3.686 \(\int \frac{(a+b x)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=164 \[ -\frac{a^{3/2} (5 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{5/2}}+\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{3/2}}-\frac{\sqrt{a+b x} (2 b c-3 a d) (b c-a d)}{c^2 d \sqrt{c+d x}}-\frac{a (a+b x)^{3/2}}{c x \sqrt{c+d x}} \]

[Out]

-(((2*b*c - 3*a*d)*(b*c - a*d)*Sqrt[a + b*x])/(c^2*d*Sqrt[c + d*x])) - (a*(a + b*x)^(3/2))/(c*x*Sqrt[c + d*x])
- (a^(3/2)*(5*b*c - 3*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/c^(5/2) + (2*b^(5/2)*Arc
Tanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/d^(3/2)

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Rubi [A]  time = 0.157656, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {98, 150, 157, 63, 217, 206, 93, 208} \[ -\frac{a^{3/2} (5 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{5/2}}+\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{3/2}}-\frac{\sqrt{a+b x} (2 b c-3 a d) (b c-a d)}{c^2 d \sqrt{c+d x}}-\frac{a (a+b x)^{3/2}}{c x \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/(x^2*(c + d*x)^(3/2)),x]

[Out]

-(((2*b*c - 3*a*d)*(b*c - a*d)*Sqrt[a + b*x])/(c^2*d*Sqrt[c + d*x])) - (a*(a + b*x)^(3/2))/(c*x*Sqrt[c + d*x])
- (a^(3/2)*(5*b*c - 3*a*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/c^(5/2) + (2*b^(5/2)*Arc
Tanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/d^(3/2)

Rule 98

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

Rule 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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)^{5/2}}{x^2 (c+d x)^{3/2}} \, dx &=-\frac{a (a+b x)^{3/2}}{c x \sqrt{c+d x}}-\frac{\int \frac{\sqrt{a+b x} \left (-\frac{1}{2} a (5 b c-3 a d)-b^2 c x\right )}{x (c+d x)^{3/2}} \, dx}{c}\\ &=-\frac{(2 b c-3 a d) (b c-a d) \sqrt{a+b x}}{c^2 d \sqrt{c+d x}}-\frac{a (a+b x)^{3/2}}{c x \sqrt{c+d x}}+\frac{2 \int \frac{\frac{1}{4} a^2 d (5 b c-3 a d)+\frac{1}{2} b^3 c^2 x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{c^2 d}\\ &=-\frac{(2 b c-3 a d) (b c-a d) \sqrt{a+b x}}{c^2 d \sqrt{c+d x}}-\frac{a (a+b x)^{3/2}}{c x \sqrt{c+d x}}+\frac{b^3 \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{d}+\frac{\left (a^2 (5 b c-3 a d)\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 c^2}\\ &=-\frac{(2 b c-3 a d) (b c-a d) \sqrt{a+b x}}{c^2 d \sqrt{c+d x}}-\frac{a (a+b x)^{3/2}}{c x \sqrt{c+d x}}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{d}+\frac{\left (a^2 (5 b c-3 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{c^2}\\ &=-\frac{(2 b c-3 a d) (b c-a d) \sqrt{a+b x}}{c^2 d \sqrt{c+d x}}-\frac{a (a+b x)^{3/2}}{c x \sqrt{c+d x}}-\frac{a^{3/2} (5 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{5/2}}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{d}\\ &=-\frac{(2 b c-3 a d) (b c-a d) \sqrt{a+b x}}{c^2 d \sqrt{c+d x}}-\frac{a (a+b x)^{3/2}}{c x \sqrt{c+d x}}-\frac{a^{3/2} (5 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{5/2}}+\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{3/2}}\\ \end{align*}

Mathematica [C]  time = 4.66447, size = 333, normalized size = 2.03 \[ \frac{-\frac{5 \sqrt{c} \sqrt{a+b x} \left (-2 a^2 b c d^2 x+2 a^3 d^2 (c+3 d x)-a b^2 c d x (7 c+3 d x)+b^3 c^2 x (3 c+d x)\right )}{d^2 x}+\frac{5 b c^{3/2} \sqrt{b c-a d} \left (9 a^2 d^2-8 a b c d+3 b^2 c^2\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{d^{5/2}}+10 a^{5/2} \sqrt{c+d x} (3 a d-5 b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{2 c^{3/2} (a+b x)^{5/2} (b c-3 a d) \left (\frac{b (c+d x)}{b c-a d}\right )^{3/2} \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};\frac{d (a+b x)}{a d-b c}\right )}{c+d x}}{10 a c^{5/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/(x^2*(c + d*x)^(3/2)),x]

[Out]

