3.18.67 \(\int \frac {(a+b x) (e+f x)^{3/2}}{c+d x} \, dx\) [1767]

Optimal. Leaf size=130 \[ -\frac {2 (b c-a d) (d e-c f) \sqrt {e+f x}}{d^3}-\frac {2 (b c-a d) (e+f x)^{3/2}}{3 d^2}+\frac {2 b (e+f x)^{5/2}}{5 d f}+\frac {2 (b c-a d) (d e-c f)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, 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)^{3/2} \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) (d e-c f)}{d^3}-\frac {2 (e+f x)^{3/2} (b c-a d)}{3 d^2}+\frac {2 b (e+f x)^{5/2}}{5 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.22, size = 129, normalized size = 0.99 \begin {gather*} \frac {2 \sqrt {e+f x} \left (5 a d f (4 d e-3 c f+d f x)+b \left (15 c^2 f^2+3 d^2 (e+f x)^2-5 c d f (4 e+f x)\right )\right )}{15 d^3 f}+\frac {2 (-b c+a d) (-d e+c f)^{3/2} \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)*(e + f*x)^(3/2))/(c + d*x),x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.11, size = 205, normalized size = 1.58

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {b \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}-\frac {a \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b c d f \left (f x +e \right )^{\frac {3}{2}}}{3}+a c d \,f^{2} \sqrt {f x +e}-a \,d^{2} e f \sqrt {f x +e}-b \,c^{2} f^{2} \sqrt {f x +e}+b c d e f \sqrt {f x +e}\right )}{d^{3}}+\frac {2 f \left (a \,c^{2} d \,f^{2}-2 a c \,d^{2} e f +a \,d^{3} e^{2}-b \,c^{3} f^{2}+2 b \,c^{2} d e f -b c \,d^{2} e^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f}\)	\(205\)
default	\(\frac {-\frac {2 \left (-\frac {b \left (f x +e \right )^{\frac {5}{2}} d^{2}}{5}-\frac {a \,d^{2} f \left (f x +e \right )^{\frac {3}{2}}}{3}+\frac {b c d f \left (f x +e \right )^{\frac {3}{2}}}{3}+a c d \,f^{2} \sqrt {f x +e}-a \,d^{2} e f \sqrt {f x +e}-b \,c^{2} f^{2} \sqrt {f x +e}+b c d e f \sqrt {f x +e}\right )}{d^{3}}+\frac {2 f \left (a \,c^{2} d \,f^{2}-2 a c \,d^{2} e f +a \,d^{3} e^{2}-b \,c^{3} f^{2}+2 b \,c^{2} d e f -b c \,d^{2} e^{2}\right ) \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right )}{d^{3} \sqrt {\left (c f -d e \right ) d}}}{f}\)	\(205\)
risch	\(-\frac {2 \left (-3 b \,f^{2} d^{2} x^{2}-5 a \,d^{2} f^{2} x +5 b c d \,f^{2} x -6 b \,d^{2} e f x +15 a c d \,f^{2}-20 a \,d^{2} e f -15 b \,c^{2} f^{2}+20 b c d e f -3 b \,d^{2} e^{2}\right ) \sqrt {f x +e}}{15 f \,d^{3}}+\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a \,c^{2} f^{2}}{d^{2} \sqrt {\left (c f -d e \right ) d}}-\frac {4 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a c e f}{d \sqrt {\left (c f -d e \right ) d}}+\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) a \,e^{2}}{\sqrt {\left (c f -d e \right ) d}}-\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) b \,c^{3} f^{2}}{d^{3} \sqrt {\left (c f -d e \right ) d}}+\frac {4 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) b \,c^{2} e f}{d^{2} \sqrt {\left (c f -d e \right ) d}}-\frac {2 \arctan \left (\frac {d \sqrt {f x +e}}{\sqrt {\left (c f -d e \right ) d}}\right ) b c \,e^{2}}{d \sqrt {\left (c f -d e \right ) d}}\)	\(363\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*(-1/d^3*(-1/5*b*(f*x+e)^(5/2)*d^2-1/3*a*d^2*f*(f*x+e)^(3/2)+1/3*b*c*d*f*(f*x+e)^(3/2)+a*c*d*f^2*(f*x+e)^(1
/2)-a*d^2*e*f*(f*x+e)^(1/2)-b*c^2*f^2*(f*x+e)^(1/2)+b*c*d*e*f*(f*x+e)^(1/2))+f*(a*c^2*d*f^2-2*a*c*d^2*e*f+a*d^
3*e^2-b*c^3*f^2+2*b*c^2*d*e*f-b*c*d^2*e^2)/d^3/((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)^(3/2)/(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 = 1.43, size = 387, normalized size = 2.98 \begin {gather*} \left [-\frac {15 \, {\left ({\left (b c^{2} - a c d\right )} f^{2} - {\left (b c d - a d^{2}\right )} f e\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 (3 \, b d^{2} f^{2} x^{2} - 5 \, {\left (b c d - a d^{2}\right )} f^{2} x + 3 \, b d^{2} e^{2} + 15 \, {\left (b c^{2} - a c d\right )} f^{2} + 2 \, {\left (3 \, b d^{2} f x - 10 \, {\left (b c d - a d^{2}\right )} f\right )} e\right )} \sqrt {f x + e}}{15 \, d^{3} f}, \frac {2 \, {\left (15 \, {\left ({\left (b c^{2} - a c d\right )} f^{2} - {\left (b c d - a d^{2}\right )} f e\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 (3 \, b d^{2} f^{2} x^{2} - 5 \, {\left (b c d - a d^{2}\right )} f^{2} x + 3 \, b d^{2} e^{2} + 15 \, {\left (b c^{2} - a c d\right )} f^{2} + 2 \, {\left (3 \, b d^{2} f x - 10 \, {\left (b c d - a d^{2}\right )} f\right )} e\right )} \sqrt {f x + e}\right )}}{15 \, d^{3} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/15*(15*((b*c^2 - a*c*d)*f^2 - (b*c*d - a*d^2)*f*e)*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*d^2*f^2*x^2 - 5*(b*c*d - a*d^2)*f^2*x + 3*b*d^2*e^2 + 15*
(b*c^2 - a*c*d)*f^2 + 2*(3*b*d^2*f*x - 10*(b*c*d - a*d^2)*f)*e)*sqrt(f*x + e))/(d^3*f), 2/15*(15*((b*c^2 - a*c
*d)*f^2 - (b*c*d - a*d^2)*f*e)*sqrt((c*f - d*e)/d)*arctan(-sqrt(f*x + e)*d*sqrt((c*f - d*e)/d)/(c*f - d*e)) +
(3*b*d^2*f^2*x^2 - 5*(b*c*d - a*d^2)*f^2*x + 3*b*d^2*e^2 + 15*(b*c^2 - a*c*d)*f^2 + 2*(3*b*d^2*f*x - 10*(b*c*d
- a*d^2)*f)*e)*sqrt(f*x + e))/(d^3*f)]

________________________________________________________________________________________

Sympy [A]
time = 29.56, size = 139, normalized size = 1.07 \begin {gather*} \frac {2 b \left (e + f x\right )^{\frac {5}{2}}}{5 d f} + \frac {\left (e + f x\right )^{\frac {3}{2}} \cdot \left (2 a d - 2 b c\right )}{3 d^{2}} + \frac {\sqrt {e + f x} \left (- 2 a c d f + 2 a d^{2} e + 2 b c^{2} f - 2 b c d e\right )}{d^{3}} + \frac {2 \left (a d - b c\right ) \left (c f - d e\right )^{2} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (117) = 234\).
time = 1.01, size = 238, normalized size = 1.83 \begin {gather*} -\frac {2 \, {\left (b c^{3} f^{2} - a c^{2} d f^{2} - 2 \, b c^{2} d f e + 2 \, a c d^{2} f e + b c d^{2} e^{2} - a d^{3} e^{2}\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 d^{4} f^{4} - 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b c d^{3} f^{5} + 5 \, {\left (f x + e\right )}^{\frac {3}{2}} a d^{4} f^{5} + 15 \, \sqrt {f x + e} b c^{2} d^{2} f^{6} - 15 \, \sqrt {f x + e} a c d^{3} f^{6} - 15 \, \sqrt {f x + e} b c d^{3} f^{5} e + 15 \, \sqrt {f x + e} a d^{4} f^{5} e\right )}}{15 \, d^{5} f^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b*c^3*f^2 - a*c^2*d*f^2 - 2*b*c^2*d*f*e + 2*a*c*d^2*f*e + b*c*d^2*e^2 - a*d^3*e^2)*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*d^4*f^4 - 5*(f*x + e)^(3/2)*b*c*d^3
*f^5 + 5*(f*x + e)^(3/2)*a*d^4*f^5 + 15*sqrt(f*x + e)*b*c^2*d^2*f^6 - 15*sqrt(f*x + e)*a*c*d^3*f^6 - 15*sqrt(f
*x + e)*b*c*d^3*f^5*e + 15*sqrt(f*x + e)*a*d^4*f^5*e)/(d^5*f^5)

________________________________________________________________________________________

Mupad [B]
time = 1.23, size = 236, normalized size = 1.82 \begin {gather*} {\left (e+f\,x\right )}^{3/2}\,\left (\frac {2\,a\,f-2\,b\,e}{3\,d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{3\,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 )}^{3/2}}{-b\,c^3\,f^2+2\,b\,c^2\,d\,e\,f+a\,c^2\,d\,f^2-b\,c\,d^2\,e^2-2\,a\,c\,d^2\,e\,f+a\,d^3\,e^2}\right )\,\left (a\,d-b\,c\right )\,{\left (c\,f-d\,e\right )}^{3/2}}{d^{7/2}}+\frac {2\,b\,{\left (e+f\,x\right )}^{5/2}}{5\,d\,f}-\frac {\sqrt {e+f\,x}\,\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 )}{d\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________