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

Optimal. Leaf size=154 \[ -\frac {15 \sqrt {c} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2}}+\frac {15 \sqrt {c+d x} (b c-a d)^2}{4 a^3 \sqrt {a+b x}}+\frac {5 (c+d x)^{3/2} (b c-a d)}{4 a^2 x \sqrt {a+b x}}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}} \]

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Rubi [A]  time = 0.07, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {94, 93, 208} \begin {gather*} \frac {5 (c+d x)^{3/2} (b c-a d)}{4 a^2 x \sqrt {a+b x}}+\frac {15 \sqrt {c+d x} (b c-a d)^2}{4 a^3 \sqrt {a+b x}}-\frac {15 \sqrt {c} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2}}-\frac {(c+d x)^{5/2}}{2 a x^2 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.11, size = 130, normalized size = 0.84 \begin {gather*} \frac {\sqrt {c+d x} \left (a^2 \left (-2 c^2-9 c d x+8 d^2 x^2\right )+5 a b c x (c-5 d x)+15 b^2 c^2 x^2\right )}{4 a^3 x^2 \sqrt {a+b x}}-\frac {15 \sqrt {c} (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.26, size = 151, normalized size = 0.98 \begin {gather*} \frac {\sqrt {c+d x} (a d-b c)^2 \left (\frac {8 a^2 (c+d x)^2}{(a+b x)^2}-\frac {25 a c (c+d x)}{a+b x}+15 c^2\right )}{4 a^3 \sqrt {a+b x} \left (\frac {a (c+d x)}{a+b x}-c\right )^2}-\frac {15 \sqrt {c} (a d-b c)^2 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{4 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-(b*c) + a*d)^2*Sqrt[c + d*x]*(15*c^2 - (25*a*c*(c + d*x))/(a + b*x) + (8*a^2*(c + d*x)^2)/(a + b*x)^2))/(4*
a^3*Sqrt[a + b*x]*(-c + (a*(c + d*x))/(a + b*x))^2) - (15*Sqrt[c]*(-(b*c) + a*d)^2*ArcTanh[(Sqrt[a]*Sqrt[c + d
*x])/(Sqrt[c]*Sqrt[a + b*x])])/(4*a^(7/2))

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fricas [A]  time = 2.44, size = 481, normalized size = 3.12 \begin {gather*} \left [\frac {15 \, {\left ({\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} + {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\right )} \sqrt {\frac {c}{a}} \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^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (2 \, a^{2} c^{2} - {\left (15 \, b^{2} c^{2} - 25 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{2} - {\left (5 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, \frac {15 \, {\left ({\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{3} + {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} x^{2}\right )} \sqrt {-\frac {c}{a}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) - 2 \, {\left (2 \, a^{2} c^{2} - {\left (15 \, b^{2} c^{2} - 25 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{2} - {\left (5 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}\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)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(15*((b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2)*x^3 + (a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x^2)*sqrt(c/a)*log((8
*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*s
qrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(2*a^2*c^2 - (15*b^2*c^2 - 25*a*b*c*d + 8*a^2*d^2)*x^2 - (5*a*b*c
^2 - 9*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b*x^3 + a^4*x^2), 1/8*(15*((b^3*c^2 - 2*a*b^2*c*d + a^2*b
*d^2)*x^3 + (a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x^2)*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x +
a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) - 2*(2*a^2*c^2 - (15*b^2*c^2 - 25*a*b*c*d
+ 8*a^2*d^2)*x^2 - (5*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^3*b*x^3 + a^4*x^2)]

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giac [B]  time = 21.61, size = 1206, normalized size = 7.83

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)^(3/2),x, algorithm="giac")

[Out]

-15/4*(sqrt(b*d)*b^2*c^3*abs(b) - 2*sqrt(b*d)*a*b*c^2*d*abs(b) + sqrt(b*d)*a^2*c*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) + 1/2*(7*sqrt(b*d)*b^8*c^6*abs(b) - 37*sqrt(b*d)*a*b^7*c^5*d*abs(b) + 78*sqrt(b*d)*a^2*b^6*c^4*d^2
*abs(b) - 82*sqrt(b*d)*a^3*b^5*c^3*d^3*abs(b) + 43*sqrt(b*d)*a^4*b^4*c^2*d^4*abs(b) - 9*sqrt(b*d)*a^5*b^3*c*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^5*abs(b) + 44*
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^4*d*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))^2*a^2*b^4*c^3*d^2*abs(b) - 52*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^2*d^3*abs(b) + 27*sqrt(b*d)*(sqrt(b*d)*sq
rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^2*c*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^4*abs(b) - 13*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^3*d*abs(b) + 3*sqrt(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^2*d^2*abs(b) - 27*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*c*d^3*abs(b) - 7*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))
^6*b^2*c^3*abs(b) + 6*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^2*d*ab
s(b) + 9*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*c*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) - sqrt(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) + 4*(sqrt(b*d)*b^3*c^3*abs(b) - 3*sqrt(b*d)*a*b^2*c^2*d*abs(b) + 3*sqrt
(b*d)*a^2*b*c*d^2*abs(b) - sqrt(b*d)*a^3*d^3*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*b)

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maple [B]  time = 0.03, size = 507, normalized size = 3.29 \begin {gather*} -\frac {\sqrt {d x +c}\, \left (15 a^{2} b c \,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 )-30 a \,b^{2} c^{2} 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^{3} 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 a^{3} c \,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 )-30 a^{2} b \,c^{2} 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^{3} 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 )-16 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} d^{2} x^{2}+50 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b c d \,x^{2}-30 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, b^{2} c^{2} x^{2}+18 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c d x -10 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a b \,c^{2} x +4 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, a^{2} c^{2}\right )}{8 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {a c}\, \sqrt {b x +a}\, a^{3} 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)^(3/2),x)

[Out]

-1/8*(d*x+c)^(1/2)*(15*a^2*b*c*d^2*x^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)-30*a*b^
2*c^2*d*x^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)+15*b^3*c^3*x^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)+15*a^3*c*d^2*x^2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*
x+c))^(1/2))/x)-30*a^2*b*c^2*d*x^2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)+15*a*b^2*c^
3*x^2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)-16*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a
^2*d^2*x^2+50*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*b*c*d*x^2-30*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*b^2*c^2*x
^2+18*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*c*d*x-10*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a*b*c^2*x+4*((b*x+a
)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*c^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 \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)^(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 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^{5/2}}{x^3\,{\left (a+b\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \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)**(3/2),x)

[Out]

Timed out

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