((-5*Sqrt[c]*Sqrt[a + b*x]*(-2*a^2*b*c*d^2*x + b^3*c^2*x*(3*c + d*x) + 2*a^3*d^2*(c + 3*d*x) - a*b^2*c*d*x*(7*
c + 3*d*x)))/(d^2*x) + (5*b*c^(3/2)*Sqrt[b*c - a*d]*(3*b^2*c^2 - 8*a*b*c*d + 9*a^2*d^2)*Sqrt[(b*(c + d*x))/(b*
c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/d^(5/2) + 10*a^(5/2)*(-5*b*c + 3*a*d)*Sqrt[c + d*x
]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + (2*c^(3/2)*(b*c - 3*a*d)*(a + b*x)^(5/2)*((b*(c +
d*x))/(b*c - a*d))^(3/2)*Hypergeometric2F1[3/2, 5/2, 7/2, (d*(a + b*x))/(-(b*c) + a*d)])/(c + d*x))/(10*a*c^(
5/2)*Sqrt[c + d*x])

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Maple [B]  time = 0.022, size = 502, normalized size = 3.1 \begin{align*}{\frac{1}{2\,{c}^{2}xd}\sqrt{bx+a} \left ( 2\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{b}^{3}{c}^{2}d\sqrt{ac}+3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{a}^{3}{d}^{3}\sqrt{bd}-5\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{a}^{2}bc{d}^{2}\sqrt{bd}+2\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) \sqrt{ac}x{b}^{3}{c}^{3}+3\,\sqrt{bd}\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) x{a}^{3}c{d}^{2}-5\,\sqrt{bd}\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) x{a}^{2}b{c}^{2}d-6\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}x{a}^{2}{d}^{2}+8\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}xabcd-4\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}x{b}^{2}{c}^{2}-2\,{a}^{2}cd\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),x)

[Out]

1/2*(b*x+a)^(1/2)*(2*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*b^3*c^2*d
*(a*c)^(1/2)+3*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a^3*d^3*(b*d)^(1/2)-5*ln((a
*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^2*a^2*b*c*d^2*(b*d)^(1/2)+2*ln(1/2*(2*b*d*x+2*((b
*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)*x*b^3*c^3+3*(b*d)^(1/2)*ln((a*d*x+b*c*x+2*(
a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x*a^3*c*d^2-5*(b*d)^(1/2)*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*
(d*x+c))^(1/2)+2*a*c)/x)*x*a^2*b*c^2*d-6*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x*a^2*d^2+8*((b*x+a)*
(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x*a*b*c*d-4*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x*b^2*c^2-2
*a^2*c*d*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/c^2/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(a*c)^(1/2)/
x/(d*x+c)^(1/2)/d

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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((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 13.7803, size = 2688, normalized size = 16.39 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(2*(b^2*c^2*d*x^2 + b^2*c^3*x)*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d^2*x
+ b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) - ((5*a*b*c*d^2 - 3*a^2*d^3
)*x^2 + (5*a*b*c^2*d - 3*a^2*c*d^2)*x)*sqrt(a/c)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 + 4*(2*a
*c^2 + (b*c^2 + a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(a^2*c*d +
(2*b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^2*d^2*x^2 + c^3*d*x), -1/4*(4*(b^2*c^2
*d*x^2 + b^2*c^3*x)*sqrt(-b/d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*
x^2 + a*b*c + (b^2*c + a*b*d)*x)) + ((5*a*b*c*d^2 - 3*a^2*d^3)*x^2 + (5*a*b*c^2*d - 3*a^2*c*d^2)*x)*sqrt(a/c)*
log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 + 4*(2*a*c^2 + (b*c^2 + a*c*d)*x)*sqrt(b*x + a)*sqrt(d*x
+ c)*sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*(a^2*c*d + (2*b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x)*sqrt(b*x
+ a)*sqrt(d*x + c))/(c^2*d^2*x^2 + c^3*d*x), 1/2*(((5*a*b*c*d^2 - 3*a^2*d^3)*x^2 + (5*a*b*c^2*d - 3*a^2*c*d^2)
*x)*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-a/c)/(a*b*d*x^2 + a^2*c +
(a*b*c + a^2*d)*x)) + (b^2*c^2*d*x^2 + b^2*c^3*x)*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2
+ 4*(2*b*d^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)*x) - 2*(a^2*c*d
+ (2*b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^2*d^2*x^2 + c^3*d*x), 1/2*(((5*a*b*c*
d^2 - 3*a^2*d^3)*x^2 + (5*a*b*c^2*d - 3*a^2*c*d^2)*x)*sqrt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x +
a)*sqrt(d*x + c)*sqrt(-a/c)/(a*b*d*x^2 + a^2*c + (a*b*c + a^2*d)*x)) - 2*(b^2*c^2*d*x^2 + b^2*c^3*x)*sqrt(-b/
d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d
)*x)) - 2*(a^2*c*d + (2*b^2*c^2 - 4*a*b*c*d + 3*a^2*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c^2*d^2*x^2 + c^3*d*
x)]

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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)**(5/2)/x**2/(d*x+c)**(3/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^2/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError