3.18.66 \(\int \frac {(a+b x) (e+f x)^{5/2}}{c+d x} \, dx\) [1766]

Optimal. Leaf size=164 \[ -\frac {2 (b c-a d) (d e-c f)^2 \sqrt {e+f x}}{d^4}-\frac {2 (b c-a d) (d e-c f) (e+f x)^{3/2}}{3 d^3}-\frac {2 (b c-a d) (e+f x)^{5/2}}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f}+\frac {2 (b c-a d) (d e-c f)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{9/2}} \]

[Out]

-2/3*(-a*d+b*c)*(-c*f+d*e)*(f*x+e)^(3/2)/d^3-2/5*(-a*d+b*c)*(f*x+e)^(5/2)/d^2+2/7*b*(f*x+e)^(7/2)/d/f+2*(-a*d+
b*c)*(-c*f+d*e)^(5/2)*arctanh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/d^(9/2)-2*(-a*d+b*c)*(-c*f+d*e)^2*(f*x+e
)^(1/2)/d^4

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {81, 52, 65, 214} \begin {gather*} \frac {2 (b c-a d) (d e-c f)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{9/2}}-\frac {2 \sqrt {e+f x} (b c-a d) (d e-c f)^2}{d^4}-\frac {2 (e+f x)^{3/2} (b c-a d) (d e-c f)}{3 d^3}-\frac {2 (e+f x)^{5/2} (b c-a d)}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*c - a*d)*(d*e - c*f)^2*Sqrt[e + f*x])/d^4 - (2*(b*c - a*d)*(d*e - c*f)*(e + f*x)^(3/2))/(3*d^3) - (2*(b
*c - a*d)*(e + f*x)^(5/2))/(5*d^2) + (2*b*(e + f*x)^(7/2))/(7*d*f) + (2*(b*c - a*d)*(d*e - c*f)^(5/2)*ArcTanh[
(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e - c*f]])/d^(9/2)

Rule 52

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 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 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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) (e+f x)^{5/2}}{c+d x} \, dx &=\frac {2 b (e+f x)^{7/2}}{7 d f}+\frac {\left (2 \left (-\frac {7}{2} b c f+\frac {7 a d f}{2}\right )\right ) \int \frac {(e+f x)^{5/2}}{c+d x} \, dx}{7 d f}\\ &=-\frac {2 (b c-a d) (e+f x)^{5/2}}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f}-\frac {((b c-a d) (d e-c f)) \int \frac {(e+f x)^{3/2}}{c+d x} \, dx}{d^2}\\ &=-\frac {2 (b c-a d) (d e-c f) (e+f x)^{3/2}}{3 d^3}-\frac {2 (b c-a d) (e+f x)^{5/2}}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f}-\frac {\left ((b c-a d) (d e-c f)^2\right ) \int \frac {\sqrt {e+f x}}{c+d x} \, dx}{d^3}\\ &=-\frac {2 (b c-a d) (d e-c f)^2 \sqrt {e+f x}}{d^4}-\frac {2 (b c-a d) (d e-c f) (e+f x)^{3/2}}{3 d^3}-\frac {2 (b c-a d) (e+f x)^{5/2}}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f}-\frac {\left ((b c-a d) (d e-c f)^3\right ) \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{d^4}\\ &=-\frac {2 (b c-a d) (d e-c f)^2 \sqrt {e+f x}}{d^4}-\frac {2 (b c-a d) (d e-c f) (e+f x)^{3/2}}{3 d^3}-\frac {2 (b c-a d) (e+f x)^{5/2}}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f}-\frac {\left (2 (b c-a d) (d e-c f)^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 )}{d^4 f}\\ &=-\frac {2 (b c-a d) (d e-c f)^2 \sqrt {e+f x}}{d^4}-\frac {2 (b c-a d) (d e-c f) (e+f x)^{3/2}}{3 d^3}-\frac {2 (b c-a d) (e+f x)^{5/2}}{5 d^2}+\frac {2 b (e+f x)^{7/2}}{7 d f}+\frac {2 (b c-a d) (d e-c f)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{9/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.29, size = 190, normalized size = 1.16 \begin {gather*} \frac {2 \sqrt {e+f x} \left (7 a d f \left (15 c^2 f^2-5 c d f (7 e+f x)+d^2 \left (23 e^2+11 e f x+3 f^2 x^2\right )\right )+b \left (-105 c^3 f^3+15 d^3 (e+f x)^3+35 c^2 d f^2 (7 e+f x)-7 c d^2 f \left (23 e^2+11 e f x+3 f^2 x^2\right )\right )\right )}{105 d^4 f}-\frac {2 (-b c+a d) (-d e+c f)^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e + f*x]*(7*a*d*f*(15*c^2*f^2 - 5*c*d*f*(7*e + f*x) + d^2*(23*e^2 + 11*e*f*x + 3*f^2*x^2)) + b*(-105*c
^3*f^3 + 15*d^3*(e + f*x)^3 + 35*c^2*d*f^2*(7*e + f*x) - 7*c*d^2*f*(23*e^2 + 11*e*f*x + 3*f^2*x^2))))/(105*d^4
*f) - (2*(-(b*c) + a*d)*(-(d*e) + c*f)^(5/2)*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/d^(9/2)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(345\) vs. \(2(140)=280\).
time = 0.10, size = 346, normalized size = 2.11

