3.1787 \(\int \frac {(a+b x)^3}{(c+d x) (e+f x)^{5/2}} \, dx\)

Optimal. Leaf size=163 \[ \frac {2 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2} (d e-c f)^{5/2}}+\frac {2 (b e-a f)^2 (a d f-3 b c f+2 b d e)}{f^3 \sqrt {e+f x} (d e-c f)^2}-\frac {2 (b e-a f)^3}{3 f^3 (e+f x)^{3/2} (d e-c f)}+\frac {2 b^3 \sqrt {e+f x}}{d f^3} \]

[Out]

-2/3*(-a*f+b*e)^3/f^3/(-c*f+d*e)/(f*x+e)^(3/2)+2*(-a*d+b*c)^3*arctanh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/
d^(3/2)/(-c*f+d*e)^(5/2)+2*(-a*f+b*e)^2*(a*d*f-3*b*c*f+2*b*d*e)/f^3/(-c*f+d*e)^2/(f*x+e)^(1/2)+2*b^3*(f*x+e)^(
1/2)/d/f^3

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Rubi [A]  time = 0.21, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {87, 63, 208} \[ \frac {2 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{3/2} (d e-c f)^{5/2}}+\frac {2 (b e-a f)^2 (a d f-3 b c f+2 b d e)}{f^3 \sqrt {e+f x} (d e-c f)^2}-\frac {2 (b e-a f)^3}{3 f^3 (e+f x)^{3/2} (d e-c f)}+\frac {2 b^3 \sqrt {e+f x}}{d f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.18, size = 165, normalized size = 1.01 \[ \frac {2 \left (-\frac {b \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{f^3}+\frac {3 b^2 d (e+f x) (-3 a d f+b c f+2 b d e)}{f^3}+\frac {(b c-a d)^3 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {d (e+f x)}{d e-c f}\right )}{c f-d e}+\frac {3 b^3 d^2 (e+f x)^2}{f^3}\right )}{3 d^3 (e+f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.92, size = 1435, normalized size = 8.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(5/2),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^5*x^2 + 2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c
*d^2 - a^3*d^3)*e*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^2*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^5 + a^3*c^2*d^2*f^5 -
2*(11*b^3*c*d^3 + 3*a*b^2*d^4)*e^4*f + (17*b^3*c^2*d^2 + 21*a*b^2*c*d^3 - 3*a^2*b*d^4)*e^3*f^2 - (3*b^3*c^3*d
+ 15*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - 4*a^3*d^4)*e^2*f^3 + (6*a^2*b*c^2*d^2 - 5*a^3*c*d^3)*e*f^4 + 3*(b^3*d^4*
e^3*f^2 - 3*b^3*c*d^3*e^2*f^3 + 3*b^3*c^2*d^2*e*f^4 - b^3*c^3*d*f^5)*x^2 + 3*(4*b^3*d^4*e^4*f - (11*b^3*c*d^3
+ 3*a*b^2*d^4)*e^3*f^2 + 9*(b^3*c^2*d^2 + a*b^2*c*d^3)*e^2*f^3 - (2*b^3*c^3*d + 6*a*b^2*c^2*d^2 + 3*a^2*b*c*d^
3 - a^3*d^4)*e*f^4 + (3*a^2*b*c^2*d^2 - a^3*c*d^3)*f^5)*x)*sqrt(f*x + e))/(d^5*e^5*f^3 - 3*c*d^4*e^4*f^4 + 3*c
^2*d^3*e^3*f^5 - c^3*d^2*e^2*f^6 + (d^5*e^3*f^5 - 3*c*d^4*e^2*f^6 + 3*c^2*d^3*e*f^7 - c^3*d^2*f^8)*x^2 + 2*(d^
5*e^4*f^4 - 3*c*d^4*e^3*f^5 + 3*c^2*d^3*e^2*f^6 - c^3*d^2*e*f^7)*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^5*x^2 + 2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*e*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^2*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^5 + a^3*c^2*d^2*f^5 - 2*(11*b^3*c*d^3 + 3*a*b^2*d^4)*e^4*f + (17*b^3*c^2*d^2 + 21*
a*b^2*c*d^3 - 3*a^2*b*d^4)*e^3*f^2 - (3*b^3*c^3*d + 15*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - 4*a^3*d^4)*e^2*f^3 + (6
*a^2*b*c^2*d^2 - 5*a^3*c*d^3)*e*f^4 + 3*(b^3*d^4*e^3*f^2 - 3*b^3*c*d^3*e^2*f^3 + 3*b^3*c^2*d^2*e*f^4 - b^3*c^3
*d*f^5)*x^2 + 3*(4*b^3*d^4*e^4*f - (11*b^3*c*d^3 + 3*a*b^2*d^4)*e^3*f^2 + 9*(b^3*c^2*d^2 + a*b^2*c*d^3)*e^2*f^
3 - (2*b^3*c^3*d + 6*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - a^3*d^4)*e*f^4 + (3*a^2*b*c^2*d^2 - a^3*c*d^3)*f^5)*x)*sq
rt(f*x + e))/(d^5*e^5*f^3 - 3*c*d^4*e^4*f^4 + 3*c^2*d^3*e^3*f^5 - c^3*d^2*e^2*f^6 + (d^5*e^3*f^5 - 3*c*d^4*e^2
*f^6 + 3*c^2*d^3*e*f^7 - c^3*d^2*f^8)*x^2 + 2*(d^5*e^4*f^4 - 3*c*d^4*e^3*f^5 + 3*c^2*d^3*e^2*f^6 - c^3*d^2*e*f
^7)*x)]

