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

Optimal. Leaf size=234 \[ -\frac{d \sqrt{a+b x} \left (105 a^2 d^2-100 a b c d+3 b^2 c^2\right )}{12 a c^4 \sqrt{c+d x} (b c-a d)}+\frac{\left (-35 a^2 d^2+10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{3/2} c^{9/2}}-\frac{d \sqrt{a+b x} (3 b c-35 a d)}{12 a c^3 (c+d x)^{3/2}}-\frac{\sqrt{a+b x} (b c-7 a d)}{4 a c^2 x (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}} \]

[Out]

-(d*(3*b*c - 35*a*d)*Sqrt[a + b*x])/(12*a*c^3*(c + d*x)^(3/2)) - Sqrt[a + b*x]/(2*c*x^2*(c + d*x)^(3/2)) - ((b
*c - 7*a*d)*Sqrt[a + b*x])/(4*a*c^2*x*(c + d*x)^(3/2)) - (d*(3*b^2*c^2 - 100*a*b*c*d + 105*a^2*d^2)*Sqrt[a + b
*x])/(12*a*c^4*(b*c - a*d)*Sqrt[c + d*x]) + ((b^2*c^2 + 10*a*b*c*d - 35*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x
])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(3/2)*c^(9/2))

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Rubi [A]  time = 0.22174, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {99, 151, 152, 12, 93, 208} \[ -\frac{d \sqrt{a+b x} \left (105 a^2 d^2-100 a b c d+3 b^2 c^2\right )}{12 a c^4 \sqrt{c+d x} (b c-a d)}+\frac{\left (-35 a^2 d^2+10 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{3/2} c^{9/2}}-\frac{d \sqrt{a+b x} (3 b c-35 a d)}{12 a c^3 (c+d x)^{3/2}}-\frac{\sqrt{a+b x} (b c-7 a d)}{4 a c^2 x (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(3*b*c - 35*a*d)*Sqrt[a + b*x])/(12*a*c^3*(c + d*x)^(3/2)) - Sqrt[a + b*x]/(2*c*x^2*(c + d*x)^(3/2)) - ((b
*c - 7*a*d)*Sqrt[a + b*x])/(4*a*c^2*x*(c + d*x)^(3/2)) - (d*(3*b^2*c^2 - 100*a*b*c*d + 105*a^2*d^2)*Sqrt[a + b
*x])/(12*a*c^4*(b*c - a*d)*Sqrt[c + d*x]) + ((b^2*c^2 + 10*a*b*c*d - 35*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x
])/(Sqrt[a]*Sqrt[c + d*x])])/(4*a^(3/2)*c^(9/2))

Rule 99

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[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

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 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{\sqrt{a+b x}}{x^3 (c+d x)^{5/2}} \, dx &=-\frac{\sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}+\frac{\int \frac{\frac{1}{2} (b c-7 a d)-3 b d x}{x^2 \sqrt{a+b x} (c+d x)^{5/2}} \, dx}{2 c}\\ &=-\frac{\sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(b c-7 a d) \sqrt{a+b x}}{4 a c^2 x (c+d x)^{3/2}}-\frac{\int \frac{\frac{1}{4} \left (b^2 c^2+10 a b c d-35 a^2 d^2\right )+b d (b c-7 a d) x}{x \sqrt{a+b x} (c+d x)^{5/2}} \, dx}{2 a c^2}\\ &=-\frac{d (3 b c-35 a d) \sqrt{a+b x}}{12 a c^3 (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(b c-7 a d) \sqrt{a+b x}}{4 a c^2 x (c+d x)^{3/2}}+\frac{\int \frac{-\frac{3}{8} (b c-a d) \left (b^2 c^2+10 a b c d-35 a^2 d^2\right )-\frac{1}{4} b d (3 b c-35 a d) (b c-a d) x}{x \sqrt{a+b x} (c+d x)^{3/2}} \, dx}{3 a c^3 (b c-a d)}\\ &=-\frac{d (3 b c-35 a d) \sqrt{a+b x}}{12 a c^3 (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(b c-7 a d) \sqrt{a+b x}}{4 a c^2 x (c+d x)^{3/2}}-\frac{d \left (3 b^2 c^2-100 a b c d+105 a^2 d^2\right ) \sqrt{a+b x}}{12 a c^4 (b c-a d) \sqrt{c+d x}}-\frac{2 \int \frac{3 (b c-a d)^2 \left (b^2 c^2+10 a b c d-35 a^2 d^2\right )}{16 x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{3 a c^4 (b c-a d)^2}\\ &=-\frac{d (3 b c-35 a d) \sqrt{a+b x}}{12 a c^3 (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(b c-7 a d) \sqrt{a+b x}}{4 a c^2 x (c+d x)^{3/2}}-\frac{d \left (3 b^2 c^2-100 a b c d+105 a^2 d^2\right ) \sqrt{a+b x}}{12 a c^4 (b c-a d) \sqrt{c+d x}}-\frac{\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 a c^4}\\ &=-\frac{d (3 b c-35 a d) \sqrt{a+b x}}{12 a c^3 (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(b c-7 a d) \sqrt{a+b x}}{4 a c^2 x (c+d x)^{3/2}}-\frac{d \left (3 b^2 c^2-100 a b c d+105 a^2 d^2\right ) \sqrt{a+b x}}{12 a c^4 (b c-a d) \sqrt{c+d x}}-\frac{\left (b^2 c^2+10 a b c d-35 a^2 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 c^4}\\ &=-\frac{d (3 b c-35 a d) \sqrt{a+b x}}{12 a c^3 (c+d x)^{3/2}}-\frac{\sqrt{a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac{(b c-7 a d) \sqrt{a+b x}}{4 a c^2 x (c+d x)^{3/2}}-\frac{d \left (3 b^2 c^2-100 a b c d+105 a^2 d^2\right ) \sqrt{a+b x}}{12 a c^4 (b c-a d) \sqrt{c+d x}}+\frac{\left (b^2 c^2+10 a b c d-35 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 a^{3/2} c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.383752, size = 236, normalized size = 1.01 \[ \frac{x^2 \left (c^{3/2} d (a+b x)^{3/2} \left (-35 a^2 d^2+24 a b c d+3 b^2 c^2\right )-3 (c+d x) (b c-a d) \left (-35 a^2 d^2+10 a b c d+b^2 c^2\right ) \left (\sqrt{c} \sqrt{a+b x}-\sqrt{a} \sqrt{c+d x} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )\right )+6 a c^{7/2} (a+b x)^{3/2} (a d-b c)+3 c^{5/2} x (a+b x)^{3/2} (b c-a d) (7 a d+b c)}{12 a^2 c^{9/2} x^2 (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*c^(7/2)*(-(b*c) + a*d)*(a + b*x)^(3/2) + 3*c^(5/2)*(b*c - a*d)*(b*c + 7*a*d)*x*(a + b*x)^(3/2) + x^2*(c^(
3/2)*d*(3*b^2*c^2 + 24*a*b*c*d - 35*a^2*d^2)*(a + b*x)^(3/2) - 3*(b*c - a*d)*(b^2*c^2 + 10*a*b*c*d - 35*a^2*d^
2)*(c + d*x)*(Sqrt[c]*Sqrt[a + b*x] - Sqrt[a]*Sqrt[c + d*x]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c +
d*x])])))/(12*a^2*c^(9/2)*(b*c - a*d)*x^2*(c + d*x)^(3/2))

