3.1786 \(\int \frac{(a+b x)^3}{(c+d x) (e+f x)^{3/2}} \, dx\)

Optimal. Leaf size=149 \[ -\frac{2 b^2 \sqrt{e+f x} (-3 a d f+b c f+2 b d e)}{d^2 f^3}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}}-\frac{2 (b e-a f)^3}{f^3 \sqrt{e+f x} (d e-c f)}+\frac{2 b^3 (e+f x)^{3/2}}{3 d f^3} \]

[Out]

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

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Rubi [A]  time = 0.214684, antiderivative size = 169, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {87, 43, 63, 208} \[ -\frac{2 b^2 \sqrt{e+f x} (-3 a d f+b c f+b d e)}{d^2 f^3}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}}-\frac{2 (b e-a f)^3}{f^3 \sqrt{e+f x} (d e-c f)}+\frac{2 b^3 (e+f x)^{3/2}}{3 d f^3}-\frac{2 b^3 e \sqrt{e+f x}}{d f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*e - a*f)^3)/(f^3*(d*e - c*f)*Sqrt[e + f*x]) - (2*b^3*e*Sqrt[e + f*x])/(d*f^3) - (2*b^2*(b*d*e + b*c*f -
3*a*d*f)*Sqrt[e + f*x])/(d^2*f^3) + (2*b^3*(e + f*x)^(3/2))/(3*d*f^3) + (2*(b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqr
t[e + f*x])/Sqrt[d*e - c*f]])/(d^(5/2)*(d*e - c*f)^(3/2))

Rule 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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{(a+b x)^3}{(c+d x) (e+f x)^{3/2}} \, dx &=\int \left (\frac{(-b e+a f)^3}{f^2 (-d e+c f) (e+f x)^{3/2}}-\frac{b^2 (b d e+b c f-3 a d f)}{d^2 f^2 \sqrt{e+f x}}+\frac{b^3 x}{d f \sqrt{e+f x}}+\frac{(-b c+a d)^3}{d^2 (d e-c f) (c+d x) \sqrt{e+f x}}\right ) \, dx\\ &=-\frac{2 (b e-a f)^3}{f^3 (d e-c f) \sqrt{e+f x}}-\frac{2 b^2 (b d e+b c f-3 a d f) \sqrt{e+f x}}{d^2 f^3}+\frac{b^3 \int \frac{x}{\sqrt{e+f x}} \, dx}{d f}-\frac{(b c-a d)^3 \int \frac{1}{(c+d x) \sqrt{e+f x}} \, dx}{d^2 (d e-c f)}\\ &=-\frac{2 (b e-a f)^3}{f^3 (d e-c f) \sqrt{e+f x}}-\frac{2 b^2 (b d e+b c f-3 a d f) \sqrt{e+f x}}{d^2 f^3}+\frac{b^3 \int \left (-\frac{e}{f \sqrt{e+f x}}+\frac{\sqrt{e+f x}}{f}\right ) \, dx}{d f}-\frac{\left (2 (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{c-\frac{d e}{f}+\frac{d x^2}{f}} \, dx,x,\sqrt{e+f x}\right )}{d^2 f (d e-c f)}\\ &=-\frac{2 (b e-a f)^3}{f^3 (d e-c f) \sqrt{e+f x}}-\frac{2 b^3 e \sqrt{e+f x}}{d f^3}-\frac{2 b^2 (b d e+b c f-3 a d f) \sqrt{e+f x}}{d^2 f^3}+\frac{2 b^3 (e+f x)^{3/2}}{3 d f^3}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{e+f x}}{\sqrt{d e-c f}}\right )}{d^{5/2} (d e-c f)^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.13299, size = 165, normalized size = 1.11 \[ \frac{2 \left (-\frac{3 b \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{f^3}-\frac{3 b^2 d (e+f x) (-3 a d f+b c f+2 b d e)}{f^3}+\frac{3 (b c-a d)^3 \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d (e+f x)}{d e-c f}\right )}{c f-d e}+\frac{b^3 d^2 (e+f x)^2}{f^3}\right )}{3 d^3 \sqrt{e+f x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((-3*b*(3*a^2*d^2*f^2 - 3*a*b*d*f*(d*e + c*f) + b^2*(d^2*e^2 + c*d*e*f + c^2*f^2)))/f^3 - (3*b^2*d*(2*b*d*e
+ b*c*f - 3*a*d*f)*(e + f*x))/f^3 + (b^3*d^2*(e + f*x)^2)/f^3 + (3*(b*c - a*d)^3*Hypergeometric2F1[-1/2, 1, 1
/2, (d*(e + f*x))/(d*e - c*f)])/(-(d*e) + c*f)))/(3*d^3*Sqrt[e + f*x])

