3.537 \(\int \frac{x^{7/2} (A+B x)}{(a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=200 \[ \frac{35 a^2 (2 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^{11/2}}+\frac{2 x^{7/2} (2 A b-3 a B)}{a b^2 \sqrt{a+b x}}-\frac{7 x^{5/2} \sqrt{a+b x} (2 A b-3 a B)}{3 a b^3}+\frac{35 x^{3/2} \sqrt{a+b x} (2 A b-3 a B)}{12 b^4}-\frac{35 a \sqrt{x} \sqrt{a+b x} (2 A b-3 a B)}{8 b^5}+\frac{2 x^{9/2} (A b-a B)}{3 a b (a+b x)^{3/2}} \]

[Out]

(2*(A*b - a*B)*x^(9/2))/(3*a*b*(a + b*x)^(3/2)) + (2*(2*A*b - 3*a*B)*x^(7/2))/(a*b^2*Sqrt[a + b*x]) - (35*a*(2
*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(8*b^5) + (35*(2*A*b - 3*a*B)*x^(3/2)*Sqrt[a + b*x])/(12*b^4) - (7*(2*A*b
 - 3*a*B)*x^(5/2)*Sqrt[a + b*x])/(3*a*b^3) + (35*a^2*(2*A*b - 3*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])
/(8*b^(11/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0838018, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {78, 47, 50, 63, 217, 206} \[ \frac{35 a^2 (2 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^{11/2}}+\frac{2 x^{7/2} (2 A b-3 a B)}{a b^2 \sqrt{a+b x}}-\frac{7 x^{5/2} \sqrt{a+b x} (2 A b-3 a B)}{3 a b^3}+\frac{35 x^{3/2} \sqrt{a+b x} (2 A b-3 a B)}{12 b^4}-\frac{35 a \sqrt{x} \sqrt{a+b x} (2 A b-3 a B)}{8 b^5}+\frac{2 x^{9/2} (A b-a B)}{3 a b (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^(7/2)*(A + B*x))/(a + b*x)^(5/2),x]

[Out]

(2*(A*b - a*B)*x^(9/2))/(3*a*b*(a + b*x)^(3/2)) + (2*(2*A*b - 3*a*B)*x^(7/2))/(a*b^2*Sqrt[a + b*x]) - (35*a*(2
*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(8*b^5) + (35*(2*A*b - 3*a*B)*x^(3/2)*Sqrt[a + b*x])/(12*b^4) - (7*(2*A*b
 - 3*a*B)*x^(5/2)*Sqrt[a + b*x])/(3*a*b^3) + (35*a^2*(2*A*b - 3*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])
/(8*b^(11/2))

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^{7/2} (A+B x)}{(a+b x)^{5/2}} \, dx &=\frac{2 (A b-a B) x^{9/2}}{3 a b (a+b x)^{3/2}}-\frac{\left (2 \left (3 A b-\frac{9 a B}{2}\right )\right ) \int \frac{x^{7/2}}{(a+b x)^{3/2}} \, dx}{3 a b}\\ &=\frac{2 (A b-a B) x^{9/2}}{3 a b (a+b x)^{3/2}}+\frac{2 (2 A b-3 a B) x^{7/2}}{a b^2 \sqrt{a+b x}}-\frac{(7 (2 A b-3 a B)) \int \frac{x^{5/2}}{\sqrt{a+b x}} \, dx}{a b^2}\\ &=\frac{2 (A b-a B) x^{9/2}}{3 a b (a+b x)^{3/2}}+\frac{2 (2 A b-3 a B) x^{7/2}}{a b^2 \sqrt{a+b x}}-\frac{7 (2 A b-3 a B) x^{5/2} \sqrt{a+b x}}{3 a b^3}+\frac{(35 (2 A b-3 a B)) \int \frac{x^{3/2}}{\sqrt{a+b x}} \, dx}{6 b^3}\\ &=\frac{2 (A b-a B) x^{9/2}}{3 a b (a+b x)^{3/2}}+\frac{2 (2 A b-3 a B) x^{7/2}}{a b^2 \sqrt{a+b x}}+\frac{35 (2 A b-3 a B) x^{3/2} \sqrt{a+b x}}{12 b^4}-\frac{7 (2 A b-3 a B) x^{5/2} \sqrt{a+b x}}{3 a b^3}-\frac{(35 a (2 A b-3 a B)) \int \frac{\sqrt{x}}{\sqrt{a+b x}} \, dx}{8 b^4}\\ &=\frac{2 (A b-a B) x^{9/2}}{3 a b (a+b x)^{3/2}}+\frac{2 (2 A b-3 a B) x^{7/2}}{a b^2 \sqrt{a+b x}}-\frac{35 a (2 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{8 b^5}+\frac{35 (2 A b-3 a B) x^{3/2} \sqrt{a+b x}}{12 b^4}-\frac{7 (2 A b-3 a B) x^{5/2} \sqrt{a+b x}}{3 a b^3}+\frac{\left (35 a^2 (2 A b-3 a B)\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx}{16 b^5}\\ &=\frac{2 (A b-a B) x^{9/2}}{3 a b (a+b x)^{3/2}}+\frac{2 (2 A b-3 a B) x^{7/2}}{a b^2 \sqrt{a+b x}}-\frac{35 a (2 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{8 b^5}+\frac{35 (2 A b-3 a B) x^{3/2} \sqrt{a+b x}}{12 b^4}-\frac{7 (2 A b-3 a B) x^{5/2} \sqrt{a+b x}}{3 a b^3}+\frac{\left (35 a^2 (2 A b-3 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )}{8 b^5}\\ &=\frac{2 (A b-a B) x^{9/2}}{3 a b (a+b x)^{3/2}}+\frac{2 (2 A b-3 a B) x^{7/2}}{a b^2 \sqrt{a+b x}}-\frac{35 a (2 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{8 b^5}+\frac{35 (2 A b-3 a B) x^{3/2} \sqrt{a+b x}}{12 b^4}-\frac{7 (2 A b-3 a B) x^{5/2} \sqrt{a+b x}}{3 a b^3}+\frac{\left (35 a^2 (2 A b-3 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^5}\\ &=\frac{2 (A b-a B) x^{9/2}}{3 a b (a+b x)^{3/2}}+\frac{2 (2 A b-3 a B) x^{7/2}}{a b^2 \sqrt{a+b x}}-\frac{35 a (2 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{8 b^5}+\frac{35 (2 A b-3 a B) x^{3/2} \sqrt{a+b x}}{12 b^4}-\frac{7 (2 A b-3 a B) x^{5/2} \sqrt{a+b x}}{3 a b^3}+\frac{35 a^2 (2 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0536763, size = 80, normalized size = 0.4 \[ \frac{2 x^{9/2} \left ((a+b x) \sqrt{\frac{b x}{a}+1} (3 a B-2 A b) \, _2F_1\left (\frac{3}{2},\frac{9}{2};\frac{11}{2};-\frac{b x}{a}\right )+3 a (A b-a B)\right )}{9 a^2 b (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(7/2)*(A + B*x))/(a + b*x)^(5/2),x]

[Out]

(2*x^(9/2)*(3*a*(A*b - a*B) + (-2*A*b + 3*a*B)*(a + b*x)*Sqrt[1 + (b*x)/a]*Hypergeometric2F1[3/2, 9/2, 11/2, -
((b*x)/a)]))/(9*a^2*b*(a + b*x)^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.018, size = 406, normalized size = 2. \begin{align*}{\frac{1}{48} \left ( 16\,B{x}^{4}{b}^{9/2}\sqrt{x \left ( bx+a \right ) }+24\,A{x}^{3}{b}^{9/2}\sqrt{x \left ( bx+a \right ) }-36\,B{x}^{3}a{b}^{7/2}\sqrt{x \left ( bx+a \right ) }+210\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){x}^{2}{a}^{2}{b}^{3}-84\,A{x}^{2}a{b}^{7/2}\sqrt{x \left ( bx+a \right ) }-315\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){x}^{2}{a}^{3}{b}^{2}+126\,B{x}^{2}{a}^{2}{b}^{5/2}\sqrt{x \left ( bx+a \right ) }+420\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) x{a}^{3}{b}^{2}-560\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}x{a}^{2}-630\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) x{a}^{4}b+840\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}x{a}^{3}+210\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{4}b-420\,A\sqrt{x \left ( bx+a \right ) }{b}^{3/2}{a}^{3}-315\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{5}+630\,B\sqrt{x \left ( bx+a \right ) }\sqrt{b}{a}^{4} \right ) \sqrt{x}{b}^{-{\frac{11}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}} \left ( bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x+A)/(b*x+a)^(5/2),x)

[Out]

1/48*(16*B*x^4*b^(9/2)*(x*(b*x+a))^(1/2)+24*A*x^3*b^(9/2)*(x*(b*x+a))^(1/2)-36*B*x^3*a*b^(7/2)*(x*(b*x+a))^(1/
2)+210*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x^2*a^2*b^3-84*A*x^2*a*b^(7/2)*(x*(b*x+a))^(1/2
)-315*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x^2*a^3*b^2+126*B*x^2*a^2*b^(5/2)*(x*(b*x+a))^(1
/2)+420*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x*a^3*b^2-560*A*(x*(b*x+a))^(1/2)*b^(5/2)*x*a^
2-630*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x*a^4*b+840*B*(x*(b*x+a))^(1/2)*b^(3/2)*x*a^3+21
0*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^4*b-420*A*(x*(b*x+a))^(1/2)*b^(3/2)*a^3-315*B*ln(1
/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^5+630*B*(x*(b*x+a))^(1/2)*b^(1/2)*a^4)/b^(11/2)*x^(1/2)/(x
*(b*x+a))^(1/2)/(b*x+a)^(3/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.74763, size = 969, normalized size = 4.84 \begin{align*} \left [-\frac{105 \,{\left (3 \, B a^{5} - 2 \, A a^{4} b +{\left (3 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x^{2} + 2 \,{\left (3 \, B a^{4} b - 2 \, A a^{3} b^{2}\right )} x\right )} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (8 \, B b^{5} x^{4} + 315 \, B a^{4} b - 210 \, A a^{3} b^{2} - 6 \,{\left (3 \, B a b^{4} - 2 \, A b^{5}\right )} x^{3} + 21 \,{\left (3 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} x^{2} + 140 \,{\left (3 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{48 \,{\left (b^{8} x^{2} + 2 \, a b^{7} x + a^{2} b^{6}\right )}}, \frac{105 \,{\left (3 \, B a^{5} - 2 \, A a^{4} b +{\left (3 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x^{2} + 2 \,{\left (3 \, B a^{4} b - 2 \, A a^{3} b^{2}\right )} x\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (8 \, B b^{5} x^{4} + 315 \, B a^{4} b - 210 \, A a^{3} b^{2} - 6 \,{\left (3 \, B a b^{4} - 2 \, A b^{5}\right )} x^{3} + 21 \,{\left (3 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} x^{2} + 140 \,{\left (3 \, B a^{3} b^{2} - 2 \, A a^{2} b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{24 \,{\left (b^{8} x^{2} + 2 \, a b^{7} x + a^{2} b^{6}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[-1/48*(105*(3*B*a^5 - 2*A*a^4*b + (3*B*a^3*b^2 - 2*A*a^2*b^3)*x^2 + 2*(3*B*a^4*b - 2*A*a^3*b^2)*x)*sqrt(b)*lo
g(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(8*B*b^5*x^4 + 315*B*a^4*b - 210*A*a^3*b^2 - 6*(3*B*a*b^4 -
 2*A*b^5)*x^3 + 21*(3*B*a^2*b^3 - 2*A*a*b^4)*x^2 + 140*(3*B*a^3*b^2 - 2*A*a^2*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(
b^8*x^2 + 2*a*b^7*x + a^2*b^6), 1/24*(105*(3*B*a^5 - 2*A*a^4*b + (3*B*a^3*b^2 - 2*A*a^2*b^3)*x^2 + 2*(3*B*a^4*
b - 2*A*a^3*b^2)*x)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*B*b^5*x^4 + 315*B*a^4*b - 210*A*a
^3*b^2 - 6*(3*B*a*b^4 - 2*A*b^5)*x^3 + 21*(3*B*a^2*b^3 - 2*A*a*b^4)*x^2 + 140*(3*B*a^3*b^2 - 2*A*a^2*b^3)*x)*s
qrt(b*x + a)*sqrt(x))/(b^8*x^2 + 2*a*b^7*x + a^2*b^6)]

________________________________________________________________________________________

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(x**(7/2)*(B*x+A)/(b*x+a)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 86.9746, size = 516, normalized size = 2.58 \begin{align*} \frac{1}{24} \, \sqrt{{\left (b x + a\right )} b - a b} \sqrt{b x + a}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )} B{\left | b \right |}}{b^{7}} - \frac{25 \, B a b^{20}{\left | b \right |} - 6 \, A b^{21}{\left | b \right |}}{b^{27}}\right )} + \frac{3 \,{\left (55 \, B a^{2} b^{20}{\left | b \right |} - 26 \, A a b^{21}{\left | b \right |}\right )}}{b^{27}}\right )} + \frac{35 \,{\left (3 \, B a^{3} \sqrt{b}{\left | b \right |} - 2 \, A a^{2} b^{\frac{3}{2}}{\left | b \right |}\right )} \log \left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{16 \, b^{7}} + \frac{4 \,{\left (15 \, B a^{4}{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{4} \sqrt{b}{\left | b \right |} + 24 \, B a^{5}{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac{3}{2}}{\left | b \right |} - 12 \, A a^{3}{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{4} b^{\frac{3}{2}}{\left | b \right |} + 13 \, B a^{6} b^{\frac{5}{2}}{\left | b \right |} - 18 \, A a^{4}{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac{5}{2}}{\left | b \right |} - 10 \, A a^{5} b^{\frac{7}{2}}{\left | b \right |}\right )}}{3 \,{\left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3} b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

1/24*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)*B*abs(b)/b^7 - (25*B*a*b^20*abs(b) - 6*A*
b^21*abs(b))/b^27) + 3*(55*B*a^2*b^20*abs(b) - 26*A*a*b^21*abs(b))/b^27) + 35/16*(3*B*a^3*sqrt(b)*abs(b) - 2*A
*a^2*b^(3/2)*abs(b))*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^7 + 4/3*(15*B*a^4*(sqrt(b*x +
a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*sqrt(b)*abs(b) + 24*B*a^5*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b -
a*b))^2*b^(3/2)*abs(b) - 12*A*a^3*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*b^(3/2)*abs(b) + 13*B*a^
6*b^(5/2)*abs(b) - 18*A*a^4*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(5/2)*abs(b) - 10*A*a^5*b^(7
/2)*abs(b))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)^3*b^6)