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

Optimal. Leaf size=178 \[ -\frac {3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2} \sqrt {c}}+\frac {3 \sqrt {c+d x} (b c-a d) (5 b c-a d)}{4 a^3 c \sqrt {a+b x}}+\frac {(c+d x)^{3/2} (5 b c-a d)}{4 a^2 c x \sqrt {a+b x}}-\frac {(c+d x)^{5/2}}{2 a c x^2 \sqrt {a+b x}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {96, 94, 93, 208} \[ \frac {(c+d x)^{3/2} (5 b c-a d)}{4 a^2 c x \sqrt {a+b x}}+\frac {3 \sqrt {c+d x} (b c-a d) (5 b c-a d)}{4 a^3 c \sqrt {a+b x}}-\frac {3 (b c-a d) (5 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2} \sqrt {c}}-\frac {(c+d x)^{5/2}}{2 a c x^2 \sqrt {a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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

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Mathematica [A]  time = 0.10, size = 130, normalized size = 0.73 \[ \frac {\sqrt {c+d x} \left (-\left (a^2 (2 c+5 d x)\right )+a b x (5 c-13 d x)+15 b^2 c x^2\right )}{4 a^3 x^2 \sqrt {a+b x}}-\frac {3 \left (a^2 d^2-6 a b c d+5 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2} \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(15*b^2*c*x^2 + a*b*x*(5*c - 13*d*x) - a^2*(2*c + 5*d*x)))/(4*a^3*x^2*Sqrt[a + b*x]) - (3*(5*b^
2*c^2 - 6*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(7/2)*Sqrt[c])

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fricas [A]  time = 1.72, size = 474, normalized size = 2.66 \[ \left [\frac {3 \, {\left ({\left (5 \, b^{3} c^{2} - 6 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} + {\left (5 \, a b^{2} c^{2} - 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\right )} \sqrt {a c} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (2 \, a^{3} c^{2} - {\left (15 \, a b^{2} c^{2} - 13 \, a^{2} b c d\right )} x^{2} - 5 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (a^{4} b c x^{3} + a^{5} c x^{2}\right )}}, \frac {3 \, {\left ({\left (5 \, b^{3} c^{2} - 6 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} + {\left (5 \, a b^{2} c^{2} - 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\right )} \sqrt {-a c} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (2 \, a^{3} c^{2} - {\left (15 \, a b^{2} c^{2} - 13 \, a^{2} b c d\right )} x^{2} - 5 \, {\left (a^{2} b c^{2} - a^{3} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (a^{4} b c x^{3} + a^{5} c x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*((5*b^3*c^2 - 6*a*b^2*c*d + a^2*b*d^2)*x^3 + (5*a*b^2*c^2 - 6*a^2*b*c*d + a^3*d^2)*x^2)*sqrt(a*c)*log
((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x
+ c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a^3*c^2 - (15*a*b^2*c^2 - 13*a^2*b*c*d)*x^2 - 5*(a^2*b*c^2 - a^3*
c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^4*b*c*x^3 + a^5*c*x^2), 1/8*(3*((5*b^3*c^2 - 6*a*b^2*c*d + a^2*b*d^2)*
x^3 + (5*a*b^2*c^2 - 6*a^2*b*c*d + a^3*d^2)*x^2)*sqrt(-a*c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt
(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) - 2*(2*a^3*c^2 - (15*a*b^2*c^2 - 13*a
^2*b*c*d)*x^2 - 5*(a^2*b*c^2 - a^3*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^4*b*c*x^3 + a^5*c*x^2)]

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giac [B]  time = 21.75, size = 1167, normalized size = 6.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*(5*sqrt(b*d)*b^2*c^2*abs(b) - 6*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b*d)*a^2*d^2*abs(b))*arctan(-1/2*(b^2*c +
a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d
)*a^3*b) + 4*(sqrt(b*d)*b^2*c^2*abs(b) - 2*sqrt(b*d)*a*b*c*d*abs(b) + sqrt(b*d)*a^2*d^2*abs(b))/((b^2*c - a*b*
d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)*a^3) + 1/2*(7*sqrt(b*d)*b^8*c^5*abs(b)
- 33*sqrt(b*d)*a*b^7*c^4*d*abs(b) + 62*sqrt(b*d)*a^2*b^6*c^3*d^2*abs(b) - 58*sqrt(b*d)*a^3*b^5*c^2*d^3*abs(b)
+ 27*sqrt(b*d)*a^4*b^4*c*d^4*abs(b) - 5*sqrt(b*d)*a^5*b^3*d^5*abs(b) - 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^6*c^4*abs(b) + 32*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^2*a*b^5*c^3*d*abs(b) + 14*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*
b*d - a*b*d))^2*a^2*b^4*c^2*d^2*abs(b) - 40*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d -
a*b*d))^2*a^3*b^3*c*d^3*abs(b) + 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^
2*a^4*b^2*d^4*abs(b) + 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^4*c^3*
abs(b) - sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^3*c^2*d*abs(b) + 11*s
qrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^2*b^2*c*d^2*abs(b) - 15*sqrt(b*d)
*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b*d^3*abs(b) - 7*sqrt(b*d)*(sqrt(b*d)*s
qrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^2*c^2*abs(b) + 2*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) -
sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b*c*d*abs(b) + 5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b
*x + a)*b*d - a*b*d))^6*a^2*d^2*abs(b))/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)*sqrt(b*x + a) - s
qrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)
)^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^2*a^3)

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maple [B]  time = 0.03, size = 464, normalized size = 2.61 \[ -\frac {\sqrt {d x +c}\, \left (3 a^{2} b \,d^{2} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-18 a \,b^{2} c d \,x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+15 b^{3} c^{2} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+3 a^{3} d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-18 a^{2} b c d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+15 a \,b^{2} c^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+26 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b d \,x^{2}-30 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c \,x^{2}+10 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d x -10 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c x +4 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c \right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, \sqrt {b x +a}\, a^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(d*x+c)^(1/2)*(3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a^2*b*d^2-18*ln((a*d
*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a*b^2*c*d+15*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*
((b*x+a)*(d*x+c))^(1/2))/x)*x^3*b^3*c^2+3*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*
a^3*d^2-18*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a^2*b*c*d+15*ln((a*d*x+b*c*x+2*
a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a*b^2*c^2+26*x^2*a*b*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-3
0*x^2*b^2*c*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+10*x*a^2*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-10*x*a*b*c*(a*c
)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*a^2*c*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/a^3/((b*x+a)*(d*x+c))^(1/2)/x^2/(
a*c)^(1/2)/(b*x+a)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)/x^3/(b*x+a)^(3/2),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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,x\right )}^{3/2}}{x^3\,{\left (a+b\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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