3.18.88 \(\int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx\) [1788]

Optimal. Leaf size=227 \[ -\frac {2 (b e-a f)^3}{5 f^3 (d e-c f) (e+f x)^{5/2}}+\frac {2 (b e-a f)^2 (2 b d e-3 b c f+a d f)}{3 f^3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )\right )}{f^3 (d e-c f)^3 \sqrt {e+f x}}+\frac {2 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{7/2}} \]

[Out]

-2/5*(-a*f+b*e)^3/f^3/(-c*f+d*e)/(f*x+e)^(5/2)+2/3*(-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)^(3/2)+2*(-a*d+b*c)^3*arctanh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/(-c*f+d*e)^(7/2)/d^(1/2)-2*(-a*f+b*e)*
(a^2*d^2*f^2+a*b*d*f*(-3*c*f+d*e)+b^2*(3*c^2*f^2-3*c*d*e*f+d^2*e^2))/f^3/(-c*f+d*e)^3/(f*x+e)^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 227, 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 = {89, 65, 214} \begin {gather*} -\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (3 c^2 f^2-3 c d e f+d^2 e^2\right )\right )}{f^3 \sqrt {e+f x} (d e-c f)^3}+\frac {2 (b e-a f)^2 (a d f-3 b c f+2 b d e)}{3 f^3 (e+f x)^{3/2} (d e-c f)^2}-\frac {2 (b e-a f)^3}{5 f^3 (e+f x)^{5/2} (d e-c f)}+\frac {2 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3/((c + d*x)*(e + f*x)^(7/2)),x]

[Out]

(-2*(b*e - a*f)^3)/(5*f^3*(d*e - c*f)*(e + f*x)^(5/2)) + (2*(b*e - a*f)^2*(2*b*d*e - 3*b*c*f + a*d*f))/(3*f^3*
(d*e - c*f)^2*(e + f*x)^(3/2)) - (2*(b*e - a*f)*(a^2*d^2*f^2 + a*b*d*f*(d*e - 3*c*f) + b^2*(d^2*e^2 - 3*c*d*e*
f + 3*c^2*f^2)))/(f^3*(d*e - c*f)^3*Sqrt[e + f*x]) + (2*(b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e
- c*f]])/(Sqrt[d]*(d*e - c*f)^(7/2))

Rule 65

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 89

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx &=\int \left (\frac {(-b e+a f)^3}{f^2 (-d e+c f) (e+f x)^{7/2}}+\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)^{5/2}}+\frac {-3 a b^2 c^2 f^3+3 a^2 b c d f^3-a^3 d^2 f^3+b^3 e \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )}{f^2 (d e-c f)^3 (e+f x)^{3/2}}-\frac {(b c-a d)^3}{(d e-c f)^3 (c+d x) \sqrt {e+f x}}\right ) \, dx\\ &=-\frac {2 (b e-a f)^3}{5 f^3 (d e-c f) (e+f x)^{5/2}}+\frac {2 (b e-a f)^2 (2 b d e-3 b c f+a d f)}{3 f^3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )\right )}{f^3 (d e-c f)^3 \sqrt {e+f x}}-\frac {(b c-a d)^3 \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{(d e-c f)^3}\\ &=-\frac {2 (b e-a f)^3}{5 f^3 (d e-c f) (e+f x)^{5/2}}+\frac {2 (b e-a f)^2 (2 b d e-3 b c f+a d f)}{3 f^3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )\right )}{f^3 (d e-c f)^3 \sqrt {e+f x}}-\frac {\left (2 (b c-a d)^3\right ) \text {Subst}\left (\int \frac {1}{c-\frac {d e}{f}+\frac {d x^2}{f}} \, dx,x,\sqrt {e+f x}\right )}{f (d e-c f)^3}\\ &=-\frac {2 (b e-a f)^3}{5 f^3 (d e-c f) (e+f x)^{5/2}}+\frac {2 (b e-a f)^2 (2 b d e-3 b c f+a d f)}{3 f^3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )\right )}{f^3 (d e-c f)^3 \sqrt {e+f x}}+\frac {2 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.63, size = 297, normalized size = 1.31 \begin {gather*} \frac {2 (b e-a f) \left (a^2 f^2 \left (3 c^2 f^2-c d f (11 e+5 f x)+d^2 \left (23 e^2+35 e f x+15 f^2 x^2\right )\right )+b^2 \left (d^2 e^2 \left (8 e^2+20 e f x+15 f^2 x^2\right )+3 c^2 f^2 \left (11 e^2+25 e f x+15 f^2 x^2\right )-c d e f \left (26 e^2+65 e f x+45 f^2 x^2\right )\right )+a b f \left (3 c^2 f^2 (3 e+5 f x)+d^2 e \left (14 e^2+35 e f x+15 f^2 x^2\right )-c d f \left (53 e^2+110 e f x+45 f^2 x^2\right )\right )\right )}{15 f^3 (-d e+c f)^3 (e+f x)^{5/2}}+\frac {2 (b c-a d)^3 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{\sqrt {d} (-d e+c f)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3/((c + d*x)*(e + f*x)^(7/2)),x]