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giac [B]  time = 1.34, size = 337, normalized size = 2.07 \[ -\frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{{\left (c^{2} d f^{2} - 2 \, c d^{2} f e + d^{3} e^{2}\right )} \sqrt {c d f - d^{2} e}} + \frac {2 \, \sqrt {f x + e} b^{3}}{d f^{3}} - \frac {2 \, {\left (9 \, {\left (f x + e\right )} a^{2} b c f^{3} - 3 \, {\left (f x + e\right )} a^{3} d f^{3} + a^{3} c f^{4} - 18 \, {\left (f x + e\right )} a b^{2} c f^{2} e - 3 \, a^{2} b c f^{3} e - a^{3} d f^{3} e + 9 \, {\left (f x + e\right )} b^{3} c f e^{2} + 9 \, {\left (f x + e\right )} a b^{2} d f e^{2} + 3 \, a b^{2} c f^{2} e^{2} + 3 \, a^{2} b d f^{2} e^{2} - 6 \, {\left (f x + e\right )} b^{3} d e^{3} - b^{3} c f e^{3} - 3 \, a b^{2} d f e^{3} + b^{3} d e^{4}\right )}}{3 \, {\left (c^{2} f^{5} - 2 \, c d f^{4} e + d^{2} f^{3} e^{2}\right )} {\left (f x + e\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(5/2),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^2*d*f^2
- 2*c*d^2*f*e + d^3*e^2)*sqrt(c*d*f - d^2*e)) + 2*sqrt(f*x + e)*b^3/(d*f^3) - 2/3*(9*(f*x + e)*a^2*b*c*f^3 -
3*(f*x + e)*a^3*d*f^3 + a^3*c*f^4 - 18*(f*x + e)*a*b^2*c*f^2*e - 3*a^2*b*c*f^3*e - a^3*d*f^3*e + 9*(f*x + e)*b
^3*c*f*e^2 + 9*(f*x + e)*a*b^2*d*f*e^2 + 3*a*b^2*c*f^2*e^2 + 3*a^2*b*d*f^2*e^2 - 6*(f*x + e)*b^3*d*e^3 - b^3*c
*f*e^3 - 3*a*b^2*d*f*e^3 + b^3*d*e^4)/((c^2*f^5 - 2*c*d*f^4*e + d^2*f^3*e^2)*(f*x + e)^(3/2))

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maple [B]  time = 0.02, size = 501, normalized size = 3.07 \[ \frac {2 a^{3} d^{2} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{2} \sqrt {\left (c f -d e \right ) d}}-\frac {6 a^{2} b c d \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{2} \sqrt {\left (c f -d e \right ) d}}+\frac {6 a \,b^{2} c^{2} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{2} \sqrt {\left (c f -d e \right ) d}}-\frac {2 b^{3} c^{3} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{2} \sqrt {\left (c f -d e \right ) d}\, d}+\frac {2 a^{3} d}{\left (c f -d e \right )^{2} \sqrt {f x +e}}-\frac {6 a^{2} b c}{\left (c f -d e \right )^{2} \sqrt {f x +e}}+\frac {12 a \,b^{2} c e}{\left (c f -d e \right )^{2} \sqrt {f x +e}\, f}-\frac {6 a \,b^{2} d \,e^{2}}{\left (c f -d e \right )^{2} \sqrt {f x +e}\, f^{2}}-\frac {6 b^{3} c \,e^{2}}{\left (c f -d e \right )^{2} \sqrt {f x +e}\, f^{2}}+\frac {4 b^{3} d \,e^{3}}{\left (c f -d e \right )^{2} \sqrt {f x +e}\, f^{3}}-\frac {2 a^{3}}{3 \left (c f -d e \right ) \left (f x +e \right )^{\frac {3}{2}}}+\frac {2 a^{2} b e}{\left (c f -d e \right ) \left (f x +e \right )^{\frac {3}{2}} f}-\frac {2 a \,b^{2} e^{2}}{\left (c f -d e \right ) \left (f x +e \right )^{\frac {3}{2}} f^{2}}+\frac {2 b^{3} e^{3}}{3 \left (c f -d e \right ) \left (f x +e \right )^{\frac {3}{2}} f^{3}}+\frac {2 \sqrt {f x +e}\, b^{3}}{d \,f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^3*(f*x+e)^(1/2)/d/f^3-2/3/(c*f-d*e)/(f*x+e)^(3/2)*a^3+2/f/(c*f-d*e)/(f*x+e)^(3/2)*a^2*b*e-2/f^2/(c*f-d*e)/
(f*x+e)^(3/2)*a*b^2*e^2+2/3/f^3/(c*f-d*e)/(f*x+e)^(3/2)*b^3*e^3+2/(c*f-d*e)^2/(f*x+e)^(1/2)*a^3*d-6/(c*f-d*e)^
2/(f*x+e)^(1/2)*a^2*b*c+12/f/(c*f-d*e)^2/(f*x+e)^(1/2)*a*b^2*c*e-6/f^2/(c*f-d*e)^2/(f*x+e)^(1/2)*a*b^2*d*e^2-6
/f^2/(c*f-d*e)^2/(f*x+e)^(1/2)*b^3*c*e^2+4/f^3/(c*f-d*e)^2/(f*x+e)^(1/2)*b^3*d*e^3+2*d^2/(c*f-d*e)^2/((c*f-d*e
)*d)^(1/2)*arctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*a^3-6*d/(c*f-d*e)^2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^
(1/2)/((c*f-d*e)*d)^(1/2)*d)*a^2*b*c+6/(c*f-d*e)^2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2
)*d)*a*b^2*c^2-2/d/(c*f-d*e)^2/((c*f-d*e)*d)^(1/2)*arctan((f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2)*d)*b^3*c^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c*f-d*e>0)', see `assume?` for
more details)Is c*f-d*e positive or negative?

