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

Optimal. Leaf size=174 \[ -\frac {\sqrt {b d-a e} (-5 a B e+3 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}+\frac {\sqrt {d+e x} (-5 a B e+3 A b e+2 b B d)}{b^3}+\frac {(d+e x)^{3/2} (-5 a B e+3 A b e+2 b B d)}{3 b^2 (b d-a e)}-\frac {(d+e x)^{5/2} (A b-a B)}{b (a+b x) (b d-a e)} \]

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Rubi [A]  time = 0.15, antiderivative size = 174, 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 = {78, 50, 63, 208} \begin {gather*} \frac {(d+e x)^{3/2} (-5 a B e+3 A b e+2 b B d)}{3 b^2 (b d-a e)}+\frac {\sqrt {d+e x} (-5 a B e+3 A b e+2 b B d)}{b^3}-\frac {\sqrt {b d-a e} (-5 a B e+3 A b e+2 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}-\frac {(d+e x)^{5/2} (A b-a B)}{b (a+b x) (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

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 63

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 78

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, 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)^2} \, dx &=-\frac {(A b-a B) (d+e x)^{5/2}}{b (b d-a e) (a+b x)}+\frac {(2 b B d+3 A b e-5 a B e) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{2 b (b d-a e)}\\ &=\frac {(2 b B d+3 A b e-5 a B e) (d+e x)^{3/2}}{3 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{5/2}}{b (b d-a e) (a+b x)}+\frac {(2 b B d+3 A b e-5 a B e) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{2 b^2}\\ &=\frac {(2 b B d+3 A b e-5 a B e) \sqrt {d+e x}}{b^3}+\frac {(2 b B d+3 A b e-5 a B e) (d+e x)^{3/2}}{3 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{5/2}}{b (b d-a e) (a+b x)}+\frac {((b d-a e) (2 b B d+3 A b e-5 a B e)) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{2 b^3}\\ &=\frac {(2 b B d+3 A b e-5 a B e) \sqrt {d+e x}}{b^3}+\frac {(2 b B d+3 A b e-5 a B e) (d+e x)^{3/2}}{3 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{5/2}}{b (b d-a e) (a+b x)}+\frac {((b d-a e) (2 b B d+3 A b e-5 a B e)) \operatorname {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 b B d+3 A b e-5 a B e) \sqrt {d+e x}}{b^3}+\frac {(2 b B d+3 A b e-5 a B e) (d+e x)^{3/2}}{3 b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{5/2}}{b (b d-a e) (a+b x)}-\frac {\sqrt {b d-a e} (2 b B d+3 A b e-5 a B e) \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 = 0.30, size = 136, normalized size = 0.78 \begin {gather*} \frac {\frac {(-5 a B e+3 A b e+2 b B d) \left (\sqrt {b} \sqrt {d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )\right )}{3 b^{5/2}}+\frac {(d+e x)^{5/2} (a B-A b)}{a+b x}}{b (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.46, size = 222, normalized size = 1.28 \begin {gather*} \frac {\left (-5 a^2 B e^2+3 a A b e^2+7 a b B d e-3 A b^2 d e-2 b^2 B d^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{b^{7/2} \sqrt {a e-b d}}+\frac {\sqrt {d+e x} \left (-15 a^2 B e^2+9 a A b e^2-10 a b B e (d+e x)+21 a b B d e+6 A b^2 e (d+e x)-9 A b^2 d e-6 b^2 B d^2+2 b^2 B (d+e x)^2+4 b^2 B d (d+e x)\right )}{3 b^3 (a e+b (d+e x)-b d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x]*(-6*b^2*B*d^2 - 9*A*b^2*d*e + 21*a*b*B*d*e + 9*a*A*b*e^2 - 15*a^2*B*e^2 + 4*b^2*B*d*(d + e*x) +
6*A*b^2*e*(d + e*x) - 10*a*b*B*e*(d + e*x) + 2*b^2*B*(d + e*x)^2))/(3*b^3*(-(b*d) + a*e + b*(d + e*x))) + ((-
2*b^2*B*d^2 - 3*A*b^2*d*e + 7*a*b*B*d*e + 3*a*A*b*e^2 - 5*a^2*B*e^2)*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*Sqrt[d
+ e*x])/(b*d - a*e)])/(b^(7/2)*Sqrt[-(b*d) + a*e])

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fricas [A]  time = 1.46, size = 392, normalized size = 2.25 \begin {gather*} \left [\frac {3 \, {\left (2 \, B a b d - {\left (5 \, B a^{2} - 3 \, A a b\right )} e + {\left (2 \, B b^{2} d - {\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (2 \, B b^{2} e x^{2} + {\left (11 \, B a b - 3 \, A b^{2}\right )} d - 3 \, {\left (5 \, B a^{2} - 3 \, A a b\right )} e + 2 \, {\left (4 \, B b^{2} d - {\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt {e x + d}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, -\frac {3 \, {\left (2 \, B a b d - {\left (5 \, B a^{2} - 3 \, A a b\right )} e + {\left (2 \, B b^{2} d - {\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (2 \, B b^{2} e x^{2} + {\left (11 \, B a b - 3 \, A b^{2}\right )} d - 3 \, {\left (5 \, B a^{2} - 3 \, A a b\right )} e + 2 \, {\left (4 \, B b^{2} d - {\left (5 \, B a b - 3 \, A b^{2}\right )} e\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (b^{4} x + a b^{3}\right )}}\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)^2,x, algorithm="fricas")

[Out]

[1/6*(3*(2*B*a*b*d - (5*B*a^2 - 3*A*a*b)*e + (2*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt((b*d - a*e)/b)*log((b
*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2*(2*B*b^2*e*x^2 + (11*B*a*b - 3*A*b^
2)*d - 3*(5*B*a^2 - 3*A*a*b)*e + 2*(4*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt(e*x + d))/(b^4*x + a*b^3), -1/3
*(3*(2*B*a*b*d - (5*B*a^2 - 3*A*a*b)*e + (2*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt(-(b*d - a*e)/b)*arctan(-s
qrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (2*B*b^2*e*x^2 + (11*B*a*b - 3*A*b^2)*d - 3*(5*B*a^2 - 3*A*
a*b)*e + 2*(4*B*b^2*d - (5*B*a*b - 3*A*b^2)*e)*x)*sqrt(e*x + d))/(b^4*x + a*b^3)]

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giac [A]  time = 1.34, size = 239, normalized size = 1.37 \begin {gather*} \frac {{\left (2 \, B b^{2} d^{2} - 7 \, B a b d e + 3 \, A b^{2} d e + 5 \, B a^{2} e^{2} - 3 \, A a 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 {\sqrt {x e + d} B a b d e - \sqrt {x e + d} A b^{2} d e - \sqrt {x e + d} B a^{2} e^{2} + \sqrt {x e + d} A a b e^{2}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{3}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} B b^{4} + 3 \, \sqrt {x e + d} B b^{4} d - 6 \, \sqrt {x e + d} B a b^{3} e + 3 \, \sqrt {x e + d} A b^{4} e\right )}}{3 \, b^{6}} \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)^2,x, algorithm="giac")

[Out]

(2*B*b^2*d^2 - 7*B*a*b*d*e + 3*A*b^2*d*e + 5*B*a^2*e^2 - 3*A*a*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) + (sqrt(x*e + d)*B*a*b*d*e - sqrt(x*e + d)*A*b^2*d*e - sqrt(x*e + d)*B*a^2*e^2
+ sqrt(x*e + d)*A*a*b*e^2)/(((x*e + d)*b - b*d + a*e)*b^3) + 2/3*((x*e + d)^(3/2)*B*b^4 + 3*sqrt(x*e + d)*B*b
^4*d - 6*sqrt(x*e + d)*B*a*b^3*e + 3*sqrt(x*e + d)*A*b^4*e)/b^6

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maple [B]  time = 0.02, size = 381, normalized size = 2.19 \begin {gather*} -\frac {3 A a \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}+\frac {3 A d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {5 B \,a^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}-\frac {7 B a d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}+\frac {2 B \,d^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {\sqrt {e x +d}\, A a \,e^{2}}{\left (b x e +a e \right ) b^{2}}-\frac {\sqrt {e x +d}\, A d e}{\left (b x e +a e \right ) b}-\frac {\sqrt {e x +d}\, B \,a^{2} e^{2}}{\left (b x e +a e \right ) b^{3}}+\frac {\sqrt {e x +d}\, B a d e}{\left (b x e +a e \right ) b^{2}}+\frac {2 \sqrt {e x +d}\, A e}{b^{2}}-\frac {4 \sqrt {e x +d}\, B a e}{b^{3}}+\frac {2 \sqrt {e x +d}\, B d}{b^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} B}{3 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/b^2*B*(e*x+d)^(3/2)+2/b^2*A*(e*x+d)^(1/2)*e-4/b^3*B*(e*x+d)^(1/2)*a*e+2/b^2*B*(e*x+d)^(1/2)*d+1/b^2*(e*x+d
)^(1/2)/(b*e*x+a*e)*A*a*e^2-1/b*(e*x+d)^(1/2)/(b*e*x+a*e)*A*d*e-1/b^3*(e*x+d)^(1/2)/(b*e*x+a*e)*B*a^2*e^2+1/b^
2*(e*x+d)^(1/2)/(b*e*x+a*e)*B*a*d*e-3/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*a*
e^2+3/b/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A*d*e+5/b^3/((a*e-b*d)*b)^(1/2)*arctan
((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*a^2*e^2-7/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(
1/2)*b)*B*a*d*e+2/b/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*d^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)^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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 1.32, size = 174, normalized size = 1.00 \begin {gather*} \left (\frac {2\,A\,e-2\,B\,d}{b^2}+\frac {2\,B\,\left (2\,b^2\,d-2\,a\,b\,e\right )}{b^4}\right )\,\sqrt {d+e\,x}-\frac {\sqrt {d+e\,x}\,\left (B\,a^2\,e^2-A\,a\,b\,e^2-B\,d\,a\,b\,e+A\,d\,b^2\,e\right )}{b^4\,\left (d+e\,x\right )-b^4\,d+a\,b^3\,e}+\frac {2\,B\,{\left (d+e\,x\right )}^{3/2}}{3\,b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,1{}\mathrm {i}}{\sqrt {b\,d-a\,e}}\right )\,\sqrt {b\,d-a\,e}\,\left (3\,A\,b\,e-5\,B\,a\,e+2\,B\,b\,d\right )\,1{}\mathrm {i}}{b^{7/2}} \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)^2,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**2,x)

[Out]

Timed out

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