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

Optimal. Leaf size=157 \[ \frac{2 \sqrt{a+b x} \left (\frac{a^2}{c^2}-\frac{b^2}{d^2}\right )}{\sqrt{c+d x}}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{5/2}}+\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{5/2}}-\frac{2 (a+b x)^{3/2} (b c-a d)}{3 c d (c+d x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.10863, antiderivative size = 157, 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} \[ \frac{2 \sqrt{a+b x} \left (\frac{a^2}{c^2}-\frac{b^2}{d^2}\right )}{\sqrt{c+d x}}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{5/2}}+\frac{2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{5/2}}-\frac{2 (a+b x)^{3/2} (b c-a d)}{3 c d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 (c+d x)^{5/2}} \, dx &=-\frac{2 (b c-a d) (a+b x)^{3/2}}{3 c d (c+d x)^{3/2}}+\frac{2 \int \frac{\sqrt{a+b x} \left (\frac{3 a^2 d}{2}+\frac{3}{2} b^2 c x\right )}{x (c+d x)^{3/2}} \, dx}{3 c d}\\ &=-\frac{2 (b c-a d) (a+b x)^{3/2}}{3 c d (c+d x)^{3/2}}+\frac{2 \left (\frac{a^2}{c^2}-\frac{b^2}{d^2}\right ) \sqrt{a+b x}}{\sqrt{c+d x}}-\frac{4 \int \frac{-\frac{3}{4} a^3 d^2-\frac{3}{4} b^3 c^2 x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{3 c^2 d^2}\\ &=-\frac{2 (b c-a d) (a+b x)^{3/2}}{3 c d (c+d x)^{3/2}}+\frac{2 \left (\frac{a^2}{c^2}-\frac{b^2}{d^2}\right ) \sqrt{a+b x}}{\sqrt{c+d x}}+\frac{a^3 \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{c^2}+\frac{b^3 \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{d^2}\\ &=-\frac{2 (b c-a d) (a+b x)^{3/2}}{3 c d (c+d x)^{3/2}}+\frac{2 \left (\frac{a^2}{c^2}-\frac{b^2}{d^2}\right ) \sqrt{a+b x}}{\sqrt{c+d x}}+\frac{\left (2 a^3\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{\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^2}\\ &=-\frac{2 (b c-a d) (a+b x)^{3/2}}{3 c d (c+d x)^{3/2}}+\frac{2 \left (\frac{a^2}{c^2}-\frac{b^2}{d^2}\right ) \sqrt{a+b x}}{\sqrt{c+d x}}-\frac{2 a^{5/2} \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^2}\\ &=-\frac{2 (b c-a d) (a+b x)^{3/2}}{3 c d (c+d x)^{3/2}}+\frac{2 \left (\frac{a^2}{c^2}-\frac{b^2}{d^2}\right ) \sqrt{a+b x}}{\sqrt{c+d x}}-\frac{2 a^{5/2} \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^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.00806, size = 206, normalized size = 1.31 \[ -\frac{2 (b c-a d) \left (a^2 d (4 c+3 d x)+a b \left (3 c^2+8 c d x+3 d^2 x^2\right )+b^2 c x (3 c+4 d x)\right )}{3 c^2 d^2 \sqrt{a+b x} (c+d x)^{3/2}}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{5/2}}+\frac{2 (b c-a d)^{5/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{d^{5/2} (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*c - a*d)*(a^2*d*(4*c + 3*d*x) + b^2*c*x*(3*c + 4*d*x) + a*b*(3*c^2 + 8*c*d*x + 3*d^2*x^2)))/(3*c^2*d^2*
Sqrt[a + b*x]*(c + d*x)^(3/2)) + (2*(b*c - a*d)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(5/2)*ArcSinh[(Sqrt[d]*Sqrt[
a + b*x])/Sqrt[b*c - a*d]])/(d^(5/2)*(c + d*x)^(5/2)) - (2*a^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sq
rt[c + d*x])])/c^(5/2)

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Maple [B]  time = 0.023, size = 566, normalized size = 3.6 \begin{align*}{\frac{1}{3\,{c}^{2}{d}^{2}}\sqrt{bx+a} \left ( 3\,\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}^{2}\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}^{4}\sqrt{bd}+6\,\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{b}^{3}{c}^{3}d\sqrt{ac}-6\,\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}^{3}\sqrt{bd}+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{3}{c}^{4}\sqrt{ac}-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){a}^{3}{c}^{2}{d}^{2}\sqrt{bd}+6\,x{a}^{2}{d}^{3}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}+2\,xabc{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}-8\,x{b}^{2}{c}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}+8\,{a}^{2}c{d}^{2}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}-2\,ab{c}^{2}d\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{bd}\sqrt{ac}-6\,{b}^{2}{c}^{3}\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}}} \left ( dx+c \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 16.6077, size = 2917, normalized size = 18.58 \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/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 3.03395, size = 494, normalized size = 3.15 \begin{align*} \frac{\sqrt{b d} c^{4} \log \left ({\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{32 \,{\left (b^{2} c d^{4} - a b d^{5}\right )}} + \frac{2 \, \sqrt{b d} a^{3} b \arctan \left (\frac{b^{2} c + a b d -{\left (\sqrt{b d} \sqrt{b x + a} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt{-a b c d} b}\right )}{\sqrt{-a b c d} c^{2}{\left | b \right |}} + \frac{\sqrt{b x + a}{\left (\frac{{\left (4 \, b^{8} c^{5} d^{2} - 5 \, a b^{7} c^{4} d^{3} - 2 \, a^{2} b^{6} c^{3} d^{4} + 3 \, a^{3} b^{5} c^{2} d^{5}\right )}{\left (b x + a\right )}}{b^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}} + \frac{3 \,{\left (b^{9} c^{6} d - 2 \, a b^{8} c^{5} d^{2} + 2 \, a^{3} b^{6} c^{3} d^{4} - a^{4} b^{5} c^{2} d^{5}\right )}}{b^{8} c^{2} d^{4} - 2 \, a b^{7} c d^{5} + a^{2} b^{6} d^{6}}\right )}}{48 \,{\left (b^{2} c +{\left (b x + a\right )} b d - a b d\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*sqrt(b*d)*c^4*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(b^2*c*d^4 - a*b*d^5
) + 2*sqrt(b*d)*a^3*b*arctan(1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*
d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*c^2*abs(b)) + 1/48*sqrt(b*x + a)*((4*b^8*c^5*d^2 - 5*a*b^7*c^4*d^3
- 2*a^2*b^6*c^3*d^4 + 3*a^3*b^5*c^2*d^5)*(b*x + a)/(b^8*c^2*d^4 - 2*a*b^7*c*d^5 + a^2*b^6*d^6) + 3*(b^9*c^6*d
- 2*a*b^8*c^5*d^2 + 2*a^3*b^6*c^3*d^4 - a^4*b^5*c^2*d^5)/(b^8*c^2*d^4 - 2*a*b^7*c*d^5 + a^2*b^6*d^6))/(b^2*c +
(b*x + a)*b*d - a*b*d)^(3/2)