[Out]

(2*(b*e - a*f)*(a^2*f^2*(3*c^2*f^2 - c*d*f*(11*e + 5*f*x) + d^2*(23*e^2 + 35*e*f*x + 15*f^2*x^2)) + b^2*(d^2*e
^2*(8*e^2 + 20*e*f*x + 15*f^2*x^2) + 3*c^2*f^2*(11*e^2 + 25*e*f*x + 15*f^2*x^2) - c*d*e*f*(26*e^2 + 65*e*f*x +
45*f^2*x^2)) + a*b*f*(3*c^2*f^2*(3*e + 5*f*x) + d^2*e*(14*e^2 + 35*e*f*x + 15*f^2*x^2) - c*d*f*(53*e^2 + 110*
e*f*x + 45*f^2*x^2))))/(15*f^3*(-(d*e) + c*f)^3*(e + f*x)^(5/2)) + (2*(b*c - a*d)^3*ArcTan[(Sqrt[d]*Sqrt[e + f
*x])/Sqrt[-(d*e) + c*f]])/(Sqrt[d]*(-(d*e) + c*f)^(7/2))

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Maple [A]
time = 0.10, size = 314, normalized size = 1.38

method	result	size
derivativedivides	\(\frac {-\frac {2 f^{3} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{3} f^{3}-3 a^{2} b e \,f^{2}+3 a \,b^{2} e^{2} f -b^{3} e^{3}\right )}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 \left (-a^{3} d \,f^{3}+3 a^{2} b c \,f^{3}-6 a \,b^{2} c e \,f^{2}+3 a \,b^{2} d \,e^{2} f +3 b^{3} c \,e^{2} f -2 b^{3} d \,e^{3}\right )}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 \left (a^{3} d^{2} f^{3}-3 a^{2} b c d \,f^{3}+3 a \,b^{2} c^{2} f^{3}-3 b^{3} c^{2} e \,f^{2}+3 b^{3} c d \,e^{2} f -b^{3} d^{2} e^{3}\right )}{\left (c f -d e \right )^{3} \sqrt {f x +e}}}{f^{3}}\)	\(314\)
default	\(\frac {-\frac {2 f^{3} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \left (a^{3} f^{3}-3 a^{2} b e \,f^{2}+3 a \,b^{2} e^{2} f -b^{3} e^{3}\right )}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}}}-\frac {2 \left (-a^{3} d \,f^{3}+3 a^{2} b c \,f^{3}-6 a \,b^{2} c e \,f^{2}+3 a \,b^{2} d \,e^{2} f +3 b^{3} c \,e^{2} f -2 b^{3} d \,e^{3}\right )}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 \left (a^{3} d^{2} f^{3}-3 a^{2} b c d \,f^{3}+3 a \,b^{2} c^{2} f^{3}-3 b^{3} c^{2} e \,f^{2}+3 b^{3} c d \,e^{2} f -b^{3} d^{2} e^{3}\right )}{\left (c f -d e \right )^{3} \sqrt {f x +e}}}{f^{3}}\)	\(314\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),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-%e*d>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1012 vs. \(2 (224) = 448\).
time = 1.24, size = 2038, normalized size = 8.98 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),x, algorithm="fricas")

[Out]

