3.18.44 \(\int \frac {(A+B x) (d+e x)^{3/2}}{a+b x} \, dx\) [1744]

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

[Out]

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

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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 (A b-a B) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}+\frac {2 \sqrt {d+e x} (A b-a B) (b d-a e)}{b^3}+\frac {2 (d+e x)^{3/2} (A b-a B)}{3 b^2}+\frac {2 B (d+e x)^{5/2}}{5 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.23, size = 129, normalized size = 0.99 \begin {gather*} \frac {2 \sqrt {d+e x} \left (15 a^2 B e^2-5 a b e (4 B d+3 A e+B e x)+b^2 \left (3 B (d+e x)^2+5 A e (4 d+e x)\right )\right )}{15 b^3 e}+\frac {2 (A b-a B) (-b d+a e)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^(3/2))/(a + b*x),x]

[Out]

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

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Maple [A]
time = 0.10, size = 205, normalized size = 1.58

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/e*(-1/b^3*(-1/5*b^2*B*(e*x+d)^(5/2)-1/3*A*b^2*e*(e*x+d)^(3/2)+1/3*B*a*b*e*(e*x+d)^(3/2)+A*a*b*e^2*(e*x+d)^(1
/2)-A*b^2*d*e*(e*x+d)^(1/2)-B*a^2*e^2*(e*x+d)^(1/2)+B*a*b*d*e*(e*x+d)^(1/2))+e*(A*a^2*b*e^2-2*A*a*b^2*d*e+A*b^
3*d^2-B*a^3*e^2+2*B*a^2*b*d*e-B*a*b^2*d^2)/b^3/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(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)*(e*x+d)^(3/2)/(b*x+a),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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 1.19, size = 372, normalized size = 2.86 \begin {gather*} \left [-\frac {{\left (15 \, {\left ({\left (B a b - A b^{2}\right )} d e - {\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {2 \, b d - 2 \, \sqrt {x e + d} b \sqrt {\frac {b d - a e}{b}} + {\left (b x - a\right )} e}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} d^{2} + {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} e^{2} + 2 \, {\left (3 \, B b^{2} d x - 10 \, {\left (B a b - A b^{2}\right )} d\right )} e\right )} \sqrt {x e + d}\right )} e^{\left (-1\right )}}{15 \, b^{3}}, \frac {2 \, {\left (15 \, {\left ({\left (B a b - A b^{2}\right )} d e - {\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {x e + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) + {\left (3 \, B b^{2} d^{2} + {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} e^{2} + 2 \, {\left (3 \, B b^{2} d x - 10 \, {\left (B a b - A b^{2}\right )} d\right )} e\right )} \sqrt {x e + d}\right )} e^{\left (-1\right )}}{15 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 21.36, size = 139, normalized size = 1.07 \begin {gather*} \frac {2 B \left (d + e x\right )^{\frac {5}{2}}}{5 b e} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (2 A b - 2 B a\right )}{3 b^{2}} + \frac {\sqrt {d + e x} \left (- 2 A a b e + 2 A b^{2} d + 2 B a^{2} e - 2 B a b d\right )}{b^{3}} - \frac {2 \left (- A b + B a\right ) \left (a e - b d\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{b^{4} \sqrt {\frac {a e - b d}{b}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(3/2)/(b*x+a),x)

[Out]

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

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Giac [A]
time = 0.66, size = 228, normalized size = 1.75 \begin {gather*} -\frac {2 \, {\left (B a b^{2} d^{2} - A b^{3} d^{2} - 2 \, B a^{2} b d e + 2 \, A a b^{2} d e + B a^{3} e^{2} - A a^{2} b e^{2}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{4} e^{4} - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{3} e^{5} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{4} e^{5} - 15 \, \sqrt {x e + d} B a b^{3} d e^{5} + 15 \, \sqrt {x e + d} A b^{4} d e^{5} + 15 \, \sqrt {x e + d} B a^{2} b^{2} e^{6} - 15 \, \sqrt {x e + d} A a b^{3} e^{6}\right )} e^{\left (-5\right )}}{15 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(B*a*b^2*d^2 - A*b^3*d^2 - 2*B*a^2*b*d*e + 2*A*a*b^2*d*e + B*a^3*e^2 - A*a^2*b*e^2)*arctan(sqrt(x*e + d)*b/
sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^3) + 2/15*(3*(x*e + d)^(5/2)*B*b^4*e^4 - 5*(x*e + d)^(3/2)*B*a*b
^3*e^5 + 5*(x*e + d)^(3/2)*A*b^4*e^5 - 15*sqrt(x*e + d)*B*a*b^3*d*e^5 + 15*sqrt(x*e + d)*A*b^4*d*e^5 + 15*sqrt
(x*e + d)*B*a^2*b^2*e^6 - 15*sqrt(x*e + d)*A*a*b^3*e^6)*e^(-5)/b^5

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Mupad [B]
time = 0.11, size = 236, normalized size = 1.82 \begin {gather*} \left (\frac {2\,A\,e-2\,B\,d}{3\,b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{3\,b^2\,e^2}\right )\,{\left (d+e\,x\right )}^{3/2}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}}{-B\,a^3\,e^2+2\,B\,a^2\,b\,d\,e+A\,a^2\,b\,e^2-B\,a\,b^2\,d^2-2\,A\,a\,b^2\,d\,e+A\,b^3\,d^2}\right )\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{3/2}}{b^{7/2}}+\frac {2\,B\,{\left (d+e\,x\right )}^{5/2}}{5\,b\,e}-\frac {\left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,\left (a\,e^2-b\,d\,e\right )\,\sqrt {d+e\,x}}{b\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^(3/2))/(a + b*x),x)

[Out]

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

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