3.1395 \(\int \frac{(c+d x)^{3/2}}{(a+b x)^3} \, dx\)

Optimal. Leaf size=100 \[ -\frac{3 d^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 b^{5/2} \sqrt{b c-a d}}-\frac{3 d \sqrt{c+d x}}{4 b^2 (a+b x)}-\frac{(c+d x)^{3/2}}{2 b (a+b x)^2} \]

[Out]

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

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Rubi [A]  time = 0.0475181, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {47, 63, 208} \[ -\frac{3 d^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 b^{5/2} \sqrt{b c-a d}}-\frac{3 d \sqrt{c+d x}}{4 b^2 (a+b x)}-\frac{(c+d x)^{3/2}}{2 b (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)/(a + b*x)^3,x]

[Out]

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

Rule 47

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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{(c+d x)^{3/2}}{(a+b x)^3} \, dx &=-\frac{(c+d x)^{3/2}}{2 b (a+b x)^2}+\frac{(3 d) \int \frac{\sqrt{c+d x}}{(a+b x)^2} \, dx}{4 b}\\ &=-\frac{3 d \sqrt{c+d x}}{4 b^2 (a+b x)}-\frac{(c+d x)^{3/2}}{2 b (a+b x)^2}+\frac{\left (3 d^2\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{8 b^2}\\ &=-\frac{3 d \sqrt{c+d x}}{4 b^2 (a+b x)}-\frac{(c+d x)^{3/2}}{2 b (a+b x)^2}+\frac{(3 d) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{4 b^2}\\ &=-\frac{3 d \sqrt{c+d x}}{4 b^2 (a+b x)}-\frac{(c+d x)^{3/2}}{2 b (a+b x)^2}-\frac{3 d^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{4 b^{5/2} \sqrt{b c-a d}}\\ \end{align*}

Mathematica [A]  time = 0.0977563, size = 90, normalized size = 0.9 \[ \frac{3 d^2 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{a d-b c}}\right )}{4 b^{5/2} \sqrt{a d-b c}}-\frac{\sqrt{c+d x} (3 a d+2 b c+5 b d x)}{4 b^2 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 121, normalized size = 1.2 \begin{align*} -{\frac{5\,{d}^{2}}{4\, \left ( bdx+ad \right ) ^{2}b} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{d}^{3}a}{4\, \left ( bdx+ad \right ) ^{2}{b}^{2}}\sqrt{dx+c}}+{\frac{3\,{d}^{2}c}{4\, \left ( bdx+ad \right ) ^{2}b}\sqrt{dx+c}}+{\frac{3\,{d}^{2}}{4\,{b}^{2}}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/4*d^2/(b*d*x+a*d)^2/b*(d*x+c)^(3/2)-3/4*d^3/(b*d*x+a*d)^2/b^2*(d*x+c)^(1/2)*a+3/4*d^2/(b*d*x+a*d)^2/b*(d*x+
c)^(1/2)*c+3/4*d^2/b^2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(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((d*x+c)^(3/2)/(b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.81, size = 795, normalized size = 7.95 \begin{align*} \left [\frac{3 \,{\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt{b^{2} c - a b d} \log \left (\frac{b d x + 2 \, b c - a d - 2 \, \sqrt{b^{2} c - a b d} \sqrt{d x + c}}{b x + a}\right ) - 2 \,{\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} + 5 \,{\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt{d x + c}}{8 \,{\left (a^{2} b^{4} c - a^{3} b^{3} d +{\left (b^{6} c - a b^{5} d\right )} x^{2} + 2 \,{\left (a b^{5} c - a^{2} b^{4} d\right )} x\right )}}, \frac{3 \,{\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \sqrt{-b^{2} c + a b d} \arctan \left (\frac{\sqrt{-b^{2} c + a b d} \sqrt{d x + c}}{b d x + b c}\right ) -{\left (2 \, b^{3} c^{2} + a b^{2} c d - 3 \, a^{2} b d^{2} + 5 \,{\left (b^{3} c d - a b^{2} d^{2}\right )} x\right )} \sqrt{d x + c}}{4 \,{\left (a^{2} b^{4} c - a^{3} b^{3} d +{\left (b^{6} c - a b^{5} d\right )} x^{2} + 2 \,{\left (a b^{5} c - a^{2} b^{4} d\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*(b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b*c - a*d - 2*sqrt(b^2*c - a*
b*d)*sqrt(d*x + c))/(b*x + a)) - 2*(2*b^3*c^2 + a*b^2*c*d - 3*a^2*b*d^2 + 5*(b^3*c*d - a*b^2*d^2)*x)*sqrt(d*x
+ c))/(a^2*b^4*c - a^3*b^3*d + (b^6*c - a*b^5*d)*x^2 + 2*(a*b^5*c - a^2*b^4*d)*x), 1/4*(3*(b^2*d^2*x^2 + 2*a*b
*d^2*x + a^2*d^2)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (2*b^3*c^2 +
a*b^2*c*d - 3*a^2*b*d^2 + 5*(b^3*c*d - a*b^2*d^2)*x)*sqrt(d*x + c))/(a^2*b^4*c - a^3*b^3*d + (b^6*c - a*b^5*d
)*x^2 + 2*(a*b^5*c - a^2*b^4*d)*x)]

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

[Out]

Timed out

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Giac [A]  time = 1.09129, size = 146, normalized size = 1.46 \begin{align*} \frac{3 \, d^{2} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{4 \, \sqrt{-b^{2} c + a b d} b^{2}} - \frac{5 \,{\left (d x + c\right )}^{\frac{3}{2}} b d^{2} - 3 \, \sqrt{d x + c} b c d^{2} + 3 \, \sqrt{d x + c} a d^{3}}{4 \,{\left ({\left (d x + c\right )} b - b c + a d\right )}^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*d^2*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^2) - 1/4*(5*(d*x + c)^(3/2)*b*d^2
- 3*sqrt(d*x + c)*b*c*d^2 + 3*sqrt(d*x + c)*a*d^3)/(((d*x + c)*b - b*c + a*d)^2*b^2)