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Maple [B]  time = 0.014, size = 395, normalized size = 2.7 \begin{align*}{\frac{2\,{b}^{3}}{3\,d{f}^{3}} \left ( fx+e \right ) ^{{\frac{3}{2}}}}+6\,{\frac{a{b}^{2}\sqrt{fx+e}}{d{f}^{2}}}-2\,{\frac{{b}^{3}c\sqrt{fx+e}}{{d}^{2}{f}^{2}}}-4\,{\frac{{b}^{3}e\sqrt{fx+e}}{d{f}^{3}}}-2\,{\frac{{a}^{3}}{ \left ( cf-de \right ) \sqrt{fx+e}}}+6\,{\frac{{a}^{2}be}{ \left ( cf-de \right ) f\sqrt{fx+e}}}-6\,{\frac{a{b}^{2}{e}^{2}}{{f}^{2} \left ( cf-de \right ) \sqrt{fx+e}}}+2\,{\frac{{b}^{3}{e}^{3}}{{f}^{3} \left ( cf-de \right ) \sqrt{fx+e}}}-2\,{\frac{d{a}^{3}}{ \left ( cf-de \right ) \sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+6\,{\frac{{a}^{2}bc}{ \left ( cf-de \right ) \sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }-6\,{\frac{a{b}^{2}{c}^{2}}{ \left ( cf-de \right ) d\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) }+2\,{\frac{{b}^{3}{c}^{3}}{ \left ( cf-de \right ){d}^{2}\sqrt{ \left ( cf-de \right ) d}}\arctan \left ({\frac{\sqrt{fx+e}d}{\sqrt{ \left ( cf-de \right ) d}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*b^3*(f*x+e)^(3/2)/d/f^3+6/f^2*b^2/d*a*(f*x+e)^(1/2)-2/f^2*b^3/d^2*c*(f*x+e)^(1/2)-4*b^3*e*(f*x+e)^(1/2)/d/
f^3-2/(c*f-d*e)/(f*x+e)^(1/2)*a^3+6/f/(c*f-d*e)/(f*x+e)^(1/2)*a^2*b*e-6/f^2/(c*f-d*e)/(f*x+e)^(1/2)*a*b^2*e^2+
2/f^3/(c*f-d*e)/(f*x+e)^(1/2)*b^3*e^3-2/(c*f-d*e)*d/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(
1/2))*a^3+6/(c*f-d*e)/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a^2*c*b-6/(c*f-d*e)/d/((
c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*a*b^2*c^2+2/(c*f-d*e)/d^2/((c*f-d*e)*d)^(1/2)*ar
ctan((f*x+e)^(1/2)*d/((c*f-d*e)*d)^(1/2))*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((b*x+a)^3/(d*x+c)/(f*x+e)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.55982, size = 1955, normalized size = 13.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^4*x + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2
- a^3*d^3)*e*f^3)*sqrt(d^2*e - c*d*f)*log((d*f*x + 2*d*e - c*f + 2*sqrt(d^2*e - c*d*f)*sqrt(f*x + e))/(d*x + c
)) - 2*(8*b^3*d^4*e^4 + 3*a^3*c*d^3*f^4 - 2*(5*b^3*c*d^3 + 9*a*b^2*d^4)*e^3*f - (b^3*c^2*d^2 - 27*a*b^2*c*d^3
- 9*a^2*b*d^4)*e^2*f^2 + 3*(b^3*c^3*d - 3*a*b^2*c^2*d^2 - 3*a^2*b*c*d^3 - a^3*d^4)*e*f^3 - (b^3*d^4*e^2*f^2 -
2*b^3*c*d^3*e*f^3 + b^3*c^2*d^2*f^4)*x^2 + (4*b^3*d^4*e^3*f - (5*b^3*c*d^3 + 9*a*b^2*d^4)*e^2*f^2 - 2*(b^3*c^2
*d^2 - 9*a*b^2*c*d^3)*e*f^3 + 3*(b^3*c^3*d - 3*a*b^2*c^2*d^2)*f^4)*x)*sqrt(f*x + e))/(d^5*e^3*f^3 - 2*c*d^4*e^
2*f^4 + c^2*d^3*e*f^5 + (d^5*e^2*f^4 - 2*c*d^4*e*f^5 + c^2*d^3*f^6)*x), -2/3*(3*((b^3*c^3 - 3*a*b^2*c^2*d + 3*
a^2*b*c*d^2 - a^3*d^3)*f^4*x + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*e*f^3)*sqrt(-d^2*e + c*d*f)
*arctan(sqrt(-d^2*e + c*d*f)*sqrt(f*x + e)/(d*f*x + d*e)) + (8*b^3*d^4*e^4 + 3*a^3*c*d^3*f^4 - 2*(5*b^3*c*d^3
+ 9*a*b^2*d^4)*e^3*f - (b^3*c^2*d^2 - 27*a*b^2*c*d^3 - 9*a^2*b*d^4)*e^2*f^2 + 3*(b^3*c^3*d - 3*a*b^2*c^2*d^2 -
3*a^2*b*c*d^3 - a^3*d^4)*e*f^3 - (b^3*d^4*e^2*f^2 - 2*b^3*c*d^3*e*f^3 + b^3*c^2*d^2*f^4)*x^2 + (4*b^3*d^4*e^3
*f - (5*b^3*c*d^3 + 9*a*b^2*d^4)*e^2*f^2 - 2*(b^3*c^2*d^2 - 9*a*b^2*c*d^3)*e*f^3 + 3*(b^3*c^3*d - 3*a*b^2*c^2*
d^2)*f^4)*x)*sqrt(f*x + e))/(d^5*e^3*f^3 - 2*c*d^4*e^2*f^4 + c^2*d^3*e*f^5 + (d^5*e^2*f^4 - 2*c*d^4*e*f^5 + c^
2*d^3*f^6)*x)]

