (c^(1/2)*log((
c + d*x)^(1/2) - c^(1/2))*((15*a^2*d^2)/8 + 3*b^2*c^2 - 5*a*b*c*d))/a^4 - (c^(1/2)*log((c + d*x)^(1/2) + c^(1/
2))*(15*a^2*d^2 + 24*b^2*c^2 - 40*a*b*c*d))/(8*a^4) + (atan((b^3*c^3*d^6*(c + d*x)^(1/2)*(b^4*c^3 - a^3*b*d^3
+ 3*a^2*b^2*c*d^2 - 3*a*b^3*c^2*d)^(1/2)*45i)/(4*((45*b^5*c^5*d^6)/4 - (291*a*b^4*c^4*d^7)/8 + 41*a^2*b^3*c^3*
d^8 - (143*a^3*b^2*c^2*d^9)/8 + 2*a^4*b*c*d^10)) - (b^2*c^2*d^7*(c + d*x)^(1/2)*(b^4*c^3 - a^3*b*d^3 + 3*a^2*b
^2*c*d^2 - 3*a*b^3*c^2*d)^(1/2)*111i)/(8*(41*a*b^3*c^3*d^8 - (291*b^4*c^4*d^7)/8 - (143*a^2*b^2*c^2*d^9)/8 + (
45*b^5*c^5*d^6)/(4*a) + 2*a^3*b*c*d^10)) + (b*c*d^8*(c + d*x)^(1/2)*(b^4*c^3 - a^3*b*d^3 + 3*a^2*b^2*c*d^2 - 3
*a*b^3*c^2*d)^(1/2)*2i)/(41*b^3*c^3*d^8 - (143*a*b^2*c^2*d^9)/8 - (291*b^4*c^4*d^7)/(8*a) + (45*b^5*c^5*d^6)/(
4*a^2) + 2*a^2*b*c*d^10))*(a*d - 6*b*c)*(-b*(a*d - b*c)^3)^(1/2)*1i)/(a^4*b)

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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)**2,x)

[Out]

Timed out

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