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

Optimal. Leaf size=165 \[ \frac{\left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{3/2} \sqrt{d}}-2 \sqrt{a} c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{1}{2} \sqrt{a+b x} (c+d x)^{3/2}+\frac{\sqrt{a+b x} \sqrt{c+d x} (a d+3 b c)}{4 b} \]

[Out]

((3*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*b) + (Sqrt[a + b*x]*(c + d*x)^(3/2))/2 - 2*Sqrt[a]*c^(3/2)*ArcT
anh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + ((3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[d]*Sqr
t[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(3/2)*Sqrt[d])

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Rubi [A]  time = 0.135259, antiderivative size = 165, 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} \[ \frac{\left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{3/2} \sqrt{d}}-2 \sqrt{a} c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{1}{2} \sqrt{a+b x} (c+d x)^{3/2}+\frac{\sqrt{a+b x} \sqrt{c+d x} (a d+3 b c)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*b*c + a*d)*Sqrt[a + b*x]*Sqrt[c + d*x])/(4*b) + (Sqrt[a + b*x]*(c + d*x)^(3/2))/2 - 2*Sqrt[a]*c^(3/2)*ArcT
anh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])] + ((3*b^2*c^2 + 6*a*b*c*d - a^2*d^2)*ArcTanh[(Sqrt[d]*Sqr
t[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(3/2)*Sqrt[d])

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{\sqrt{a+b x} (c+d x)^{3/2}}{x} \, dx &=\frac{1}{2} \sqrt{a+b x} (c+d x)^{3/2}-\frac{1}{2} \int \frac{\sqrt{c+d x} \left (-2 a c+\frac{1}{2} (-3 b c-a d) x\right )}{x \sqrt{a+b x}} \, dx\\ &=\frac{(3 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b}+\frac{1}{2} \sqrt{a+b x} (c+d x)^{3/2}-\frac{\int \frac{-2 a b c^2+\frac{1}{4} \left (-3 b^2 c^2-6 a b c d+a^2 d^2\right ) x}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 b}\\ &=\frac{(3 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b}+\frac{1}{2} \sqrt{a+b x} (c+d x)^{3/2}+\left (a c^2\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx+\frac{\left (3 b^2 c^2+6 a b c d-a^2 d^2\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 b}\\ &=\frac{(3 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b}+\frac{1}{2} \sqrt{a+b x} (c+d x)^{3/2}+\left (2 a c^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )+\frac{\left (3 b^2 c^2+6 a b c d-a^2 d^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 )}{4 b^2}\\ &=\frac{(3 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b}+\frac{1}{2} \sqrt{a+b x} (c+d x)^{3/2}-2 \sqrt{a} c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{\left (3 b^2 c^2+6 a b c d-a^2 d^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 )}{4 b^2}\\ &=\frac{(3 b c+a d) \sqrt{a+b x} \sqrt{c+d x}}{4 b}+\frac{1}{2} \sqrt{a+b x} (c+d x)^{3/2}-2 \sqrt{a} c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\frac{\left (3 b^2 c^2+6 a b c d-a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 b^{3/2} \sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 0.717632, size = 195, normalized size = 1.18 \[ \frac{\sqrt{b c-a d} \left (-a^2 d^2+6 a b c d+3 b^2 c^2\right ) \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )+b \sqrt{d} \left (\sqrt{a+b x} (c+d x) (a d+5 b c+2 b d x)-8 \sqrt{a} b c^{3/2} \sqrt{c+d x} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )}{4 b^2 \sqrt{d} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*c - a*d]*(3*b^2*c^2 + 6*a*b*c*d - 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)*(5*b*c + a*d + 2*b*d*x) - 8*Sqrt[a]*b*c^(3/2)*Sqrt[
c + d*x]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(4*b^2*Sqrt[d]*Sqrt[c + d*x])

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Maple [B]  time = 0.011, size = 388, normalized size = 2.4 \begin{align*} -{\frac{1}{8\,b}\sqrt{bx+a}\sqrt{dx+c} \left ( \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}{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-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}{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 ) ab{c}^{2}-4\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}xbd-2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}\sqrt{ac}ad-10\,\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((d*x+c)^(3/2)*(b*x+a)^(1/2)/x,x)

[Out]

-1/8*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(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-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)*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*b*
c^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-2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1
/2)*(a*c)^(1/2)*a*d-10*(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)/b/(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((d*x+c)^(3/2)*(b*x+a)^(1/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 17.4695, size = 2272, normalized size = 13.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)*(b*x+a)^(1/2)/x,x, algorithm="fricas")

[Out]

[1/16*(8*sqrt(a*c)*b^2*c*d*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) - (3*b^2*c^2 + 6*a*b*c*d - 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 + 5*b^2*c*d + a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d
), 1/8*(4*sqrt(a*c)*b^2*c*d*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) - (3*b^2*c^2 + 6*a*b*c*d - 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 + 5*b^2*c*d + a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d), 1/16*(16*sqrt(-
a*c)*b^2*c*d*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)) - (3*b^2*c^2 + 6*a*b*c*d - 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 + 5*b^2*c*d + a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^2*d), 1/8*(8*sqrt(-a*c)*b^2*c*d*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))
- (3*b^2*c^2 + 6*a*b*c*d - 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 + 5*b^2*c*d + a*b*d^2)*sqrt(b*x + a
)*sqrt(d*x + c))/(b^2*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.42483, size = 346, normalized size = 2.1 \begin{align*} -\frac{2 \, \sqrt{b d} a c^{2}{\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 )} d{\left | b \right |}}{b^{3}} + \frac{5 \, b^{4} c d^{2}{\left | b \right |} - a b^{3} d^{3}{\left | b \right |}}{b^{6} d^{2}}\right )} - \frac{{\left (3 \, \sqrt{b d} b^{2} c^{2}{\left | b \right |} + 6 \, \sqrt{b d} a b c d{\left | b \right |} - \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^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)*(b*x+a)^(1/2)/x,x, algorithm="giac")

[Out]

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