3.601 \(\int \frac{(a+b x)^{3/2} \sqrt{c+d x}}{x} \, dx\)
Optimal. Leaf size=164 \[ -\frac{\left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} d^{3/2}}-2 a^{3/2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{1}{2} (a+b x)^{3/2} \sqrt{c+d x}+\frac{\sqrt{a+b x} \sqrt{c+d x} (3 a d+b c)}{4 d} \]
[Out]
((b*c + 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*d) + ((a + b*x)^(3/2)*Sqrt[c + d*x])/2 - 2*a^(3/2)*Sqrt[c]*ArcT
anh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] - ((b^2*c^2 - 6*a*b*c*d - 3*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqr
t[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*Sqrt[b]*d^(3/2))
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Rubi [A] time = 0.132889, 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 =
{101, 154, 157, 63, 217, 206, 93, 208} \[ -\frac{\left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} d^{3/2}}-2 a^{3/2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{1}{2} (a+b x)^{3/2} \sqrt{c+d x}+\frac{\sqrt{a+b x} \sqrt{c+d x} (3 a d+b c)}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*x)^(3/2)*Sqrt[c + d*x])/x,x]
[Out]
((b*c + 3*a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*d) + ((a + b*x)^(3/2)*Sqrt[c + d*x])/2 - 2*a^(3/2)*Sqrt[c]*ArcT
anh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] - ((b^2*c^2 - 6*a*b*c*d - 3*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqr
t[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*Sqrt[b]*d^(3/2))
Rule 101
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))
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)^{3/2} \sqrt{c+d x}}{x} \, dx &=\frac{1}{2} (a+b x)^{3/2} \sqrt{c+d x}-\frac{1}{2} \int \frac{\sqrt{a+b x} \left (-2 a c+\frac{1}{2} (-b c-3 a d) x\right )}{x \sqrt{c+d x}} \, dx\\ &=\frac{(b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d}+\frac{1}{2} (a+b x)^{3/2} \sqrt{c+d x}-\frac{\int \frac{-2 a^2 c d+\frac{1}{4} \left (b^2 c^2-6 a b c d-3 a^2 d^2\right ) x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 d}\\ &=\frac{(b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d}+\frac{1}{2} (a+b x)^{3/2} \sqrt{c+d x}+\left (a^2 c\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx-\frac{\left (b^2 c^2-6 a b c d-3 a^2 d^2\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 d}\\ &=\frac{(b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d}+\frac{1}{2} (a+b x)^{3/2} \sqrt{c+d x}+\left (2 a^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )-\frac{1}{4} \left (-6 a c+\frac{b c^2}{d}-\frac{3 a^2 d}{b}\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 )\\ &=\frac{(b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d}+\frac{1}{2} (a+b x)^{3/2} \sqrt{c+d x}-2 a^{3/2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )-\frac{1}{4} \left (-6 a c+\frac{b c^2}{d}-\frac{3 a^2 d}{b}\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 )\\ &=\frac{(b c+3 a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d}+\frac{1}{2} (a+b x)^{3/2} \sqrt{c+d x}-2 a^{3/2} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )-\frac{\left (b^2 c^2-6 a b c d-3 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} d^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.737274, size = 194, normalized size = 1.18 \[ \frac{\sqrt{d} \left (\sqrt{a+b x} (c+d x) (5 a d+b c+2 b d x)-8 a^{3/2} \sqrt{c} d \sqrt{c+d x} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )-\frac{\sqrt{b c-a d} \left (-3 a^2 d^2-6 a b c d+b^2 c^2\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{b}}{4 d^{3/2} \sqrt{c+d x}} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*x)^(3/2)*Sqrt[c + d*x])/x,x]
[Out]
(-((Sqrt[b*c - a*d]*(b^2*c^2 - 6*a*b*c*d - 3*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]])/b) + Sqrt[d]*(Sqrt[a + b*x]*(c + d*x)*(b*c + 5*a*d + 2*b*d*x) - 8*a^(3/2)*Sqrt[c]*d*
Sqrt[c + d*x]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(4*d^(3/2)*Sqrt[c + d*x])
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Maple [B] time = 0.012, size = 389, normalized size = 2.4 \begin{align*}{\frac{1}{8\,d}\sqrt{bx+a}\sqrt{dx+c} \left ( 4\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}xbd+3\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) \sqrt{ac}{a}^{2}{d}^{2}+6\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) \sqrt{ac}abcd-\ln \left ({\frac{1}{2} \left ( 2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) \sqrt{ac}{b}^{2}{c}^{2}-8\,\sqrt{bd}\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{d{x}^{2}b+adx+bcx+ac}+2\,ac}{x}} \right ){a}^{2}cd+10\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}ad+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}bc \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)^(3/2)*(d*x+c)^(1/2)/x,x)
[Out]
1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(4*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*x*b*d+3*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))*(a*c)^(1/2)*a^2*d^2+6*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))*(a*c)^(1/2)*a*b*c*d-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))*(a*c)^(1/2)*b^2*c^2-8*(b*d)^(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)*a^2*c*d+10*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1
/2)*(a*c)^(1/2)*a*d+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)*(a*c)^(1/2)*b*c)/(b*d*x^2+a*d*x+b*c*x+a*c)^(
1/2)/d/(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)^(3/2)*(d*x+c)^(1/2)/x,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 16.0239, size = 2272, normalized size = 13.85 \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)^(3/2)*(d*x+c)^(1/2)/x,x, algorithm="fricas")
[Out]
[1/16*(8*sqrt(a*c)*a*b*d^2*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 + (b*c + a*d)*x)*sq
rt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - (b^2*c^2 - 6*a*b*c*d - 3*a^2*d^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*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d
*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(2*b^2*d^2*x + b^2*c*d + 5*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^2
), 1/8*(4*sqrt(a*c)*a*b*d^2*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 + (b*c + a*d)*x)*s
qrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + (b^2*c^2 - 6*a*b*c*d - 3*a^2*d^2)*sqrt(
-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*
d + a*b*d^2)*x)) + 2*(2*b^2*d^2*x + b^2*c*d + 5*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^2), 1/16*(16*sqrt(-
a*c)*a*b*d^2*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2
+ (a*b*c^2 + a^2*c*d)*x)) - (b^2*c^2 - 6*a*b*c*d - 3*a^2*d^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*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(2*
b^2*d^2*x + b^2*c*d + 5*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b*d^2), 1/8*(8*sqrt(-a*c)*a*b*d^2*arctan(1/2*(2
*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x + c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x))
+ (b^2*c^2 - 6*a*b*c*d - 3*a^2*d^2)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(
d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(2*b^2*d^2*x + b^2*c*d + 5*a*b*d^2)*sqrt(b*x + a
)*sqrt(d*x + c))/(b*d^2)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{\frac{3}{2}} \sqrt{c + d x}}{x}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)**(3/2)*(d*x+c)**(1/2)/x,x)
[Out]
Integral((a + b*x)**(3/2)*sqrt(c + d*x)/x, x)
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Giac [A] time = 1.4367, size = 336, normalized size = 2.05 \begin{align*} -\frac{2 \, \sqrt{b d} a^{2} c{\left | b \right |} \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} b} + \frac{1}{4} \, \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )}{\left | b \right |}}{b^{2}} + \frac{b^{2} c d{\left | b \right |} + 3 \, a b d^{2}{\left | b \right |}}{b^{3} d^{2}}\right )} + \frac{{\left (\sqrt{b d} b^{2} c^{2}{\left | b \right |} - 6 \, \sqrt{b d} a b c d{\left | b \right |} - 3 \, \sqrt{b d} a^{2} d^{2}{\left | b \right |}\right )} \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 )}{8 \, b^{2} d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)^(3/2)*(d*x+c)^(1/2)/x,x, algorithm="giac")
[Out]
-2*sqrt(b*d)*a^2*c*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) + 1/4*sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*
(b*x + a)*abs(b)/b^2 + (b^2*c*d*abs(b) + 3*a*b*d^2*abs(b))/(b^3*d^2)) + 1/8*(sqrt(b*d)*b^2*c^2*abs(b) - 6*sqrt
(b*d)*a*b*c*d*abs(b) - 3*sqrt(b*d)*a^2*d^2*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d -
a*b*d))^2)/(b^2*d^2)