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\frac {b \left (f x +e \right )^{\frac {7}{2}} d^{3}}{7}+\frac {a \,d^{3} f \left (f x +e \right )^{\frac {5}{2}}}{5}-\frac {b c \,d^{2} f \left (f x +e \right )^{\frac {5}{2}}}{5}-\frac {a c \,d^{2} f^{2} \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {a \,d^{3} e f \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b \,c^{2} d \,f^{2} \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b c \,d^{2} e f \left (f x +e \right )^{\frac {3}{2}}}{3}+a \,c^{2} d \,f^{3} \sqrt {f x +e}-2 a c \,d^{2} e \,f^{2} \sqrt {f x +e}+a \,d^{3} e^{2} f \sqrt {f x +e}-b \,c^{3} f^{3} \sqrt {f x +e}+2 b \,c^{2} d e \,f^{2} \sqrt {f x +e}-b c \,d^{2} e^{2} f \sqrt {f x +e}\right )}{d^{4}}-\frac {2 f \left (a \,c^{3} d \,f^{3}-3 a \,c^{2} d^{2} e \,f^{2}+3 a c \,d^{3} e^{2} f -a \,d^{4} e^{3}-b \,c^{4} f^{3}+3 b \,c^{3} d e \,f^{2}-3 b \,c^{2} d^{2} e^{2} f +b c \,d^{3} e^{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{4} \sqrt {\left (c f -d e \right ) d}}}{f}\)	\(346\)
default	\(\frac {\frac {2 \left (\frac {b \left (f x +e \right )^{\frac {7}{2}} d^{3}}{7}+\frac {a \,d^{3} f \left (f x +e \right )^{\frac {5}{2}}}{5}-\frac {b c \,d^{2} f \left (f x +e \right )^{\frac {5}{2}}}{5}-\frac {a c \,d^{2} f^{2} \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {a \,d^{3} e f \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b \,c^{2} d \,f^{2} \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b c \,d^{2} e f \left (f x +e \right )^{\frac {3}{2}}}{3}+a \,c^{2} d \,f^{3} \sqrt {f x +e}-2 a c \,d^{2} e \,f^{2} \sqrt {f x +e}+a \,d^{3} e^{2} f \sqrt {f x +e}-b \,c^{3} f^{3} \sqrt {f x +e}+2 b \,c^{2} d e \,f^{2} \sqrt {f x +e}-b c \,d^{2} e^{2} f \sqrt {f x +e}\right )}{d^{4}}-\frac {2 f \left (a \,c^{3} d \,f^{3}-3 a \,c^{2} d^{2} e \,f^{2}+3 a c \,d^{3} e^{2} f -a \,d^{4} e^{3}-b \,c^{4} f^{3}+3 b \,c^{3} d e \,f^{2}-3 b \,c^{2} d^{2} e^{2} f +b c \,d^{3} e^{3}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{4} \sqrt {\left (c f -d e \right ) d}}}{f}\)	\(346\)
risch	\(\frac {2 \left (15 b \,f^{3} d^{3} x^{3}+21 a \,d^{3} f^{3} x^{2}-21 b c \,d^{2} f^{3} x^{2}+45 b \,d^{3} e \,f^{2} x^{2}-35 a c \,d^{2} f^{3} x +77 a \,d^{3} e \,f^{2} x +35 b \,c^{2} d \,f^{3} x -77 b c \,d^{2} e \,f^{2} x +45 b \,d^{3} e^{2} f x +105 a \,c^{2} d \,f^{3}-245 a c \,d^{2} e \,f^{2}+161 a \,d^{3} e^{2} f -105 b \,c^{3} f^{3}+245 b \,c^{2} d e \,f^{2}-161 b c \,d^{2} e^{2} f +15 b \,d^{3} e^{3}\right ) \sqrt {f x +e}}{105 f \,d^{4}}-\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a \,c^{3} f^{3}}{d^{3} \sqrt {\left (c f -d e \right ) d}}+\frac {6 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a \,c^{2} e \,f^{2}}{d^{2} \sqrt {\left (c f -d e \right ) d}}-\frac {6 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a c \,e^{2} f}{d \sqrt {\left (c f -d e \right ) d}}+\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a \,e^{3}}{\sqrt {\left (c f -d e \right ) d}}+\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) b \,c^{4} f^{3}}{d^{4} \sqrt {\left (c f -d e \right ) d}}-\frac {6 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) b \,c^{3} e \,f^{2}}{d^{3} \sqrt {\left (c f -d e \right ) d}}+\frac {6 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) b \,c^{2} e^{2} f}{d^{2} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) b c \,e^{3}}{d \sqrt {\left (c f -d e \right ) d}}\)	\(557\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*(1/d^4*(1/7*b*(f*x+e)^(7/2)*d^3+1/5*a*d^3*f*(f*x+e)^(5/2)-1/5*b*c*d^2*f*(f*x+e)^(5/2)-1/3*a*c*d^2*f^2*(f*x
+e)^(3/2)+1/3*a*d^3*e*f*(f*x+e)^(3/2)+1/3*b*c^2*d*f^2*(f*x+e)^(3/2)-1/3*b*c*d^2*e*f*(f*x+e)^(3/2)+a*c^2*d*f^3*
(f*x+e)^(1/2)-2*a*c*d^2*e*f^2*(f*x+e)^(1/2)+a*d^3*e^2*f*(f*x+e)^(1/2)-b*c^3*f^3*(f*x+e)^(1/2)+2*b*c^2*d*e*f^2*
(f*x+e)^(1/2)-b*c*d^2*e^2*f*(f*x+e)^(1/2))-f*(a*c^3*d*f^3-3*a*c^2*d^2*e*f^2+3*a*c*d^3*e^2*f-a*d^4*e^3-b*c^4*f^
3+3*b*c^3*d*e*f^2-3*b*c^2*d^2*e^2*f+b*c*d^3*e^3)/d^4/((c*f-d*e)*d)^(1/2)*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^
(1/2)))

________________________________________________________________________________________

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)*(f*x+e)^(5/2)/(d*x+c),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

________________________________________________________________________________________

Fricas [A]
time = 1.03, size = 598, normalized size = 3.65 \begin {gather*} \left [\frac {105 \, {\left ({\left (b c^{3} - a c^{2} d\right )} f^{3} - 2 \, {\left (b c^{2} d - a c d^{2}\right )} f^{2} e + {\left (b c d^{2} - a d^{3}\right )} f e^{2}\right )} \sqrt {-\frac {c f - d e}{d}} \log \left (\frac {d f x - c f + 2 \, \sqrt {f x + e} d \sqrt {-\frac {c f - d e}{d}} + 2 \, d e}{d x + c}\right ) + 2 \, {\left (15 \, b d^{3} f^{3} x^{3} - 21 \, {\left (b c d^{2} - a d^{3}\right )} f^{3} x^{2} + 35 \, {\left (b c^{2} d - a c d^{2}\right )} f^{3} x + 15 \, b d^{3} e^{3} - 105 \, {\left (b c^{3} - a c^{2} d\right )} f^{3} + {\left (45 \, b d^{3} f x - 161 \, {\left (b c d^{2} - a d^{3}\right )} f\right )} e^{2} + {\left (45 \, b d^{3} f^{2} x^{2} - 77 \, {\left (b c d^{2} - a d^{3}\right )} f^{2} x + 245 \, {\left (b c^{2} d - a c d^{2}\right )} f^{2}\right )} e\right )} \sqrt {f x + e}}{105 \, d^{4} f}, -\frac {2 \, {\left (105 \, {\left ({\left (b c^{3} - a c^{2} d\right )} f^{3} - 2 \, {\left (b c^{2} d - a c d^{2}\right )} f^{2} e + {\left (b c d^{2} - a d^{3}\right )} f e^{2}\right )} \sqrt {\frac {c f - d e}{d}} \arctan \left (-\frac {\sqrt {f x + e} d \sqrt {\frac {c f - d e}{d}}}{c f - d e}\right ) - {\left (15 \, b d^{3} f^{3} x^{3} - 21 \, {\left (b c d^{2} - a d^{3}\right )} f^{3} x^{2} + 35 \, {\left (b c^{2} d - a c d^{2}\right )} f^{3} x + 15 \, b d^{3} e^{3} - 105 \, {\left (b c^{3} - a c^{2} d\right )} f^{3} + {\left (45 \, b d^{3} f x - 161 \, {\left (b c d^{2} - a d^{3}\right )} f\right )} e^{2} + {\left (45 \, b d^{3} f^{2} x^{2} - 77 \, {\left (b c d^{2} - a d^{3}\right )} f^{2} x + 245 \, {\left (b c^{2} d - a c d^{2}\right )} f^{2}\right )} e\right )} \sqrt {f x + e}\right )}}{105 \, d^{4} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/105*(105*((b*c^3 - a*c^2*d)*f^3 - 2*(b*c^2*d - a*c*d^2)*f^2*e + (b*c*d^2 - a*d^3)*f*e^2)*sqrt(-(c*f - d*e)/
d)*log((d*f*x - c*f + 2*sqrt(f*x + e)*d*sqrt(-(c*f - d*e)/d) + 2*d*e)/(d*x + c)) + 2*(15*b*d^3*f^3*x^3 - 21*(b
*c*d^2 - a*d^3)*f^3*x^2 + 35*(b*c^2*d - a*c*d^2)*f^3*x + 15*b*d^3*e^3 - 105*(b*c^3 - a*c^2*d)*f^3 + (45*b*d^3*
f*x - 161*(b*c*d^2 - a*d^3)*f)*e^2 + (45*b*d^3*f^2*x^2 - 77*(b*c*d^2 - a*d^3)*f^2*x + 245*(b*c^2*d - a*c*d^2)*
f^2)*e)*sqrt(f*x + e))/(d^4*f), -2/105*(105*((b*c^3 - a*c^2*d)*f^3 - 2*(b*c^2*d - a*c*d^2)*f^2*e + (b*c*d^2 -
a*d^3)*f*e^2)*sqrt((c*f - d*e)/d)*arctan(-sqrt(f*x + e)*d*sqrt((c*f - d*e)/d)/(c*f - d*e)) - (15*b*d^3*f^3*x^3
- 21*(b*c*d^2 - a*d^3)*f^3*x^2 + 35*(b*c^2*d - a*c*d^2)*f^3*x + 15*b*d^3*e^3 - 105*(b*c^3 - a*c^2*d)*f^3 + (4
5*b*d^3*f*x - 161*(b*c*d^2 - a*d^3)*f)*e^2 + (45*b*d^3*f^2*x^2 - 77*(b*c*d^2 - a*d^3)*f^2*x + 245*(b*c^2*d - a
*c*d^2)*f^2)*e)*sqrt(f*x + e))/(d^4*f)]

________________________________________________________________________________________

Sympy [A]
time = 54.15, size = 221, normalized size = 1.35 \begin {gather*} \frac {2 b \left (e + f x\right )^{\frac {7}{2}}}{7 d f} + \frac {\left (e + f x\right )^{\frac {5}{2}} \cdot \left (2 a d - 2 b c\right )}{5 d^{2}} + \frac {\left (e + f x\right )^{\frac {3}{2}} \left (- 2 a c d f + 2 a d^{2} e + 2 b c^{2} f - 2 b c d e\right )}{3 d^{3}} + \frac {\sqrt {e + f x} \left (2 a c^{2} d f^{2} - 4 a c d^{2} e f + 2 a d^{3} e^{2} - 2 b c^{3} f^{2} + 4 b c^{2} d e f - 2 b c d^{2} e^{2}\right )}{d^{4}} - \frac {2 \left (a d - b c\right ) \left (c f - d e\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{5} \sqrt {\frac {c f - d e}{d}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*(e + f*x)**(7/2)/(7*d*f) + (e + f*x)**(5/2)*(2*a*d - 2*b*c)/(5*d**2) + (e + f*x)**(3/2)*(-2*a*c*d*f + 2*a*
d**2*e + 2*b*c**2*f - 2*b*c*d*e)/(3*d**3) + sqrt(e + f*x)*(2*a*c**2*d*f**2 - 4*a*c*d**2*e*f + 2*a*d**3*e**2 -
2*b*c**3*f**2 + 4*b*c**2*d*e*f - 2*b*c*d**2*e**2)/d**4 - 2*(a*d - b*c)*(c*f - d*e)**3*atan(sqrt(e + f*x)/sqrt(
(c*f - d*e)/d))/(d**5*sqrt((c*f - d*e)/d))

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (149) = 298\).
time = 1.11, size = 387, normalized size = 2.36 \begin {gather*} \frac {2 \, {\left (b c^{4} f^{3} - a c^{3} d f^{3} - 3 \, b c^{3} d f^{2} e + 3 \, a c^{2} d^{2} f^{2} e + 3 \, b c^{2} d^{2} f e^{2} - 3 \, a c d^{3} f e^{2} - b c d^{3} e^{3} + a d^{4} e^{3}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{\sqrt {c d f - d^{2} e} d^{4}} + \frac {2 \, {\left (15 \, {\left (f x + e\right )}^{\frac {7}{2}} b d^{6} f^{6} - 21 \, {\left (f x + e\right )}^{\frac {5}{2}} b c d^{5} f^{7} + 21 \, {\left (f x + e\right )}^{\frac {5}{2}} a d^{6} f^{7} + 35 \, {\left (f x + e\right )}^{\frac {3}{2}} b c^{2} d^{4} f^{8} - 35 \, {\left (f x + e\right )}^{\frac {3}{2}} a c d^{5} f^{8} - 105 \, \sqrt {f x + e} b c^{3} d^{3} f^{9} + 105 \, \sqrt {f x + e} a c^{2} d^{4} f^{9} - 35 \, {\left (f x + e\right )}^{\frac {3}{2}} b c d^{5} f^{7} e + 35 \, {\left (f x + e\right )}^{\frac {3}{2}} a d^{6} f^{7} e + 210 \, \sqrt {f x + e} b c^{2} d^{4} f^{8} e - 210 \, \sqrt {f x + e} a c d^{5} f^{8} e - 105 \, \sqrt {f x + e} b c d^{5} f^{7} e^{2} + 105 \, \sqrt {f x + e} a d^{6} f^{7} e^{2}\right )}}{105 \, d^{7} f^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b*c^4*f^3 - a*c^3*d*f^3 - 3*b*c^3*d*f^2*e + 3*a*c^2*d^2*f^2*e + 3*b*c^2*d^2*f*e^2 - 3*a*c*d^3*f*e^2 - b*c*d
^3*e^3 + a*d^4*e^3)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/(sqrt(c*d*f - d^2*e)*d^4) + 2/105*(15*(f*x + e
)^(7/2)*b*d^6*f^6 - 21*(f*x + e)^(5/2)*b*c*d^5*f^7 + 21*(f*x + e)^(5/2)*a*d^6*f^7 + 35*(f*x + e)^(3/2)*b*c^2*d
^4*f^8 - 35*(f*x + e)^(3/2)*a*c*d^5*f^8 - 105*sqrt(f*x + e)*b*c^3*d^3*f^9 + 105*sqrt(f*x + e)*a*c^2*d^4*f^9 -
35*(f*x + e)^(3/2)*b*c*d^5*f^7*e + 35*(f*x + e)^(3/2)*a*d^6*f^7*e + 210*sqrt(f*x + e)*b*c^2*d^4*f^8*e - 210*sq
rt(f*x + e)*a*c*d^5*f^8*e - 105*sqrt(f*x + e)*b*c*d^5*f^7*e^2 + 105*sqrt(f*x + e)*a*d^6*f^7*e^2)/(d^7*f^7)

