3.1859 \(\int \frac{(A+B x) (d+e x)^{5/2}}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=289 \[ \frac{2 (a+b x) (d+e x)^{5/2} (A b-a B)}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (d+e x)^{3/2} (A b-a B) (b d-a e)}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) \sqrt{d+e x} (A b-a B) (b d-a e)^2}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (a+b x) (A b-a B) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{7/2}}{7 b e \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.222815, antiderivative size = 289, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {770, 80, 50, 63, 208} \[ \frac{2 (a+b x) (d+e x)^{5/2} (A b-a B)}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) (d+e x)^{3/2} (A b-a B) (b d-a e)}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (a+b x) \sqrt{d+e x} (A b-a B) (b d-a e)^2}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (a+b x) (A b-a B) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{7/2}}{7 b e \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^(5/2))/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 80

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 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 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)^{5/2}}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{(A+B x) (d+e x)^{5/2}}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 B (a+b x) (d+e x)^{7/2}}{7 b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (\frac{7}{2} A b^2 e-\frac{7}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{5/2}}{a b+b^2 x} \, dx}{7 b^2 e \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (A b-a B) (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{7/2}}{7 b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (b^2 d-a b e\right ) \left (\frac{7}{2} A b^2 e-\frac{7}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{a b+b^2 x} \, dx}{7 b^4 e \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (A b-a B) (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{7/2}}{7 b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (b^2 d-a b e\right )^2 \left (\frac{7}{2} A b^2 e-\frac{7}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{a b+b^2 x} \, dx}{7 b^6 e \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (A b-a B) (b d-a e)^2 (a+b x) \sqrt{d+e x}}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{7/2}}{7 b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (b^2 d-a b e\right )^3 \left (\frac{7}{2} A b^2 e-\frac{7}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{7 b^8 e \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (A b-a B) (b d-a e)^2 (a+b x) \sqrt{d+e x}}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{7/2}}{7 b e \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (4 \left (b^2 d-a b e\right )^3 \left (\frac{7}{2} A b^2 e-\frac{7}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b-\frac{b^2 d}{e}+\frac{b^2 x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{7 b^8 e^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{2 (A b-a B) (b d-a e)^2 (a+b x) \sqrt{d+e x}}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 B (a+b x) (d+e x)^{7/2}}{7 b e \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 (A b-a B) (b d-a e)^{5/2} (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.328412, size = 154, normalized size = 0.53 \[ \frac{2 (a+b x) \left (\frac{7 e (A b-a B) \left (5 (b d-a e) \left (\sqrt{b} \sqrt{d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )\right )+3 b^{5/2} (d+e x)^{5/2}\right )}{15 b^{7/2}}+B (d+e x)^{7/2}\right )}{7 b e \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^(5/2))/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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Maple [B]  time = 0.013, size = 671, normalized size = 2.3 \begin{align*} -{\frac{2\,bx+2\,a}{105\,e{b}^{4}} \left ( -15\,B\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{7/2}{b}^{3}-21\,A\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}{b}^{3}e+21\,B\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}a{b}^{2}e+105\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{3}b{e}^{4}-315\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}{b}^{2}d{e}^{3}+315\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) a{b}^{3}{d}^{2}{e}^{2}-105\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){b}^{4}{d}^{3}e+35\,A\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}a{b}^{2}{e}^{2}-35\,A\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{b}^{3}de-105\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{4}{e}^{4}+315\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{3}bd{e}^{3}-315\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}{b}^{2}{d}^{2}{e}^{2}+105\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) a{b}^{3}{d}^{3}e-35\,B\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{a}^{2}b{e}^{2}+35\,B\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}a{b}^{2}de-105\,A\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}b{e}^{3}+210\,A\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}a{b}^{2}d{e}^{2}-105\,A\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{3}{d}^{2}e+105\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{3}{e}^{3}-210\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}bd{e}^{2}+105\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}a{b}^{2}{d}^{2}e \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(5/2)/((b*x+a)^2)^(1/2),x)

[Out]

-2/105*(b*x+a)*(-15*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(7/2)*b^3-21*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*b^3*e+21*B*
((a*e-b*d)*b)^(1/2)*(e*x+d)^(5/2)*a*b^2*e+105*A*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^3*b*e^4-315*A*ar
ctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^2*b^2*d*e^3+315*A*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a*b^
3*d^2*e^2-105*A*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*b^4*d^3*e+35*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a
*b^2*e^2-35*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b^3*d*e-105*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^4*
e^4+315*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^3*b*d*e^3-315*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(
1/2))*a^2*b^2*d^2*e^2+105*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a*b^3*d^3*e-35*B*((a*e-b*d)*b)^(1/2)*(
e*x+d)^(3/2)*a^2*b*e^2+35*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b^2*d*e-105*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2
)*a^2*b*e^3+210*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b^2*d*e^2-105*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^3*d^
2*e+105*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^3*e^3-210*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*b*d*e^2+105*B*
((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b^2*d^2*e)/((b*x+a)^2)^(1/2)/e/b^4/((a*e-b*d)*b)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (e x + d\right )}^{\frac{5}{2}}}{\sqrt{{\left (b x + a\right )}^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)*(e*x + d)^(5/2)/sqrt((b*x + a)^2), x)

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Fricas [A]  time = 1.52296, size = 1247, normalized size = 4.31 \begin{align*} \left [\frac{105 \,{\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} e - 2 \,{\left (B a^{2} b - A a b^{2}\right )} d e^{2} +{\left (B a^{3} - A a^{2} b\right )} e^{3}\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 (15 \, B b^{3} e^{3} x^{3} + 15 \, B b^{3} d^{3} - 161 \,{\left (B a b^{2} - A b^{3}\right )} d^{2} e + 245 \,{\left (B a^{2} b - A a b^{2}\right )} d e^{2} - 105 \,{\left (B a^{3} - A a^{2} b\right )} e^{3} + 3 \,{\left (15 \, B b^{3} d e^{2} - 7 \,{\left (B a b^{2} - A b^{3}\right )} e^{3}\right )} x^{2} +{\left (45 \, B b^{3} d^{2} e - 77 \,{\left (B a b^{2} - A b^{3}\right )} d e^{2} + 35 \,{\left (B a^{2} b - A a b^{2}\right )} e^{3}\right )} x\right )} \sqrt{e x + d}}{105 \, b^{4} e}, \frac{2 \,{\left (105 \,{\left ({\left (B a b^{2} - A b^{3}\right )} d^{2} e - 2 \,{\left (B a^{2} b - A a b^{2}\right )} d e^{2} +{\left (B a^{3} - A a^{2} b\right )} e^{3}\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 (15 \, B b^{3} e^{3} x^{3} + 15 \, B b^{3} d^{3} - 161 \,{\left (B a b^{2} - A b^{3}\right )} d^{2} e + 245 \,{\left (B a^{2} b - A a b^{2}\right )} d e^{2} - 105 \,{\left (B a^{3} - A a^{2} b\right )} e^{3} + 3 \,{\left (15 \, B b^{3} d e^{2} - 7 \,{\left (B a b^{2} - A b^{3}\right )} e^{3}\right )} x^{2} +{\left (45 \, B b^{3} d^{2} e - 77 \,{\left (B a b^{2} - A b^{3}\right )} d e^{2} + 35 \,{\left (B a^{2} b - A a b^{2}\right )} e^{3}\right )} x\right )} \sqrt{e x + d}\right )}}{105 \, b^{4} e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/105*(105*((B*a*b^2 - A*b^3)*d^2*e - 2*(B*a^2*b - A*a*b^2)*d*e^2 + (B*a^3 - A*a^2*b)*e^3)*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*(15*B*b^3*e^3*x^3 + 15*B*b^
3*d^3 - 161*(B*a*b^2 - A*b^3)*d^2*e + 245*(B*a^2*b - A*a*b^2)*d*e^2 - 105*(B*a^3 - A*a^2*b)*e^3 + 3*(15*B*b^3*
d*e^2 - 7*(B*a*b^2 - A*b^3)*e^3)*x^2 + (45*B*b^3*d^2*e - 77*(B*a*b^2 - A*b^3)*d*e^2 + 35*(B*a^2*b - A*a*b^2)*e
^3)*x)*sqrt(e*x + d))/(b^4*e), 2/105*(105*((B*a*b^2 - A*b^3)*d^2*e - 2*(B*a^2*b - A*a*b^2)*d*e^2 + (B*a^3 - A*
a^2*b)*e^3)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) + (15*B*b^3*e^3*x^3
+ 15*B*b^3*d^3 - 161*(B*a*b^2 - A*b^3)*d^2*e + 245*(B*a^2*b - A*a*b^2)*d*e^2 - 105*(B*a^3 - A*a^2*b)*e^3 + 3*
(15*B*b^3*d*e^2 - 7*(B*a*b^2 - A*b^3)*e^3)*x^2 + (45*B*b^3*d^2*e - 77*(B*a*b^2 - A*b^3)*d*e^2 + 35*(B*a^2*b -
A*a*b^2)*e^3)*x)*sqrt(e*x + d))/(b^4*e)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(5/2)/((b*x+a)**2)**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 1.21447, size = 671, normalized size = 2.32 \begin{align*} -\frac{2 \,{\left (B a b^{3} d^{3} \mathrm{sgn}\left (b x + a\right ) - A b^{4} d^{3} \mathrm{sgn}\left (b x + a\right ) - 3 \, B a^{2} b^{2} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 3 \, A a b^{3} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 3 \, B a^{3} b d e^{2} \mathrm{sgn}\left (b x + a\right ) - 3 \, A a^{2} b^{2} d e^{2} \mathrm{sgn}\left (b x + a\right ) - B a^{4} e^{3} \mathrm{sgn}\left (b x + a\right ) + A a^{3} b e^{3} \mathrm{sgn}\left (b x + a\right )\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^{4}} + \frac{2 \,{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} B b^{6} e^{6} \mathrm{sgn}\left (b x + a\right ) - 21 \,{\left (x e + d\right )}^{\frac{5}{2}} B a b^{5} e^{7} \mathrm{sgn}\left (b x + a\right ) + 21 \,{\left (x e + d\right )}^{\frac{5}{2}} A b^{6} e^{7} \mathrm{sgn}\left (b x + a\right ) - 35 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{5} d e^{7} \mathrm{sgn}\left (b x + a\right ) + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{6} d e^{7} \mathrm{sgn}\left (b x + a\right ) - 105 \, \sqrt{x e + d} B a b^{5} d^{2} e^{7} \mathrm{sgn}\left (b x + a\right ) + 105 \, \sqrt{x e + d} A b^{6} d^{2} e^{7} \mathrm{sgn}\left (b x + a\right ) + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{2} b^{4} e^{8} \mathrm{sgn}\left (b x + a\right ) - 35 \,{\left (x e + d\right )}^{\frac{3}{2}} A a b^{5} e^{8} \mathrm{sgn}\left (b x + a\right ) + 210 \, \sqrt{x e + d} B a^{2} b^{4} d e^{8} \mathrm{sgn}\left (b x + a\right ) - 210 \, \sqrt{x e + d} A a b^{5} d e^{8} \mathrm{sgn}\left (b x + a\right ) - 105 \, \sqrt{x e + d} B a^{3} b^{3} e^{9} \mathrm{sgn}\left (b x + a\right ) + 105 \, \sqrt{x e + d} A a^{2} b^{4} e^{9} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-7\right )}}{105 \, b^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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