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mupad [B]  time = 0.21, size = 295, normalized size = 1.81 \[ \frac {2\,b^3\,\sqrt {e+f\,x}}{d\,f^3}-\frac {\frac {2\,\left (d\,a^3\,f^3-3\,d\,a^2\,b\,e\,f^2+3\,d\,a\,b^2\,e^2\,f-d\,b^3\,e^3\right )}{3\,\left (c\,f-d\,e\right )}-\frac {2\,\left (e+f\,x\right )\,\left (a^3\,d^2\,f^3-3\,c\,a^2\,b\,d\,f^3-3\,a\,b^2\,d^2\,e^2\,f+6\,c\,a\,b^2\,d\,e\,f^2+2\,b^3\,d^2\,e^3-3\,c\,b^3\,d\,e^2\,f\right )}{{\left (c\,f-d\,e\right )}^2}}{d\,f^3\,{\left (e+f\,x\right )}^{3/2}}+\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^3\,\left (c^2\,d\,f^2-2\,c\,d^2\,e\,f+d^3\,e^2\right )}{\sqrt {d}\,{\left (c\,f-d\,e\right )}^{5/2}\,\left (2\,a^3\,d^3-6\,a^2\,b\,c\,d^2+6\,a\,b^2\,c^2\,d-2\,b^3\,c^3\right )}\right )\,{\left (a\,d-b\,c\right )}^3}{d^{3/2}\,{\left (c\,f-d\,e\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^3*(e + f*x)^(1/2))/(d*f^3) - ((2*(a^3*d*f^3 - b^3*d*e^3 + 3*a*b^2*d*e^2*f - 3*a^2*b*d*e*f^2))/(3*(c*f - d
*e)) - (2*(e + f*x)*(a^3*d^2*f^3 + 2*b^3*d^2*e^3 - 3*a^2*b*c*d*f^3 - 3*b^3*c*d*e^2*f - 3*a*b^2*d^2*e^2*f + 6*a
*b^2*c*d*e*f^2))/(c*f - d*e)^2)/(d*f^3*(e + f*x)^(3/2)) + (2*atan((2*(e + f*x)^(1/2)*(a*d - b*c)^3*(d^3*e^2 +
c^2*d*f^2 - 2*c*d^2*e*f))/(d^(1/2)*(c*f - d*e)^(5/2)*(2*a^3*d^3 - 2*b^3*c^3 + 6*a*b^2*c^2*d - 6*a^2*b*c*d^2)))
*(a*d - b*c)^3)/(d^(3/2)*(c*f - d*e)^(5/2))

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sympy [A]  time = 154.33, size = 153, normalized size = 0.94 \[ \frac {2 b^{3} \sqrt {e + f x}}{d f^{3}} + \frac {2 \left (a f - b e\right )^{2} \left (a d f - 3 b c f + 2 b d e\right )}{f^{3} \sqrt {e + f x} \left (c f - d e\right )^{2}} - \frac {2 \left (a f - b e\right )^{3}}{3 f^{3} \left (e + f x\right )^{\frac {3}{2}} \left (c f - d e\right )} + \frac {2 \left (a d - b c\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{2} \sqrt {\frac {c f - d e}{d}} \left (c f - d e\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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