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

Optimal. Leaf size=212 \[ -\frac{\sqrt{a} \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{3/2}}+\frac{b^{3/2} (5 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{d}}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{2 x^2}-\frac{(a+b x)^{3/2} \sqrt{c+d x} (a d+5 b c)}{4 c x}+\frac{b \sqrt{a+b x} \sqrt{c+d x} (a d+11 b c)}{4 c} \]

[Out]

(b*(11*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*c) - ((5*b*c + a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(4*c*x) -
((a + b*x)^(5/2)*Sqrt[c + d*x])/(2*x^2) - (Sqrt[a]*(15*b^2*c^2 + 10*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[
a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*c^(3/2)) + (b^(3/2)*(b*c + 5*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt
[b]*Sqrt[c + d*x])])/Sqrt[d]

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Rubi [A]  time = 0.183564, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {97, 149, 154, 157, 63, 217, 206, 93, 208} \[ -\frac{\sqrt{a} \left (-a^2 d^2+10 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{3/2}}+\frac{b^{3/2} (5 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{d}}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{2 x^2}-\frac{(a+b x)^{3/2} \sqrt{c+d x} (a d+5 b c)}{4 c x}+\frac{b \sqrt{a+b x} \sqrt{c+d x} (a d+11 b c)}{4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(11*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*c) - ((5*b*c + a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(4*c*x) -
((a + b*x)^(5/2)*Sqrt[c + d*x])/(2*x^2) - (Sqrt[a]*(15*b^2*c^2 + 10*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[
a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/(4*c^(3/2)) + (b^(3/2)*(b*c + 5*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt
[b]*Sqrt[c + d*x])])/Sqrt[d]

Rule 97

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

Rule 149

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] && IntegerQ[m]

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 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} \sqrt{c+d x}}{x^3} \, dx &=-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{2 x^2}+\frac{1}{2} \int \frac{(a+b x)^{3/2} \left (\frac{1}{2} (5 b c+a d)+3 b d x\right )}{x^2 \sqrt{c+d x}} \, dx\\ &=-\frac{(5 b c+a d) (a+b x)^{3/2} \sqrt{c+d x}}{4 c x}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{2 x^2}+\frac{\int \frac{\sqrt{a+b x} \left (\frac{1}{4} \left (15 b^2 c^2+10 a b c d-a^2 d^2\right )+\frac{1}{2} b d (11 b c+a d) x\right )}{x \sqrt{c+d x}} \, dx}{2 c}\\ &=\frac{b (11 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c}-\frac{(5 b c+a d) (a+b x)^{3/2} \sqrt{c+d x}}{4 c x}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{2 x^2}+\frac{\int \frac{\frac{1}{4} a d \left (15 b^2 c^2+10 a b c d-a^2 d^2\right )+b^2 c d (b c+5 a d) x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 c d}\\ &=\frac{b (11 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c}-\frac{(5 b c+a d) (a+b x)^{3/2} \sqrt{c+d x}}{4 c x}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{2 x^2}+\frac{1}{2} \left (b^2 (b c+5 a d)\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx+\frac{\left (a \left (15 b^2 c^2+10 a b c d-a^2 d^2\right )\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 c}\\ &=\frac{b (11 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c}-\frac{(5 b c+a d) (a+b x)^{3/2} \sqrt{c+d x}}{4 c x}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{2 x^2}+(b (b c+5 a d)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )+\frac{\left (a \left (15 b^2 c^2+10 a b c d-a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{4 c}\\ &=\frac{b (11 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c}-\frac{(5 b c+a d) (a+b x)^{3/2} \sqrt{c+d x}}{4 c x}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{2 x^2}-\frac{\sqrt{a} \left (15 b^2 c^2+10 a b c d-a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{3/2}}+(b (b c+5 a d)) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )\\ &=\frac{b (11 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 c}-\frac{(5 b c+a d) (a+b x)^{3/2} \sqrt{c+d x}}{4 c x}-\frac{(a+b x)^{5/2} \sqrt{c+d x}}{2 x^2}-\frac{\sqrt{a} \left (15 b^2 c^2+10 a b c d-a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{3/2}}+\frac{b^{3/2} (b c+5 a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{\sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 1.8038, size = 206, normalized size = 0.97 \[ \frac{\sqrt{a} \left (a^2 d^2-10 a b c d-15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{4 c^{3/2}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (a^2 (2 c+d x)+9 a b c x-4 b^2 c x^2\right )}{4 c x^2}+\frac{(b c-a d)^{3/2} (5 a d+b c) \left (\frac{b (c+d x)}{b c-a d}\right )^{3/2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{\sqrt{d} (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a + b*x]*Sqrt[c + d*x]*(9*a*b*c*x - 4*b^2*c*x^2 + a^2*(2*c + d*x)))/(4*c*x^2) + ((b*c - a*d)^(3/2)*(b*c
+ 5*a*d)*((b*(c + d*x))/(b*c - a*d))^(3/2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(Sqrt[d]*(c + d*
x)^(3/2)) + (Sqrt[a]*(-15*b^2*c^2 - 10*a*b*c*d + a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*
x])])/(4*c^(3/2))