[-1/15*(15*((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^6*x^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*e + 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*e^2 + (b^3*c^3 - 3*a
*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^3*e^3)*sqrt(-c*d*f + d^2*e)*log((d*f*x - c*f + 2*d*e - 2*sqrt(-c*d*f +
d^2*e)*sqrt(f*x + e))/(d*x + c)) + 2*(3*a^3*c^3*d*f^6 + 15*(3*a*b^2*c^3*d - 3*a^2*b*c^2*d^2 + a^3*c*d^3)*f^6*
x^2 + 5*(3*a^2*b*c^3*d - a^3*c^2*d^2)*f^6*x + 8*b^3*d^4*e^6 + 2*(10*b^3*d^4*f*x - (17*b^3*c*d^3 - 3*a*b^2*d^4)
*f)*e^5 + (15*b^3*d^4*f^2*x^2 - 5*(17*b^3*c*d^3 - 3*a*b^2*d^4)*f^2*x + (59*b^3*c^2*d^2 - 33*a*b^2*c*d^3 + 9*a^
2*b*d^4)*f^2)*e^4 - (60*b^3*c*d^3*f^3*x^2 - 20*(7*b^3*c^2*d^2 - 3*a*b^2*c*d^3)*f^3*x + (33*b^3*c^3*d - 3*a*b^2
*c^2*d^2 - 33*a^2*b*c*d^3 + 23*a^3*d^4)*f^3)*e^3 + (90*b^3*c^2*d^2*f^4*x^2 - 5*(15*b^3*c^3*d + 3*a*b^2*c^2*d^2
- 21*a^2*b*c*d^3 + 7*a^3*d^4)*f^4*x + 2*(12*a*b^2*c^3*d - 24*a^2*b*c^2*d^2 + 17*a^3*c*d^3)*f^4)*e^2 - (15*(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)*f^5*x^2 - 20*(3*a*b^2*c^3*d - 6*a^2*b*c^2*d^2 + 2*a^3*c
*d^3)*f^5*x - 2*(3*a^2*b*c^3*d - 7*a^3*c^2*d^2)*f^5)*e)*sqrt(f*x + e))/(c^4*d*f^10*x^3 + d^5*f^3*e^7 + (3*d^5*
f^4*x - 4*c*d^4*f^4)*e^6 + 3*(d^5*f^5*x^2 - 4*c*d^4*f^5*x + 2*c^2*d^3*f^5)*e^5 + (d^5*f^6*x^3 - 12*c*d^4*f^6*x
^2 + 18*c^2*d^3*f^6*x - 4*c^3*d^2*f^6)*e^4 - (4*c*d^4*f^7*x^3 - 18*c^2*d^3*f^7*x^2 + 12*c^3*d^2*f^7*x - c^4*d*
f^7)*e^3 + 3*(2*c^2*d^3*f^8*x^3 - 4*c^3*d^2*f^8*x^2 + c^4*d*f^8*x)*e^2 - (4*c^3*d^2*f^9*x^3 - 3*c^4*d*f^9*x^2)
*e), -2/15*(15*((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^6*x^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*e + 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*e^2 + (b^3*c^3 -
3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*f^3*e^3)*sqrt(c*d*f - d^2*e)*arctan(sqrt(c*d*f - d^2*e)*sqrt(f*x + e
)/(d*f*x + d*e)) + (3*a^3*c^3*d*f^6 + 15*(3*a*b^2*c^3*d - 3*a^2*b*c^2*d^2 + a^3*c*d^3)*f^6*x^2 + 5*(3*a^2*b*c^
3*d - a^3*c^2*d^2)*f^6*x + 8*b^3*d^4*e^6 + 2*(10*b^3*d^4*f*x - (17*b^3*c*d^3 - 3*a*b^2*d^4)*f)*e^5 + (15*b^3*d
^4*f^2*x^2 - 5*(17*b^3*c*d^3 - 3*a*b^2*d^4)*f^2*x + (59*b^3*c^2*d^2 - 33*a*b^2*c*d^3 + 9*a^2*b*d^4)*f^2)*e^4 -
(60*b^3*c*d^3*f^3*x^2 - 20*(7*b^3*c^2*d^2 - 3*a*b^2*c*d^3)*f^3*x + (33*b^3*c^3*d - 3*a*b^2*c^2*d^2 - 33*a^2*b
*c*d^3 + 23*a^3*d^4)*f^3)*e^3 + (90*b^3*c^2*d^2*f^4*x^2 - 5*(15*b^3*c^3*d + 3*a*b^2*c^2*d^2 - 21*a^2*b*c*d^3 +
7*a^3*d^4)*f^4*x + 2*(12*a*b^2*c^3*d - 24*a^2*b*c^2*d^2 + 17*a^3*c*d^3)*f^4)*e^2 - (15*(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)*f^5*x^2 - 20*(3*a*b^2*c^3*d - 6*a^2*b*c^2*d^2 + 2*a^3*c*d^3)*f^5*x - 2*(3*
a^2*b*c^3*d - 7*a^3*c^2*d^2)*f^5)*e)*sqrt(f*x + e))/(c^4*d*f^10*x^3 + d^5*f^3*e^7 + (3*d^5*f^4*x - 4*c*d^4*f^4
)*e^6 + 3*(d^5*f^5*x^2 - 4*c*d^4*f^5*x + 2*c^2*d^3*f^5)*e^5 + (d^5*f^6*x^3 - 12*c*d^4*f^6*x^2 + 18*c^2*d^3*f^6
*x - 4*c^3*d^2*f^6)*e^4 - (4*c*d^4*f^7*x^3 - 18*c^2*d^3*f^7*x^2 + 12*c^3*d^2*f^7*x - c^4*d*f^7)*e^3 + 3*(2*c^2
*d^3*f^8*x^3 - 4*c^3*d^2*f^8*x^2 + c^4*d*f^8*x)*e^2 - (4*c^3*d^2*f^9*x^3 - 3*c^4*d*f^9*x^2)*e)]

