3.754 \(\int \frac{1}{x^2 \sqrt{a+b x} (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=189 \[ -\frac{d \sqrt{a+b x} \left (15 a^2 d^2-22 a b c d+3 b^2 c^2\right )}{3 a c^3 \sqrt{c+d x} (b c-a d)^2}+\frac{(5 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{3/2} c^{7/2}}-\frac{d \sqrt{a+b x} (3 b c-5 a d)}{3 a c^2 (c+d x)^{3/2} (b c-a d)}-\frac{\sqrt{a+b x}}{a c x (c+d x)^{3/2}} \]

[Out]

-(d*(3*b*c - 5*a*d)*Sqrt[a + b*x])/(3*a*c^2*(b*c - a*d)*(c + d*x)^(3/2)) - Sqrt[a + b*x]/(a*c*x*(c + d*x)^(3/2
)) - (d*(3*b^2*c^2 - 22*a*b*c*d + 15*a^2*d^2)*Sqrt[a + b*x])/(3*a*c^3*(b*c - a*d)^2*Sqrt[c + d*x]) + ((b*c + 5
*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(a^(3/2)*c^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.184367, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {103, 152, 12, 93, 208} \[ -\frac{d \sqrt{a+b x} \left (15 a^2 d^2-22 a b c d+3 b^2 c^2\right )}{3 a c^3 \sqrt{c+d x} (b c-a d)^2}+\frac{(5 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{3/2} c^{7/2}}-\frac{d \sqrt{a+b x} (3 b c-5 a d)}{3 a c^2 (c+d x)^{3/2} (b c-a d)}-\frac{\sqrt{a+b x}}{a c x (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[a + b*x]*(c + d*x)^(5/2)),x]

[Out]

-(d*(3*b*c - 5*a*d)*Sqrt[a + b*x])/(3*a*c^2*(b*c - a*d)*(c + d*x)^(3/2)) - Sqrt[a + b*x]/(a*c*x*(c + d*x)^(3/2
)) - (d*(3*b^2*c^2 - 22*a*b*c*d + 15*a^2*d^2)*Sqrt[a + b*x])/(3*a*c^3*(b*c - a*d)^2*Sqrt[c + d*x]) + ((b*c + 5
*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(a^(3/2)*c^(7/2))

Rule 103

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[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*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 152

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] && IntegersQ[2*m, 2*n, 2*p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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{1}{x^2 \sqrt{a+b x} (c+d x)^{5/2}} \, dx &=-\frac{\sqrt{a+b x}}{a c x (c+d x)^{3/2}}-\frac{\int \frac{\frac{1}{2} (b c+5 a d)+2 b d x}{x \sqrt{a+b x} (c+d x)^{5/2}} \, dx}{a c}\\ &=-\frac{d (3 b c-5 a d) \sqrt{a+b x}}{3 a c^2 (b c-a d) (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{a c x (c+d x)^{3/2}}+\frac{2 \int \frac{-\frac{3}{4} (b c-a d) (b c+5 a d)-\frac{1}{2} b d (3 b c-5 a d) x}{x \sqrt{a+b x} (c+d x)^{3/2}} \, dx}{3 a c^2 (b c-a d)}\\ &=-\frac{d (3 b c-5 a d) \sqrt{a+b x}}{3 a c^2 (b c-a d) (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{a c x (c+d x)^{3/2}}-\frac{d \left (3 b^2 c^2-22 a b c d+15 a^2 d^2\right ) \sqrt{a+b x}}{3 a c^3 (b c-a d)^2 \sqrt{c+d x}}-\frac{4 \int \frac{3 (b c-a d)^2 (b c+5 a d)}{8 x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{3 a c^3 (b c-a d)^2}\\ &=-\frac{d (3 b c-5 a d) \sqrt{a+b x}}{3 a c^2 (b c-a d) (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{a c x (c+d x)^{3/2}}-\frac{d \left (3 b^2 c^2-22 a b c d+15 a^2 d^2\right ) \sqrt{a+b x}}{3 a c^3 (b c-a d)^2 \sqrt{c+d x}}-\frac{(b c+5 a d) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 a c^3}\\ &=-\frac{d (3 b c-5 a d) \sqrt{a+b x}}{3 a c^2 (b c-a d) (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{a c x (c+d x)^{3/2}}-\frac{d \left (3 b^2 c^2-22 a b c d+15 a^2 d^2\right ) \sqrt{a+b x}}{3 a c^3 (b c-a d)^2 \sqrt{c+d x}}-\frac{(b c+5 a d) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{a c^3}\\ &=-\frac{d (3 b c-5 a d) \sqrt{a+b x}}{3 a c^2 (b c-a d) (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{a c x (c+d x)^{3/2}}-\frac{d \left (3 b^2 c^2-22 a b c d+15 a^2 d^2\right ) \sqrt{a+b x}}{3 a c^3 (b c-a d)^2 \sqrt{c+d x}}+\frac{(b c+5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{3/2} c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.349796, size = 198, normalized size = 1.05 \[ \frac{\frac{(c+d x) \left (\sqrt{a} \sqrt{c} d \sqrt{a+b x} \left (-15 a^2 d^2+22 a b c d-3 b^2 c^2\right )+3 \sqrt{c+d x} (5 a d+b c) (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )}{\sqrt{a} c^{5/2} (b c-a d)^2}-\frac{d \sqrt{a+b x} (3 b c-5 a d)}{c (b c-a d)}-\frac{3 \sqrt{a+b x}}{x}}{3 a c (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[a + b*x]*(c + d*x)^(5/2)),x]

