3.1394 \(\int \frac{(c+d x)^{3/2}}{(a+b x)^2} \, dx\)
Optimal. Leaf size=85 \[ -\frac{3 d \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{5/2}}-\frac{(c+d x)^{3/2}}{b (a+b x)}+\frac{3 d \sqrt{c+d x}}{b^2} \]
[Out]
(3*d*Sqrt[c + d*x])/b^2 - (c + d*x)^(3/2)/(b*(a + b*x)) - (3*d*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])
/Sqrt[b*c - a*d]])/b^(5/2)
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Rubi [A] time = 0.0381649, antiderivative size = 85, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used =
{47, 50, 63, 208} \[ -\frac{3 d \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{5/2}}-\frac{(c+d x)^{3/2}}{b (a+b x)}+\frac{3 d \sqrt{c+d x}}{b^2} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*x)^(3/2)/(a + b*x)^2,x]
[Out]
(3*d*Sqrt[c + d*x])/b^2 - (c + d*x)^(3/2)/(b*(a + b*x)) - (3*d*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])
/Sqrt[b*c - a*d]])/b^(5/2)
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 50
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 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)^2} \, dx &=-\frac{(c+d x)^{3/2}}{b (a+b x)}+\frac{(3 d) \int \frac{\sqrt{c+d x}}{a+b x} \, dx}{2 b}\\ &=\frac{3 d \sqrt{c+d x}}{b^2}-\frac{(c+d x)^{3/2}}{b (a+b x)}+\frac{(3 d (b c-a d)) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{2 b^2}\\ &=\frac{3 d \sqrt{c+d x}}{b^2}-\frac{(c+d x)^{3/2}}{b (a+b x)}+\frac{(3 (b c-a d)) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{b^2}\\ &=\frac{3 d \sqrt{c+d x}}{b^2}-\frac{(c+d x)^{3/2}}{b (a+b x)}-\frac{3 d \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{5/2}}\\ \end{align*}
Mathematica [C] time = 0.0135204, size = 50, normalized size = 0.59 \[ \frac{2 d (c+d x)^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};-\frac{b (c+d x)}{a d-b c}\right )}{5 (a d-b c)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*x)^(3/2)/(a + b*x)^2,x]
[Out]
(2*d*(c + d*x)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(5*(-(b*c) + a*d)^2)
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Maple [B] time = 0.011, size = 148, normalized size = 1.7 \begin{align*} 2\,{\frac{d\sqrt{dx+c}}{{b}^{2}}}+{\frac{a{d}^{2}}{{b}^{2} \left ( bdx+ad \right ) }\sqrt{dx+c}}-{\frac{dc}{b \left ( bdx+ad \right ) }\sqrt{dx+c}}-3\,{\frac{a{d}^{2}}{{b}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+3\,{\frac{dc}{b\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*x+c)^(3/2)/(b*x+a)^2,x)
[Out]
2*d*(d*x+c)^(1/2)/b^2+1/b^2*(d*x+c)^(1/2)/(b*d*x+a*d)*a*d^2-d/b*(d*x+c)^(1/2)/(b*d*x+a*d)*c-3/b^2/((a*d-b*c)*b
)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*a*d^2+3*d/b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a
*d-b*c)*b)^(1/2))*c
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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)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.8096, size = 455, normalized size = 5.35 \begin{align*} \left [\frac{3 \,{\left (b d x + a d\right )} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x + 2 \, b c - a d - 2 \, \sqrt{d x + c} b \sqrt{\frac{b c - a d}{b}}}{b x + a}\right ) + 2 \,{\left (2 \, b d x - b c + 3 \, a d\right )} \sqrt{d x + c}}{2 \,{\left (b^{3} x + a b^{2}\right )}}, -\frac{3 \,{\left (b d x + a d\right )} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) -{\left (2 \, b d x - b c + 3 \, a d\right )} \sqrt{d x + c}}{b^{3} x + a b^{2}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^(3/2)/(b*x+a)^2,x, algorithm="fricas")
[Out]
[1/2*(3*(b*d*x + a*d)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d - 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b
*x + a)) + 2*(2*b*d*x - b*c + 3*a*d)*sqrt(d*x + c))/(b^3*x + a*b^2), -(3*(b*d*x + a*d)*sqrt(-(b*c - a*d)/b)*ar
ctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (2*b*d*x - b*c + 3*a*d)*sqrt(d*x + c))/(b^3*x + a*b^
2)]
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Sympy [B] time = 83.9816, size = 923, normalized size = 10.86 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)**(3/2)/(b*x+a)**2,x)
[Out]
2*a**2*d**3*sqrt(c + d*x)/(2*a**2*b**2*d**2 - 2*a*b**3*c*d + 2*a*b**3*d**2*x - 2*b**4*c*d*x) - a**2*d**3*sqrt(
-1/(b*(a*d - b*c)**3))*log(-a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) + 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) - b*
*2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/(2*b**2) + a**2*d**3*sqrt(-1/(b*(a*d - b*c)**3))*log(a**2
*d**2*sqrt(-1/(b*(a*d - b*c)**3)) - 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) + b**2*c**2*sqrt(-1/(b*(a*d - b*c)**
3)) + sqrt(c + d*x))/(2*b**2) - 4*a*c*d**2*sqrt(c + d*x)/(2*a**2*b*d**2 - 2*a*b**2*c*d + 2*a*b**2*d**2*x - 2*b
**3*c*d*x) + a*c*d**2*sqrt(-1/(b*(a*d - b*c)**3))*log(-a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) + 2*a*b*c*d*sqrt(
-1/(b*(a*d - b*c)**3)) - b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/b - a*c*d**2*sqrt(-1/(b*(a*d -
b*c)**3))*log(a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) - 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) + b**2*c**2*sqrt(
-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/b - 4*a*d**2*atan(sqrt(c + d*x)/sqrt(a*d/b - c))/(b**3*sqrt(a*d/b - c)
) - c**2*d*sqrt(-1/(b*(a*d - b*c)**3))*log(-a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) + 2*a*b*c*d*sqrt(-1/(b*(a*d
- b*c)**3)) - b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/2 + c**2*d*sqrt(-1/(b*(a*d - b*c)**3))*lo
g(a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) - 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) + b**2*c**2*sqrt(-1/(b*(a*d -
b*c)**3)) + sqrt(c + d*x))/2 + 2*c**2*d*sqrt(c + d*x)/(2*a**2*d**2 - 2*a*b*c*d + 2*a*b*d**2*x - 2*b**2*c*d*x)
+ 4*c*d*atan(sqrt(c + d*x)/sqrt(a*d/b - c))/(b**2*sqrt(a*d/b - c)) + 2*d*sqrt(c + d*x)/b**2
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Giac [A] time = 1.08262, size = 153, normalized size = 1.8 \begin{align*} \frac{2 \, \sqrt{d x + c} d}{b^{2}} + \frac{3 \,{\left (b c d - a d^{2}\right )} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} b^{2}} - \frac{\sqrt{d x + c} b c d - \sqrt{d x + c} a d^{2}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*x+c)^(3/2)/(b*x+a)^2,x, algorithm="giac")
[Out]
2*sqrt(d*x + c)*d/b^2 + 3*(b*c*d - a*d^2)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b
^2) - (sqrt(d*x + c)*b*c*d - sqrt(d*x + c)*a*d^2)/(((d*x + c)*b - b*c + a*d)*b^2)