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Sympy [A]
time = 185.56, size = 230, normalized size = 1.01 \begin {gather*} - \frac {2 \left (a f - b e\right ) \left (a^{2} d^{2} f^{2} - 3 a b c d f^{2} + a b d^{2} e f + 3 b^{2} c^{2} f^{2} - 3 b^{2} c d e f + b^{2} d^{2} e^{2}\right )}{f^{3} \sqrt {e + f x} \left (c f - d e\right )^{3}} + \frac {2 \left (a f - b e\right )^{2} \left (a d f - 3 b c f + 2 b d e\right )}{3 f^{3} \left (e + f x\right )^{\frac {3}{2}} \left (c f - d e\right )^{2}} - \frac {2 \left (a f - b e\right )^{3}}{5 f^{3} \left (e + f x\right )^{\frac {5}{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 \sqrt {\frac {c f - d e}{d}} \left (c f - d e\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3/(d*x+c)/(f*x+e)**(7/2),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 612 vs. \(2 (224) = 448\).
time = 0.66, size = 612, normalized size = 2.70 \begin {gather*} \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^{3} f^{3} - 3 \, c^{2} d f^{2} e + 3 \, c d^{2} f e^{2} - d^{3} e^{3}\right )} \sqrt {c d f - d^{2} e}} - \frac {2 \, {\left (45 \, {\left (f x + e\right )}^{2} a b^{2} c^{2} f^{3} - 45 \, {\left (f x + e\right )}^{2} a^{2} b c d f^{3} + 15 \, {\left (f x + e\right )}^{2} a^{3} d^{2} f^{3} + 15 \, {\left (f x + e\right )} a^{2} b c^{2} f^{4} - 5 \, {\left (f x + e\right )} a^{3} c d f^{4} + 3 \, a^{3} c^{2} f^{5} - 45 \, {\left (f x + e\right )}^{2} b^{3} c^{2} f^{2} e - 30 \, {\left (f x + e\right )} a b^{2} c^{2} f^{3} e - 15 \, {\left (f x + e\right )} a^{2} b c d f^{3} e + 5 \, {\left (f x + e\right )} a^{3} d^{2} f^{3} e - 9 \, a^{2} b c^{2} f^{4} e - 6 \, a^{3} c d f^{4} e + 45 \, {\left (f x + e\right )}^{2} b^{3} c d f e^{2} + 15 \, {\left (f x + e\right )} b^{3} c^{2} f^{2} e^{2} + 45 \, {\left (f x + e\right )} a b^{2} c d f^{2} e^{2} + 9 \, a b^{2} c^{2} f^{3} e^{2} + 18 \, a^{2} b c d f^{3} e^{2} + 3 \, a^{3} d^{2} f^{3} e^{2} - 15 \, {\left (f x + e\right )}^{2} b^{3} d^{2} e^{3} - 25 \, {\left (f x + e\right )} b^{3} c d f e^{3} - 15 \, {\left (f x + e\right )} a b^{2} d^{2} f e^{3} - 3 \, b^{3} c^{2} f^{2} e^{3} - 18 \, a b^{2} c d f^{2} e^{3} - 9 \, a^{2} b d^{2} f^{2} e^{3} + 10 \, {\left (f x + e\right )} b^{3} d^{2} e^{4} + 6 \, b^{3} c d f e^{4} + 9 \, a b^{2} d^{2} f e^{4} - 3 \, b^{3} d^{2} e^{5}\right )}}{15 \, {\left (c^{3} f^{6} - 3 \, c^{2} d f^{5} e + 3 \, c d^{2} f^{4} e^{2} - d^{3} f^{3} e^{3}\right )} {\left (f x + e\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/(d*x+c)/(f*x+e)^(7/2),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^3*f^3 -
3*c^2*d*f^2*e + 3*c*d^2*f*e^2 - d^3*e^3)*sqrt(c*d*f - d^2*e)) - 2/15*(45*(f*x + e)^2*a*b^2*c^2*f^3 - 45*(f*x +
e)^2*a^2*b*c*d*f^3 + 15*(f*x + e)^2*a^3*d^2*f^3 + 15*(f*x + e)*a^2*b*c^2*f^4 - 5*(f*x + e)*a^3*c*d*f^4 + 3*a^
3*c^2*f^5 - 45*(f*x + e)^2*b^3*c^2*f^2*e - 30*(f*x + e)*a*b^2*c^2*f^3*e - 15*(f*x + e)*a^2*b*c*d*f^3*e + 5*(f*
x + e)*a^3*d^2*f^3*e - 9*a^2*b*c^2*f^4*e - 6*a^3*c*d*f^4*e + 45*(f*x + e)^2*b^3*c*d*f*e^2 + 15*(f*x + e)*b^3*c
^2*f^2*e^2 + 45*(f*x + e)*a*b^2*c*d*f^2*e^2 + 9*a*b^2*c^2*f^3*e^2 + 18*a^2*b*c*d*f^3*e^2 + 3*a^3*d^2*f^3*e^2 -
15*(f*x + e)^2*b^3*d^2*e^3 - 25*(f*x + e)*b^3*c*d*f*e^3 - 15*(f*x + e)*a*b^2*d^2*f*e^3 - 3*b^3*c^2*f^2*e^3 -
18*a*b^2*c*d*f^2*e^3 - 9*a^2*b*d^2*f^2*e^3 + 10*(f*x + e)*b^3*d^2*e^4 + 6*b^3*c*d*f*e^4 + 9*a*b^2*d^2*f*e^4 -
3*b^3*d^2*e^5)/((c^3*f^6 - 3*c^2*d*f^5*e + 3*c*d^2*f^4*e^2 - d^3*f^3*e^3)*(f*x + e)^(5/2))