[Out]

(-((d*(3*b*c - 5*a*d)*Sqrt[a + b*x])/(c*(b*c - a*d))) - (3*Sqrt[a + b*x])/x + ((c + d*x)*(Sqrt[a]*Sqrt[c]*d*(-
3*b^2*c^2 + 22*a*b*c*d - 15*a^2*d^2)*Sqrt[a + b*x] + 3*(b*c - a*d)^2*(b*c + 5*a*d)*Sqrt[c + d*x]*ArcTanh[(Sqrt
[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(Sqrt[a]*c^(5/2)*(b*c - a*d)^2))/(3*a*c*(c + d*x)^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.032, size = 919, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(d*x+c)^(5/2)/(b*x+a)^(1/2),x)

[Out]

1/6*(b*x+a)^(1/2)/a/c^3*(15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^3*a^3*d^5-27*ln(
(a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^3*a^2*b*c*d^4+9*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(
(b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^3*a*b^2*c^2*d^3+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a
*c)/x)*x^3*b^3*c^3*d^2+30*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a^3*c*d^4-54*ln(
(a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a^2*b*c^2*d^3+18*ln((a*d*x+b*c*x+2*(a*c)^(1/2
)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a*b^2*c^3*d^2+6*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+
2*a*c)/x)*x^2*b^3*c^4*d+15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a^3*c^2*d^3-27*ln
((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a^2*b*c^3*d^2+9*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*
((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a*b^2*c^4*d+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)
/x)*x*b^3*c^5-30*x^2*a^2*d^4*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+44*x^2*a*b*c*d^3*(a*c)^(1/2)*((b*x+a)*(d*x+c)
)^(1/2)-6*x^2*b^2*c^2*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-40*x*a^2*c*d^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/
2)+60*x*a*b*c^2*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-12*x*b^2*c^3*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-6*a
^2*c^2*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+12*a*b*c^3*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-6*b^2*c^4*(a*c
)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x/(a*c)^(1/2)/(a*d-b*c)^2/((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(3/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x + a}{\left (d x + c\right )}^{\frac{5}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x + a)*(d*x + c)^(5/2)*x^2), x)

________________________________________________________________________________________

Fricas [B]  time = 10.5498, size = 1926, normalized size = 10.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/12*(3*((b^3*c^3*d^2 + 3*a*b^2*c^2*d^3 - 9*a^2*b*c*d^4 + 5*a^3*d^5)*x^3 + 2*(b^3*c^4*d + 3*a*b^2*c^3*d^2 - 9
*a^2*b*c^2*d^3 + 5*a^3*c*d^4)*x^2 + (b^3*c^5 + 3*a*b^2*c^4*d - 9*a^2*b*c^3*d^2 + 5*a^3*c^2*d^3)*x)*sqrt(a*c)*l
og((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*(3*a*b^2*c^5 - 6*a^2*b*c^4*d + 3*a^3*c^3*d^2 + (3*a*b^2*c^3*d^2 -
22*a^2*b*c^2*d^3 + 15*a^3*c*d^4)*x^2 + 2*(3*a*b^2*c^4*d - 15*a^2*b*c^3*d^2 + 10*a^3*c^2*d^3)*x)*sqrt(b*x + a)*
sqrt(d*x + c))/((a^2*b^2*c^6*d^2 - 2*a^3*b*c^5*d^3 + a^4*c^4*d^4)*x^3 + 2*(a^2*b^2*c^7*d - 2*a^3*b*c^6*d^2 + a
^4*c^5*d^3)*x^2 + (a^2*b^2*c^8 - 2*a^3*b*c^7*d + a^4*c^6*d^2)*x), -1/6*(3*((b^3*c^3*d^2 + 3*a*b^2*c^2*d^3 - 9*
a^2*b*c*d^4 + 5*a^3*d^5)*x^3 + 2*(b^3*c^4*d + 3*a*b^2*c^3*d^2 - 9*a^2*b*c^2*d^3 + 5*a^3*c*d^4)*x^2 + (b^3*c^5
+ 3*a*b^2*c^4*d - 9*a^2*b*c^3*d^2 + 5*a^3*c^2*d^3)*x)*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*(3*a*b^2*c^5 - 6*a^2*b*c^4*d
+ 3*a^3*c^3*d^2 + (3*a*b^2*c^3*d^2 - 22*a^2*b*c^2*d^3 + 15*a^3*c*d^4)*x^2 + 2*(3*a*b^2*c^4*d - 15*a^2*b*c^3*d
^2 + 10*a^3*c^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/((a^2*b^2*c^6*d^2 - 2*a^3*b*c^5*d^3 + a^4*c^4*d^4)*x^3 +
2*(a^2*b^2*c^7*d - 2*a^3*b*c^6*d^2 + a^4*c^5*d^3)*x^2 + (a^2*b^2*c^8 - 2*a^3*b*c^7*d + a^4*c^6*d^2)*x)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \sqrt{a + b x} \left (c + d x\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(d*x+c)**(5/2)/(b*x+a)**(1/2),x)

[Out]

Integral(1/(x**2*sqrt(a + b*x)*(c + d*x)**(5/2)), x)

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError