3.18.76 \(\int \frac {(a+b x)^2 \sqrt {e+f x}}{c+d x} \, dx\) [1776]

Optimal. Leaf size=138 \[ \frac {2 (b c-a d)^2 \sqrt {e+f x}}{d^3}-\frac {2 b (b d e+b c f-2 a d f) (e+f x)^{3/2}}{3 d^2 f^2}+\frac {2 b^2 (e+f x)^{5/2}}{5 d f^2}-\frac {2 (b c-a d)^2 \sqrt {d e-c f} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {90, 52, 65, 214} \begin {gather*} -\frac {2 (b c-a d)^2 \sqrt {d e-c f} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2}}+\frac {2 \sqrt {e+f x} (b c-a d)^2}{d^3}-\frac {2 b (e+f x)^{3/2} (-2 a d f+b c f+b d e)}{3 d^2 f^2}+\frac {2 b^2 (e+f x)^{5/2}}{5 d f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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

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Mathematica [A]
time = 0.23, size = 152, normalized size = 1.10 \begin {gather*} \frac {2 \sqrt {e+f x} \left (15 a^2 d^2 f^2+10 a b d f (-3 c f+d (e+f x))+b^2 \left (15 c^2 f^2-5 c d f (e+f x)+d^2 \left (-2 e^2+e f x+3 f^2 x^2\right )\right )\right )}{15 d^3 f^2}-\frac {2 (b c-a d)^2 \sqrt {-d e+c f} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e + f*x]*(15*a^2*d^2*f^2 + 10*a*b*d*f*(-3*c*f + d*(e + f*x)) + b^2*(15*c^2*f^2 - 5*c*d*f*(e + f*x) + d
^2*(-2*e^2 + e*f*x + 3*f^2*x^2))))/(15*d^3*f^2) - (2*(b*c - a*d)^2*Sqrt[-(d*e) + c*f]*ArcTan[(Sqrt[d]*Sqrt[e +
f*x])/Sqrt[-(d*e) + c*f]])/d^(7/2)

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Maple [A]
time = 0.10, size = 220, normalized size = 1.59

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\frac {b^{2} \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}+\frac {2 a b \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b^{2} c d f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b^{2} d^{2} e \left (f x +e \right )^{\frac {3}{2}}}{3}+a^{2} d^{2} f^{2} \sqrt {f x +e}-2 a b c d \,f^{2} \sqrt {f x +e}+b^{2} c^{2} f^{2} \sqrt {f x +e}\right )}{d^{3}}-\frac {2 f^{2} \left (a^{2} c \,d^{2} f -a^{2} d^{3} e -2 a b \,c^{2} d f +2 a b c \,d^{2} e +b^{2} c^{3} f -b^{2} c^{2} d e \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f^{2}}\)	\(220\)
default	\(\frac {\frac {2 \left (\frac {b^{2} \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}+\frac {2 a b \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b^{2} c d f \left (f x +e \right )^{\frac {3}{2}}}{3}-\frac {b^{2} d^{2} e \left (f x +e \right )^{\frac {3}{2}}}{3}+a^{2} d^{2} f^{2} \sqrt {f x +e}-2 a b c d \,f^{2} \sqrt {f x +e}+b^{2} c^{2} f^{2} \sqrt {f x +e}\right )}{d^{3}}-\frac {2 f^{2} \left (a^{2} c \,d^{2} f -a^{2} d^{3} e -2 a b \,c^{2} d f +2 a b c \,d^{2} e +b^{2} c^{3} f -b^{2} c^{2} d e \right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f^{2}}\)	\(220\)
risch	\(\frac {2 \left (3 b^{2} d^{2} f^{2} x^{2}+10 a b \,d^{2} f^{2} x -5 b^{2} c d \,f^{2} x +b^{2} d^{2} e f x +15 a^{2} d^{2} f^{2}-30 a b c d \,f^{2}+10 a b \,d^{2} e f +15 b^{2} c^{2} f^{2}-5 b^{2} c d e f -2 b^{2} d^{2} e^{2}\right ) \sqrt {f x +e}}{15 f^{2} d^{3}}-\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a^{2} c f}{d \sqrt {\left (c f -d e \right ) d}}+\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a^{2} e}{\sqrt {\left (c f -d e \right ) d}}+\frac {4 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a b \,c^{2} f}{d^{2} \sqrt {\left (c f -d e \right ) d}}-\frac {4 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a b c e}{d \sqrt {\left (c f -d e \right ) d}}-\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) b^{2} c^{3} f}{d^{3} \sqrt {\left (c f -d e \right ) d}}+\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) b^{2} c^{2} e}{d^{2} \sqrt {\left (c f -d e \right ) d}}\)	\(388\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*(f*x+e)^(1/2)/(d*x+c),x,method=_RETURNVERBOSE)

[Out]

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

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Fricas [A]
time = 0.98, size = 411, normalized size = 2.98 \begin {gather*} \left [\frac {15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} \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 (3 \, b^{2} d^{2} f^{2} x^{2} - 2 \, b^{2} d^{2} e^{2} - 5 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} f^{2} x + 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} + {\left (b^{2} d^{2} f x - 5 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} f\right )} e\right )} \sqrt {f x + e}}{15 \, d^{3} f^{2}}, \frac {2 \, {\left (15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} \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 (3 \, b^{2} d^{2} f^{2} x^{2} - 2 \, b^{2} d^{2} e^{2} - 5 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} f^{2} x + 15 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} f^{2} + {\left (b^{2} d^{2} f x - 5 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} f\right )} e\right )} \sqrt {f x + e}\right )}}{15 \, d^{3} f^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*(f*x+e)^(1/2)/(d*x+c),x, algorithm="fricas")

[Out]

[1/15*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^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*(3*b^2*d^2*f^2*x^2 - 2*b^2*d^2*e^2 - 5*(b^2*c*d - 2*a*b*d^2)*f^2*x + 15
*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2 + (b^2*d^2*f*x - 5*(b^2*c*d - 2*a*b*d^2)*f)*e)*sqrt(f*x + e))/(d^3*f^2),
2/15*(15*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*f^2*sqrt((c*f - d*e)/d)*arctan(-sqrt(f*x + e)*d*sqrt((c*f - d*e)/d)/(
c*f - d*e)) + (3*b^2*d^2*f^2*x^2 - 2*b^2*d^2*e^2 - 5*(b^2*c*d - 2*a*b*d^2)*f^2*x + 15*(b^2*c^2 - 2*a*b*c*d + a
^2*d^2)*f^2 + (b^2*d^2*f*x - 5*(b^2*c*d - 2*a*b*d^2)*f)*e)*sqrt(f*x + e))/(d^3*f^2)]

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Sympy [A]
time = 7.18, size = 155, normalized size = 1.12 \begin {gather*} \frac {2 \left (\frac {b^{2} \left (e + f x\right )^{\frac {5}{2}}}{5 d f} + \frac {\left (e + f x\right )^{\frac {3}{2}} \cdot \left (2 a b d f - b^{2} c f - b^{2} d e\right )}{3 d^{2} f} + \frac {\sqrt {e + f x} \left (a^{2} d^{2} f - 2 a b c d f + b^{2} c^{2} f\right )}{d^{3}} - \frac {f \left (a d - b c\right )^{2} \left (c f - d e\right ) \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{4} \sqrt {\frac {c f - d e}{d}}}\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2*(f*x+e)**(1/2)/(d*x+c),x)

[Out]

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

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Giac [A]
time = 0.60, size = 250, normalized size = 1.81 \begin {gather*} -\frac {2 \, {\left (b^{2} c^{3} f - 2 \, a b c^{2} d f + a^{2} c d^{2} f - b^{2} c^{2} d e + 2 \, a b c d^{2} e - a^{2} d^{3} e\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^{3}} + \frac {2 \, {\left (3 \, {\left (f x + e\right )}^{\frac {5}{2}} b^{2} d^{4} f^{8} - 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{2} c d^{3} f^{9} + 10 \, {\left (f x + e\right )}^{\frac {3}{2}} a b d^{4} f^{9} + 15 \, \sqrt {f x + e} b^{2} c^{2} d^{2} f^{10} - 30 \, \sqrt {f x + e} a b c d^{3} f^{10} + 15 \, \sqrt {f x + e} a^{2} d^{4} f^{10} - 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b^{2} d^{4} f^{8} e\right )}}{15 \, d^{5} f^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*(f*x+e)^(1/2)/(d*x+c),x, algorithm="giac")

[Out]

-2*(b^2*c^3*f - 2*a*b*c^2*d*f + a^2*c*d^2*f - b^2*c^2*d*e + 2*a*b*c*d^2*e - a^2*d^3*e)*arctan(sqrt(f*x + e)*d/
sqrt(c*d*f - d^2*e))/(sqrt(c*d*f - d^2*e)*d^3) + 2/15*(3*(f*x + e)^(5/2)*b^2*d^4*f^8 - 5*(f*x + e)^(3/2)*b^2*c
*d^3*f^9 + 10*(f*x + e)^(3/2)*a*b*d^4*f^9 + 15*sqrt(f*x + e)*b^2*c^2*d^2*f^10 - 30*sqrt(f*x + e)*a*b*c*d^3*f^1
0 + 15*sqrt(f*x + e)*a^2*d^4*f^10 - 5*(f*x + e)^(3/2)*b^2*d^4*f^8*e)/(d^5*f^10)

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Mupad [B]
time = 0.17, size = 281, normalized size = 2.04 \begin {gather*} \sqrt {e+f\,x}\,\left (\frac {2\,{\left (a\,f-b\,e\right )}^2}{d\,f^2}+\frac {\left (\frac {4\,b^2\,e-4\,a\,b\,f}{d\,f^2}+\frac {2\,b^2\,\left (c\,f^3-d\,e\,f^2\right )}{d^2\,f^4}\right )\,\left (c\,f^3-d\,e\,f^2\right )}{d\,f^2}\right )-{\left (e+f\,x\right )}^{3/2}\,\left (\frac {4\,b^2\,e-4\,a\,b\,f}{3\,d\,f^2}+\frac {2\,b^2\,\left (c\,f^3-d\,e\,f^2\right )}{3\,d^2\,f^4}\right )+\frac {2\,b^2\,{\left (e+f\,x\right )}^{5/2}}{5\,d\,f^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^2\,\sqrt {d\,e-c\,f}\,1{}\mathrm {i}}{-f\,a^2\,c\,d^2+e\,a^2\,d^3+2\,f\,a\,b\,c^2\,d-2\,e\,a\,b\,c\,d^2-f\,b^2\,c^3+e\,b^2\,c^2\,d}\right )\,{\left (a\,d-b\,c\right )}^2\,\sqrt {d\,e-c\,f}\,2{}\mathrm {i}}{d^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e + f*x)^(1/2)*(a + b*x)^2)/(c + d*x),x)

[Out]

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

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