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Maple [B]  time = 0.028, size = 988, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)/x^3/(d*x+c)^(5/2),x)

[Out]

-1/24*(105*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^4*a^3*d^5-135*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^4*a^2*b*c*d^4+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^4*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^4*b^3*
c^3*d^2+210*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*c*d^4-270*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^2*d^3+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^3*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^3
*b^3*c^4*d+105*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^2*d^3-135*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^3*d^2+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^2*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^2*b^3*c^5-210*x^3*a^2*d^4*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+200*x^3*a*b*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(a*c
)^(1/2)-6*x^3*b^2*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-280*x^2*a^2*c*d^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^
(1/2)+276*x^2*a*b*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-12*x^2*b^2*c^3*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(
1/2)-42*x*a^2*c^2*d^2*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+48*x*a*b*c^3*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-6
*x*b^2*c^4*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+12*a^2*c^3*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-12*a*b*c^4*((b
*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2))/c^4/a*(b*x+a)^(1/2)/x^2/(a*d-b*c)/(a*c)^(1/2)/((b*x+a)*(d*x+c))^(1/2)/(d*x+c
)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)/x^3/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 26.3352, size = 1891, normalized size = 8.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)/x^3/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[-1/48*(3*((b^3*c^3*d^2 + 9*a*b^2*c^2*d^3 - 45*a^2*b*c*d^4 + 35*a^3*d^5)*x^4 + 2*(b^3*c^4*d + 9*a*b^2*c^3*d^2
- 45*a^2*b*c^2*d^3 + 35*a^3*c*d^4)*x^3 + (b^3*c^5 + 9*a*b^2*c^4*d - 45*a^2*b*c^3*d^2 + 35*a^3*c^2*d^3)*x^2)*sq
rt(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*(6*a^2*b*c^5 - 6*a^3*c^4*d + (3*a*b^2*c^3*d^2 - 100*a^2*b
*c^2*d^3 + 105*a^3*c*d^4)*x^3 + 2*(3*a*b^2*c^4*d - 69*a^2*b*c^3*d^2 + 70*a^3*c^2*d^3)*x^2 + 3*(a*b^2*c^5 - 8*a
^2*b*c^4*d + 7*a^3*c^3*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/((a^2*b*c^6*d^2 - a^3*c^5*d^3)*x^4 + 2*(a^2*b*c^7*
d - a^3*c^6*d^2)*x^3 + (a^2*b*c^8 - a^3*c^7*d)*x^2), -1/24*(3*((b^3*c^3*d^2 + 9*a*b^2*c^2*d^3 - 45*a^2*b*c*d^4
+ 35*a^3*d^5)*x^4 + 2*(b^3*c^4*d + 9*a*b^2*c^3*d^2 - 45*a^2*b*c^2*d^3 + 35*a^3*c*d^4)*x^3 + (b^3*c^5 + 9*a*b^
2*c^4*d - 45*a^2*b*c^3*d^2 + 35*a^3*c^2*d^3)*x^2)*sqrt(-a*c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqr
t(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*(6*a^2*b*c^5 - 6*a^3*c^4*d + (3*
a*b^2*c^3*d^2 - 100*a^2*b*c^2*d^3 + 105*a^3*c*d^4)*x^3 + 2*(3*a*b^2*c^4*d - 69*a^2*b*c^3*d^2 + 70*a^3*c^2*d^3)
*x^2 + 3*(a*b^2*c^5 - 8*a^2*b*c^4*d + 7*a^3*c^3*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/((a^2*b*c^6*d^2 - a^3*c^5
*d^3)*x^4 + 2*(a^2*b*c^7*d - a^3*c^6*d^2)*x^3 + (a^2*b*c^8 - a^3*c^7*d)*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)/x**3/(d*x+c)**(5/2),x)

[Out]

Timed out

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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((b*x+a)^(1/2)/x^3/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError