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

Optimal. Leaf size=118 \[ \frac{2 a}{b \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}+\frac{4 \sqrt{a+b x} (3 a d+b c)}{3 \sqrt{c+d x} (b c-a d)^3}+\frac{2 \sqrt{a+b x} (3 a d+b c)}{3 b (c+d x)^{3/2} (b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.0375537, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {78, 45, 37} \[ \frac{2 a}{b \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}+\frac{4 \sqrt{a+b x} (3 a d+b c)}{3 \sqrt{c+d x} (b c-a d)^3}+\frac{2 \sqrt{a+b x} (3 a d+b c)}{3 b (c+d x)^{3/2} (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0283627, size = 83, normalized size = 0.7 \[ \frac{2 \left (a^2 d (2 c+3 d x)+2 a b \left (3 c^2+5 c d x+3 d^2 x^2\right )+b^2 c x (3 c+2 d x)\right )}{3 \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 115, normalized size = 1. \begin{align*} -{\frac{12\,ab{d}^{2}{x}^{2}+4\,{b}^{2}cd{x}^{2}+6\,{a}^{2}{d}^{2}x+20\,abcdx+6\,{b}^{2}{c}^{2}x+4\,{a}^{2}cd+12\,ab{c}^{2}}{3\,{a}^{3}{d}^{3}-9\,{a}^{2}cb{d}^{2}+9\,a{b}^{2}{c}^{2}d-3\,{b}^{3}{c}^{3}}{\frac{1}{\sqrt{bx+a}}} \left ( dx+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(6*a*b*d^2*x^2+2*b^2*c*d*x^2+3*a^2*d^2*x+10*a*b*c*d*x+3*b^2*c^2*x+2*a^2*c*d+6*a*b*c^2)/(b*x+a)^(1/2)/(d*x
+c)^(3/2)/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.62413, size = 567, normalized size = 4.81 \begin{align*} \frac{2 \,{\left (6 \, a b c^{2} + 2 \, a^{2} c d + 2 \,{\left (b^{2} c d + 3 \, a b d^{2}\right )} x^{2} +{\left (3 \, b^{2} c^{2} + 10 \, a b c d + 3 \, a^{2} d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{3 \,{\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} +{\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} +{\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} +{\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(6*a*b*c^2 + 2*a^2*c*d + 2*(b^2*c*d + 3*a*b*d^2)*x^2 + (3*b^2*c^2 + 10*a*b*c*d + 3*a^2*d^2)*x)*sqrt(b*x +
a)*sqrt(d*x + c)/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3
+ 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d
^5)*x^2 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.87349, size = 421, normalized size = 3.57 \begin{align*} \frac{\frac{48 \, \sqrt{b d} a b^{3}}{{\left (b^{2} c^{2}{\left | b \right |} - 2 \, a b c d{\left | b \right |} + a^{2} d^{2}{\left | b \right |}\right )}{\left (b^{2} c - a b d -{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}} - \frac{\sqrt{b x + a}{\left (\frac{{\left (2 \, b^{7} c^{3} d^{2}{\left | b \right |} - a b^{6} c^{2} d^{3}{\left | b \right |} - 4 \, a^{2} b^{5} c d^{4}{\left | b \right |} + 3 \, a^{3} b^{4} d^{5}{\left | b \right |}\right )}{\left (b x + a\right )}}{b^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}} + \frac{3 \,{\left (b^{8} c^{4} d{\left | b \right |} - 2 \, a b^{7} c^{3} d^{2}{\left | b \right |} + 2 \, a^{3} b^{5} c d^{4}{\left | b \right |} - a^{4} b^{4} d^{5}{\left | b \right |}\right )}}{b^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}}\right )}}{{\left (b^{2} c +{\left (b x + a\right )} b d - a b d\right )}^{\frac{3}{2}}}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(48*sqrt(b*d)*a*b^3/((b^2*c^2*abs(b) - 2*a*b*c*d*abs(b) + a^2*d^2*abs(b))*(b^2*c - a*b*d - (sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)) - sqrt(b*x + a)*((2*b^7*c^3*d^2*abs(b) - a*b^6*c^2*d^3*a
bs(b) - 4*a^2*b^5*c*d^4*abs(b) + 3*a^3*b^4*d^5*abs(b))*(b*x + a)/(b^8*c^2*d^4 - 2*a*b^7*c*d^5 + a^2*b^6*d^6) +
3*(b^8*c^4*d*abs(b) - 2*a*b^7*c^3*d^2*abs(b) + 2*a^3*b^5*c*d^4*abs(b) - a^4*b^4*d^5*abs(b))/(b^8*c^2*d^4 - 2*
a*b^7*c*d^5 + a^2*b^6*d^6))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2))/b