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Mupad [B]
time = 1.48, size = 360, normalized size = 1.59 \begin {gather*} -\frac {\frac {2\,\left (a^3\,f^3-3\,a^2\,b\,e\,f^2+3\,a\,b^2\,e^2\,f-b^3\,e^3\right )}{5\,\left (c\,f-d\,e\right )}+\frac {2\,{\left (e+f\,x\right )}^2\,\left (a^3\,d^2\,f^3-3\,a^2\,b\,c\,d\,f^3+3\,a\,b^2\,c^2\,f^3-3\,b^3\,c^2\,e\,f^2+3\,b^3\,c\,d\,e^2\,f-b^3\,d^2\,e^3\right )}{{\left (c\,f-d\,e\right )}^3}-\frac {2\,\left (e+f\,x\right )\,\left (d\,a^3\,f^3-3\,c\,a^2\,b\,f^3-3\,d\,a\,b^2\,e^2\,f+6\,c\,a\,b^2\,e\,f^2+2\,d\,b^3\,e^3-3\,c\,b^3\,e^2\,f\right )}{3\,{\left (c\,f-d\,e\right )}^2}}{f^3\,{\left (e+f\,x\right )}^{5/2}}-\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^3\,\left (c^3\,f^3-3\,c^2\,d\,e\,f^2+3\,c\,d^2\,e^2\,f-d^3\,e^3\right )}{{\left (c\,f-d\,e\right )}^{7/2}\,\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}{\sqrt {d}\,{\left (c\,f-d\,e\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^3/((e + f*x)^(7/2)*(c + d*x)),x)

[Out]

- ((2*(a^3*f^3 - b^3*e^3 + 3*a*b^2*e^2*f - 3*a^2*b*e*f^2))/(5*(c*f - d*e)) + (2*(e + f*x)^2*(a^3*d^2*f^3 - b^3
*d^2*e^3 + 3*a*b^2*c^2*f^3 - 3*b^3*c^2*e*f^2 - 3*a^2*b*c*d*f^3 + 3*b^3*c*d*e^2*f))/(c*f - d*e)^3 - (2*(e + f*x
)*(a^3*d*f^3 + 2*b^3*d*e^3 - 3*a^2*b*c*f^3 - 3*b^3*c*e^2*f + 6*a*b^2*c*e*f^2 - 3*a*b^2*d*e^2*f))/(3*(c*f - d*e
)^2))/(f^3*(e + f*x)^(5/2)) - (2*atan((2*d^(1/2)*(e + f*x)^(1/2)*(a*d - b*c)^3*(c^3*f^3 - d^3*e^3 + 3*c*d^2*e^
2*f - 3*c^2*d*e*f^2))/((c*f - d*e)^(7/2)*(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^(1/2)*(c*f - d*e)^(7/2))

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