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

Optimal. Leaf size=201 \[ \frac{3 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^3}{64 a^2 c^2 x}-\frac{3 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{64 a^{5/2} c^{5/2}}-\frac{\sqrt{a+b x} (c+d x)^{3/2} (b c-a d)^2}{32 a c^2 x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2} (b c-a d)}{8 c^2 x^3}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4} \]

[Out]

(3*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*a^2*c^2*x) - ((b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(3/2))/(
32*a*c^2*x^2) - ((b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(5/2))/(8*c^2*x^3) - ((a + b*x)^(3/2)*(c + d*x)^(5/2))/(4
*c*x^4) - (3*(b*c - a*d)^4*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(64*a^(5/2)*c^(5/2))

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Rubi [A]  time = 0.0989714, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {94, 93, 208} \[ \frac{3 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^3}{64 a^2 c^2 x}-\frac{3 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{64 a^{5/2} c^{5/2}}-\frac{\sqrt{a+b x} (c+d x)^{3/2} (b c-a d)^2}{32 a c^2 x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2} (b c-a d)}{8 c^2 x^3}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*a^2*c^2*x) - ((b*c - a*d)^2*Sqrt[a + b*x]*(c + d*x)^(3/2))/(
32*a*c^2*x^2) - ((b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(5/2))/(8*c^2*x^3) - ((a + b*x)^(3/2)*(c + d*x)^(5/2))/(4
*c*x^4) - (3*(b*c - a*d)^4*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(64*a^(5/2)*c^(5/2))

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} (c+d x)^{3/2}}{x^5} \, dx &=-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}+\frac{(3 (b c-a d)) \int \frac{\sqrt{a+b x} (c+d x)^{3/2}}{x^4} \, dx}{8 c}\\ &=-\frac{(b c-a d) \sqrt{a+b x} (c+d x)^{5/2}}{8 c^2 x^3}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}+\frac{(b c-a d)^2 \int \frac{(c+d x)^{3/2}}{x^3 \sqrt{a+b x}} \, dx}{16 c^2}\\ &=-\frac{(b c-a d)^2 \sqrt{a+b x} (c+d x)^{3/2}}{32 a c^2 x^2}-\frac{(b c-a d) \sqrt{a+b x} (c+d x)^{5/2}}{8 c^2 x^3}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}-\frac{\left (3 (b c-a d)^3\right ) \int \frac{\sqrt{c+d x}}{x^2 \sqrt{a+b x}} \, dx}{64 a c^2}\\ &=\frac{3 (b c-a d)^3 \sqrt{a+b x} \sqrt{c+d x}}{64 a^2 c^2 x}-\frac{(b c-a d)^2 \sqrt{a+b x} (c+d x)^{3/2}}{32 a c^2 x^2}-\frac{(b c-a d) \sqrt{a+b x} (c+d x)^{5/2}}{8 c^2 x^3}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}+\frac{\left (3 (b c-a d)^4\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{128 a^2 c^2}\\ &=\frac{3 (b c-a d)^3 \sqrt{a+b x} \sqrt{c+d x}}{64 a^2 c^2 x}-\frac{(b c-a d)^2 \sqrt{a+b x} (c+d x)^{3/2}}{32 a c^2 x^2}-\frac{(b c-a d) \sqrt{a+b x} (c+d x)^{5/2}}{8 c^2 x^3}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}+\frac{\left (3 (b c-a d)^4\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^2}\\ &=\frac{3 (b c-a d)^3 \sqrt{a+b x} \sqrt{c+d x}}{64 a^2 c^2 x}-\frac{(b c-a d)^2 \sqrt{a+b x} (c+d x)^{3/2}}{32 a c^2 x^2}-\frac{(b c-a d) \sqrt{a+b x} (c+d x)^{5/2}}{8 c^2 x^3}-\frac{(a+b x)^{3/2} (c+d x)^{5/2}}{4 c x^4}-\frac{3 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{64 a^{5/2} c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.498495, size = 173, normalized size = 0.86 \[ -\frac{x (b c-a d) \left (\frac{x (b c-a d) \left (3 x^2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\sqrt{a} \sqrt{c} \sqrt{a+b x} \sqrt{c+d x} (2 a c+5 a d x-3 b c x)\right )}{a^{5/2} \sqrt{c}}+8 \sqrt{a+b x} (c+d x)^{5/2}\right )+16 c (a+b x)^{3/2} (c+d x)^{5/2}}{64 c^2 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(16*c*(a + b*x)^(3/2)*(c + d*x)^(5/2) + (b*c - a*d)*x*(8*Sqrt[a + b*x]*(c + d*x)^(5/2) + ((b*c - a*d)*x*(Sqrt
[a]*Sqrt[c]*Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*c - 3*b*c*x + 5*a*d*x) + 3*(b*c - a*d)^2*x^2*ArcTanh[(Sqrt[c]*Sqr
t[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(a^(5/2)*Sqrt[c])))/(64*c^2*x^4)

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Maple [B]  time = 0.016, size = 705, normalized size = 3.5 \begin{align*} -{\frac{1}{128\,{a}^{2}{c}^{2}{x}^{4}}\sqrt{bx+a}\sqrt{dx+c} \left ( 3\,\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}-12\,\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+3\,\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}-6\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{3}{a}^{3}{d}^{3}+22\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{3}{a}^{2}bc{d}^{2}+22\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{3}a{b}^{2}{c}^{2}d-6\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{3}{b}^{3}{c}^{3}+4\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{2}{a}^{3}c{d}^{2}+88\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{2}{a}^{2}b{c}^{2}d+4\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}{x}^{2}a{b}^{2}{c}^{3}+48\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}x{a}^{3}{c}^{2}d+48\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}x{a}^{2}b{c}^{3}+32\,\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)^(3/2)/x^5,x)

[Out]

-1/128*(b*x+a)^(1/2)*(d*x+c)^(1/2)/a^2/c^2*(3*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^4-12*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^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+3*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^4-6*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x^3*a^3*d^3
+22*(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+22*(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-6*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x^3*b^3*c^3+4*(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+88*(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+4*(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+48*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^3*c^2*d+48*(a*c)^(1/2
)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^2*b*c^3+32*(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)^(3/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 25.6635, size = 1219, normalized size = 6.06 \begin{align*} \left [\frac{3 \,{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + 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 (16 \, a^{4} c^{4} -{\left (3 \, a b^{3} c^{4} - 11 \, a^{2} b^{2} c^{3} d - 11 \, a^{3} b c^{2} d^{2} + 3 \, a^{4} c d^{3}\right )} x^{3} + 2 \,{\left (a^{2} b^{2} c^{4} + 22 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2}\right )} x^{2} + 24 \,{\left (a^{3} b c^{4} + a^{4} c^{3} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{256 \, a^{3} c^{3} x^{4}}, \frac{3 \,{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + 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 (16 \, a^{4} c^{4} -{\left (3 \, a b^{3} c^{4} - 11 \, a^{2} b^{2} c^{3} d - 11 \, a^{3} b c^{2} d^{2} + 3 \, a^{4} c d^{3}\right )} x^{3} + 2 \,{\left (a^{2} b^{2} c^{4} + 22 \, a^{3} b c^{3} d + a^{4} c^{2} d^{2}\right )} x^{2} + 24 \,{\left (a^{3} b c^{4} + a^{4} c^{3} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{128 \, a^{3} c^{3} 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)^(3/2)/x^5,x, algorithm="fricas")

[Out]

[1/256*(3*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + 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*(16*a^4*c^4 - (3*a*b^3*c^4 - 11*a^2*b^2*c^3*d - 11*a^3*b*c^2*d^2 + 3*a^4*c*d^3)
*x^3 + 2*(a^2*b^2*c^4 + 22*a^3*b*c^3*d + a^4*c^2*d^2)*x^2 + 24*(a^3*b*c^4 + a^4*c^3*d)*x)*sqrt(b*x + a)*sqrt(d
*x + c))/(a^3*c^3*x^4), 1/128*(3*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + 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*(16*a^4*c^4 - (3*a*b^3*c^4 - 11*a^2*b^2*c^3*d - 11*a^3*b*c^2*d^2 + 3*a^4*c*d^3)*x^3
+ 2*(a^2*b^2*c^4 + 22*a^3*b*c^3*d + a^4*c^2*d^2)*x^2 + 24*(a^3*b*c^4 + a^4*c^3*d)*x)*sqrt(b*x + a)*sqrt(d*x +
c))/(a^3*c^3*x^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{\frac{3}{2}} \left (c + d x\right )^{\frac{3}{2}}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(3/2)/x^5,x, algorithm="giac")

[Out]

Exception raised: TypeError