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Maple [B]  time = 0.014, size = 511, normalized size = 2.4 \begin{align*}{\frac{1}{8\,c{x}^{2}}\sqrt{bx+a}\sqrt{dx+c} \left ( 20\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}a{b}^{2}cd\sqrt{ac}+4\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{b}^{3}{c}^{2}\sqrt{ac}+\ln \left ({\frac{1}{x} \left ( adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac \right ) } \right ){x}^{2}{a}^{3}{d}^{2}\sqrt{bd}-10\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{2}{a}^{2}bcd\sqrt{bd}-15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){x}^{2}a{b}^{2}{c}^{2}\sqrt{bd}+8\,{x}^{2}{b}^{2}c\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}-2\,x{a}^{2}d\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}-18\,xabc\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}-4\,{a}^{2}c\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac} \right ){\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(d*x+c)^(1/2)/x^3,x)

[Out]

1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)/c*(20*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(
b*d)^(1/2))*x^2*a*b^2*c*d*(a*c)^(1/2)+4*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)
/(b*d)^(1/2))*x^2*b^3*c^2*(a*c)^(1/2)+ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*
x^2*a^3*d^2*(b*d)^(1/2)-10*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^2*a^2*b*c
*d*(b*d)^(1/2)-15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)+2*a*c)/x)*x^2*a*b^2*c^2*(b*d)^
(1/2)+8*x^2*b^2*c*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-2*x*a^2*d*(b*d*x^2+a*d*x+b*c*x+a*c)^
(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-18*x*a*b*c*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)-4*a^2*c*(b*d*
x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)/x^2/(b*d)^(1/2)/(a*c)^(1/2
)

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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)*(d*x+c)^(1/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 16.3228, size = 2445, normalized size = 11.53 \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)*(d*x+c)^(1/2)/x^3,x, algorithm="fricas")

[Out]

[1/16*(4*(b^2*c^2 + 5*a*b*c*d)*x^2*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) - (15*b^2*c^2 + 10*a*b*c*d -
a^2*d^2)*x^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*(4*b^2*c*x^2 - 2*a^2*c - (9*a*b*c +
a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c*x^2), -1/16*(8*(b^2*c^2 + 5*a*b*c*d)*x^2*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)) + (15*b^2*c^
2 + 10*a*b*c*d - a^2*d^2)*x^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*(4*b^2*c*x^2 - 2*a^
2*c - (9*a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c*x^2), 1/8*((15*b^2*c^2 + 10*a*b*c*d - a^2*d^2)*x^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)) + 2*(b^2*c^2 + 5*a*b*c*d)*x^2*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*(4*b^2*c*x^2
- 2*a^2*c - (9*a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c*x^2), 1/8*((15*b^2*c^2 + 10*a*b*c*d - a^2*d^2
)*x^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)) - 4*(b^2*c^2 + 5*a*b*c*d)*x^2*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*(4*b^2*c*x^2 - 2*a^2*c - (9*a*b*c + a^2*
d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(c*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{\frac{5}{2}} \sqrt{c + d x}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(d*x+c)**(1/2)/x**3,x)

[Out]

Integral((a + b*x)**(5/2)*sqrt(c + d*x)/x**3, x)

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Giac [B]  time = 5.59824, size = 1616, normalized size = 7.62 \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)*(d*x+c)^(1/2)/x^3,x, algorithm="giac")

[Out]

1/4*(4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*b*abs(b) - 2*(sqrt(b*d)*b^2*c*abs(b) + 5*sqrt(b*d)*a*
b*d*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/d - (15*sqrt(b*d)*a*b^3*c^2
*abs(b) + 10*sqrt(b*d)*a^2*b^2*c*d*abs(b) - sqrt(b*d)*a^3*b*d^2*abs(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)*b*c) - 2*(9*sqrt
(b*d)*a*b^9*c^5*abs(b) - 35*sqrt(b*d)*a^2*b^8*c^4*d*abs(b) + 50*sqrt(b*d)*a^3*b^7*c^3*d^2*abs(b) - 30*sqrt(b*d
)*a^4*b^6*c^2*d^3*abs(b) + 5*sqrt(b*d)*a^5*b^5*c*d^4*abs(b) + sqrt(b*d)*a^6*b^4*d^5*abs(b) - 27*sqrt(b*d)*(sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^7*c^4*abs(b) + 28*sqrt(b*d)*(sqrt(b*d)*sqrt(
b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^6*c^3*d*abs(b) + 22*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^5*c^2*d^2*abs(b) - 20*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr
t(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^4*c*d^3*abs(b) - 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^2*a^5*b^3*d^4*abs(b) + 27*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)
*b*d - a*b*d))^4*a*b^5*c^3*abs(b) + 21*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d
))^4*a^2*b^4*c^2*d*abs(b) + 29*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3
*b^3*c*d^2*abs(b) + 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^4*b^2*d^3*
abs(b) - 9*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^3*c^2*abs(b) - 14*s
qrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b^2*c*d*abs(b) - sqrt(b*d)*(sqr
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^3*b*d^2*abs(b))/((b^4*c^2 - 2*a*b^3*c*d + a^2*
b^2*d^2 - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*x +
a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*
b*d))^4)^2*c))/b