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

Optimal. Leaf size=233 \[ \frac{\sqrt{a+b x} \sqrt{c+d x} (5 a d+3 b c) (b c-a d)^2}{64 a^2 c^3 x}-\frac{(5 a d+3 b c) (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{64 a^{5/2} c^{7/2}}+\frac{\sqrt{a+b x} (c+d x)^{3/2} (5 a d+3 b c) (b c-a d)}{32 a c^3 x^2}+\frac{(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{24 a c^2 x^3}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{4 a c x^4} \]

[Out]

((b*c - a*d)^2*(3*b*c + 5*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*a^2*c^3*x) + ((b*c - a*d)*(3*b*c + 5*a*d)*Sqrt
[a + b*x]*(c + d*x)^(3/2))/(32*a*c^3*x^2) + ((3*b*c + 5*a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(24*a*c^2*x^3) -
((a + b*x)^(5/2)*(c + d*x)^(3/2))/(4*a*c*x^4) - ((b*c - a*d)^3*(3*b*c + 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x]
)/(Sqrt[a]*Sqrt[c + d*x])])/(64*a^(5/2)*c^(7/2))

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Rubi [A]  time = 0.115154, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {96, 94, 93, 208} \[ \frac{\sqrt{a+b x} \sqrt{c+d x} (5 a d+3 b c) (b c-a d)^2}{64 a^2 c^3 x}-\frac{(5 a d+3 b c) (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{64 a^{5/2} c^{7/2}}+\frac{\sqrt{a+b x} (c+d x)^{3/2} (5 a d+3 b c) (b c-a d)}{32 a c^3 x^2}+\frac{(a+b x)^{3/2} (c+d x)^{3/2} (5 a d+3 b c)}{24 a c^2 x^3}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{4 a c x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*(3*b*c + 5*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*a^2*c^3*x) + ((b*c - a*d)*(3*b*c + 5*a*d)*Sqrt
[a + b*x]*(c + d*x)^(3/2))/(32*a*c^3*x^2) + ((3*b*c + 5*a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(24*a*c^2*x^3) -
((a + b*x)^(5/2)*(c + d*x)^(3/2))/(4*a*c*x^4) - ((b*c - a*d)^3*(3*b*c + 5*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x]
)/(Sqrt[a]*Sqrt[c + d*x])])/(64*a^(5/2)*c^(7/2))

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 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 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{(a+b x)^{3/2} \sqrt{c+d x}}{x^5} \, dx &=-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{4 a c x^4}-\frac{\left (\frac{3 b c}{2}+\frac{5 a d}{2}\right ) \int \frac{(a+b x)^{3/2} \sqrt{c+d x}}{x^4} \, dx}{4 a c}\\ &=\frac{(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 a c^2 x^3}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{4 a c x^4}-\frac{((b c-a d) (3 b c+5 a d)) \int \frac{\sqrt{a+b x} \sqrt{c+d x}}{x^3} \, dx}{16 a c^2}\\ &=\frac{(b c-a d) (3 b c+5 a d) \sqrt{a+b x} (c+d x)^{3/2}}{32 a c^3 x^2}+\frac{(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 a c^2 x^3}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{4 a c x^4}-\frac{\left ((b c-a d)^2 (3 b c+5 a d)\right ) \int \frac{\sqrt{c+d x}}{x^2 \sqrt{a+b x}} \, dx}{64 a c^3}\\ &=\frac{(b c-a d)^2 (3 b c+5 a d) \sqrt{a+b x} \sqrt{c+d x}}{64 a^2 c^3 x}+\frac{(b c-a d) (3 b c+5 a d) \sqrt{a+b x} (c+d x)^{3/2}}{32 a c^3 x^2}+\frac{(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 a c^2 x^3}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{4 a c x^4}+\frac{\left ((b c-a d)^3 (3 b c+5 a d)\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{128 a^2 c^3}\\ &=\frac{(b c-a d)^2 (3 b c+5 a d) \sqrt{a+b x} \sqrt{c+d x}}{64 a^2 c^3 x}+\frac{(b c-a d) (3 b c+5 a d) \sqrt{a+b x} (c+d x)^{3/2}}{32 a c^3 x^2}+\frac{(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 a c^2 x^3}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{4 a c x^4}+\frac{\left ((b c-a d)^3 (3 b c+5 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{64 a^2 c^3}\\ &=\frac{(b c-a d)^2 (3 b c+5 a d) \sqrt{a+b x} \sqrt{c+d x}}{64 a^2 c^3 x}+\frac{(b c-a d) (3 b c+5 a d) \sqrt{a+b x} (c+d x)^{3/2}}{32 a c^3 x^2}+\frac{(3 b c+5 a d) (a+b x)^{3/2} (c+d x)^{3/2}}{24 a c^2 x^3}-\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{4 a c x^4}-\frac{(b c-a d)^3 (3 b c+5 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{64 a^{5/2} c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.276181, size = 182, normalized size = 0.78 \[ -\frac{\frac{(5 a d+3 b c) \left (\frac{3 x (b c-a d) \left (x^2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )-\sqrt{a} \sqrt{c} \sqrt{a+b x} \sqrt{c+d x} (2 a c+a d x+b c x)\right )}{a^{3/2} c^{3/2}}-8 (a+b x)^{3/2} (c+d x)^{3/2}\right )}{48 c x^3}+\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{x^4}}{4 a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b*x)^(5/2)*(c + d*x)^(3/2))/x^4 + ((3*b*c + 5*a*d)*(-8*(a + b*x)^(3/2)*(c + d*x)^(3/2) + (3*(b*c - a*d
)*x*(-(Sqrt[a]*Sqrt[c]*Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*c + b*c*x + a*d*x)) + (b*c - a*d)^2*x^2*ArcTanh[(Sqrt[
c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(a^(3/2)*c^(3/2))))/(48*c*x^3))/(4*a*c)

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Maple [B]  time = 0.017, size = 705, normalized size = 3. \begin{align*}{\frac{1}{384\,{a}^{2}{c}^{3}{x}^{4}}\sqrt{bx+a}\sqrt{dx+c} \left ( 15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{4}{a}^{4}{d}^{4}-36\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{4}{a}^{3}bc{d}^{3}+18\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{4}{a}^{2}{b}^{2}{c}^{2}{d}^{2}+12\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{4}a{b}^{3}{c}^{3}d-9\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{4}{b}^{4}{c}^{4}-30\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{3}{a}^{3}{d}^{3}+62\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{3}{a}^{2}bc{d}^{2}-18\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{3}a{b}^{2}{c}^{2}d+18\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{3}{b}^{3}{c}^{3}+20\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{2}{a}^{3}c{d}^{2}-40\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{2}{a}^{2}b{c}^{2}d-12\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{2}a{b}^{2}{c}^{3}-16\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}x{a}^{3}{c}^{2}d-144\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}x{a}^{2}b{c}^{3}-96\,\sqrt{d{x}^{2}b+adx+bcx+ac}{a}^{3}{c}^{3}\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2/c^3*(15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*
a*c)/x)*x^4*a^4*d^4-36*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^4*a^3*b*c*d^3
+18*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^4*a^2*b^2*c^2*d^2+12*ln((a*d*x+b
*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^4*a*b^3*c^3*d-9*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(
b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^4*b^4*c^4-30*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x^3*a^3*d^
3+62*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x^3*a^2*b*c*d^2-18*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2
)*x^3*a*b^2*c^2*d+18*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x^3*b^3*c^3+20*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c
*x+a*c)^(1/2)*x^2*a^3*c*d^2-40*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x^2*a^2*b*c^2*d-12*(a*c)^(1/2)*(b*d
*x^2+a*d*x+b*c*x+a*c)^(1/2)*x^2*a*b^2*c^3-16*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^3*c^2*d-144*(a*c)
^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^2*b*c^3-96*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^3*c^3*(a*c)^(1/2))/(b*
d*x^2+a*d*x+b*c*x+a*c)^(1/2)/x^4/(a*c)^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 27.0739, size = 1247, normalized size = 5.35 \begin{align*} \left [-\frac{3 \,{\left (3 \, b^{4} c^{4} - 4 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} + 12 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4}\right )} \sqrt{a c} x^{4} \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 (48 \, a^{4} c^{4} -{\left (9 \, a b^{3} c^{4} - 9 \, a^{2} b^{2} c^{3} d + 31 \, a^{3} b c^{2} d^{2} - 15 \, a^{4} c d^{3}\right )} x^{3} + 2 \,{\left (3 \, a^{2} b^{2} c^{4} + 10 \, a^{3} b c^{3} d - 5 \, a^{4} c^{2} d^{2}\right )} x^{2} + 8 \,{\left (9 \, a^{3} b c^{4} + a^{4} c^{3} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{768 \, a^{3} c^{4} x^{4}}, \frac{3 \,{\left (3 \, b^{4} c^{4} - 4 \, a b^{3} c^{3} d - 6 \, a^{2} b^{2} c^{2} d^{2} + 12 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4}\right )} \sqrt{-a c} x^{4} \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 (48 \, a^{4} c^{4} -{\left (9 \, a b^{3} c^{4} - 9 \, a^{2} b^{2} c^{3} d + 31 \, a^{3} b c^{2} d^{2} - 15 \, a^{4} c d^{3}\right )} x^{3} + 2 \,{\left (3 \, a^{2} b^{2} c^{4} + 10 \, a^{3} b c^{3} d - 5 \, a^{4} c^{2} d^{2}\right )} x^{2} + 8 \,{\left (9 \, a^{3} b c^{4} + a^{4} c^{3} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{384 \, a^{3} c^{4} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/768*(3*(3*b^4*c^4 - 4*a*b^3*c^3*d - 6*a^2*b^2*c^2*d^2 + 12*a^3*b*c*d^3 - 5*a^4*d^4)*sqrt(a*c)*x^4*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*(48*a^4*c^4 - (9*a*b^3*c^4 - 9*a^2*b^2*c^3*d + 31*a^3*b*c^2*d^2 - 15*a^4*
c*d^3)*x^3 + 2*(3*a^2*b^2*c^4 + 10*a^3*b*c^3*d - 5*a^4*c^2*d^2)*x^2 + 8*(9*a^3*b*c^4 + a^4*c^3*d)*x)*sqrt(b*x
+ a)*sqrt(d*x + c))/(a^3*c^4*x^4), 1/384*(3*(3*b^4*c^4 - 4*a*b^3*c^3*d - 6*a^2*b^2*c^2*d^2 + 12*a^3*b*c*d^3 -
5*a^4*d^4)*sqrt(-a*c)*x^4*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*(48*a^4*c^4 - (9*a*b^3*c^4 - 9*a^2*b^2*c^3*d + 31*a^3*b*c^2*d^2 - 1
5*a^4*c*d^3)*x^3 + 2*(3*a^2*b^2*c^4 + 10*a^3*b*c^3*d - 5*a^4*c^2*d^2)*x^2 + 8*(9*a^3*b*c^4 + a^4*c^3*d)*x)*sqr
t(b*x + a)*sqrt(d*x + c))/(a^3*c^4*x^4)]

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

[Out]

Exception raised: TypeError