________________________________________________________________________________________

Mupad [B]
time = 0.13, size = 330, normalized size = 2.01 \begin {gather*} {\left (e+f\,x\right )}^{5/2}\,\left (\frac {2\,a\,f-2\,b\,e}{5\,d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{5\,d^2\,f^2}\right )+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,\left (a\,d-b\,c\right )\,{\left (c\,f-d\,e\right )}^{5/2}}{b\,c^4\,f^3-3\,b\,c^3\,d\,e\,f^2-a\,c^3\,d\,f^3+3\,b\,c^2\,d^2\,e^2\,f+3\,a\,c^2\,d^2\,e\,f^2-b\,c\,d^3\,e^3-3\,a\,c\,d^3\,e^2\,f+a\,d^4\,e^3}\right )\,\left (a\,d-b\,c\right )\,{\left (c\,f-d\,e\right )}^{5/2}}{d^{9/2}}+\frac {2\,b\,{\left (e+f\,x\right )}^{7/2}}{7\,d\,f}-\frac {{\left (e+f\,x\right )}^{3/2}\,\left (c\,f^2-d\,e\,f\right )\,\left (\frac {2\,a\,f-2\,b\,e}{d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{d^2\,f^2}\right )}{3\,d\,f}+\frac {\sqrt {e+f\,x}\,{\left (c\,f^2-d\,e\,f\right )}^2\,\left (\frac {2\,a\,f-2\,b\,e}{d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{d^2\,f^2}\right )}{d^2\,f^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e + f*x)^(5/2)*((2*a*f - 2*b*e)/(5*d*f) - (2*b*(c*f^2 - d*e*f))/(5*d^2*f^2)) + (2*atan((d^(1/2)*(e + f*x)^(1/
2)*(a*d - b*c)*(c*f - d*e)^(5/2))/(a*d^4*e^3 + b*c^4*f^3 - a*c^3*d*f^3 - b*c*d^3*e^3 - 3*a*c*d^3*e^2*f - 3*b*c
^3*d*e*f^2 + 3*a*c^2*d^2*e*f^2 + 3*b*c^2*d^2*e^2*f))*(a*d - b*c)*(c*f - d*e)^(5/2))/d^(9/2) + (2*b*(e + f*x)^(
7/2))/(7*d*f) - ((e + f*x)^(3/2)*(c*f^2 - d*e*f)*((2*a*f - 2*b*e)/(d*f) - (2*b*(c*f^2 - d*e*f))/(d^2*f^2)))/(3
*d*f) + ((e + f*x)^(1/2)*(c*f^2 - d*e*f)^2*((2*a*f - 2*b*e)/(d*f) - (2*b*(c*f^2 - d*e*f))/(d^2*f^2)))/(d^2*f^2
)

________________________________________________________________________________________