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Sympy [A]  time = 45.2917, size = 144, normalized size = 0.97 \begin{align*} \frac{2 b^{3} \left (e + f x\right )^{\frac{3}{2}}}{3 d f^{3}} - \frac{2 \left (a f - b e\right )^{3}}{f^{3} \sqrt{e + f x} \left (c f - d e\right )} + \frac{\sqrt{e + f x} \left (6 a b^{2} d f - 2 b^{3} c f - 4 b^{3} d e\right )}{d^{2} f^{3}} - \frac{2 \left (a d - b c\right )^{3} \operatorname{atan}{\left (\frac{\sqrt{e + f x}}{\sqrt{\frac{c f - d e}{d}}} \right )}}{d^{3} \sqrt{\frac{c f - d e}{d}} \left (c f - d e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b**3*(e + f*x)**(3/2)/(3*d*f**3) - 2*(a*f - b*e)**3/(f**3*sqrt(e + f*x)*(c*f - d*e)) + sqrt(e + f*x)*(6*a*b*
*2*d*f - 2*b**3*c*f - 4*b**3*d*e)/(d**2*f**3) - 2*(a*d - b*c)**3*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))/(d**3
*sqrt((c*f - d*e)/d)*(c*f - d*e))

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Giac [A]  time = 2.23375, size = 325, normalized size = 2.18 \begin{align*} \frac{2 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac{\sqrt{f x + e} d}{\sqrt{c d f - d^{2} e}}\right )}{{\left (c d^{2} f - d^{3} e\right )} \sqrt{c d f - d^{2} e}} - \frac{2 \,{\left (a^{3} f^{3} - 3 \, a^{2} b f^{2} e + 3 \, a b^{2} f e^{2} - b^{3} e^{3}\right )}}{{\left (c f^{4} - d f^{3} e\right )} \sqrt{f x + e}} + \frac{2 \,{\left ({\left (f x + e\right )}^{\frac{3}{2}} b^{3} d^{2} f^{6} - 3 \, \sqrt{f x + e} b^{3} c d f^{7} + 9 \, \sqrt{f x + e} a b^{2} d^{2} f^{7} - 6 \, \sqrt{f x + e} b^{3} d^{2} f^{6} e\right )}}{3 \, d^{3} f^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/((c*d^2*f -
d^3*e)*sqrt(c*d*f - d^2*e)) - 2*(a^3*f^3 - 3*a^2*b*f^2*e + 3*a*b^2*f*e^2 - b^3*e^3)/((c*f^4 - d*f^3*e)*sqrt(f*
x + e)) + 2/3*((f*x + e)^(3/2)*b^3*d^2*f^6 - 3*sqrt(f*x + e)*b^3*c*d*f^7 + 9*sqrt(f*x + e)*a*b^2*d^2*f^7 - 6*s
qrt(f*x + e)*b^3*d^2*f^6*e)/(d^3*f^9)