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

Optimal. Leaf size=113 \[ -\frac{3 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)}{4 d^2}+\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} d^{5/2}}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0542366, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {50, 63, 217, 206} \[ -\frac{3 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)}{4 d^2}+\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} d^{5/2}}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/Sqrt[c + d*x],x]

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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])

Rubi steps

\begin{align*} \int \frac{(a+b x)^{3/2}}{\sqrt{c+d x}} \, dx &=\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 d}-\frac{(3 (b c-a d)) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{4 d}\\ &=-\frac{3 (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^2}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 d}+\frac{\left (3 (b c-a d)^2\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{8 d^2}\\ &=-\frac{3 (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^2}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 d}+\frac{\left (3 (b c-a 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 d^2}\\ &=-\frac{3 (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^2}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 d}+\frac{\left (3 (b c-a 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 d^2}\\ &=-\frac{3 (b c-a d) \sqrt{a+b x} \sqrt{c+d x}}{4 d^2}+\frac{(a+b x)^{3/2} \sqrt{c+d x}}{2 d}+\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{4 \sqrt{b} d^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.348681, size = 119, normalized size = 1.05 \[ \frac{\sqrt{d} \sqrt{a+b x} (c+d x) (5 a d-3 b c+2 b d x)+\frac{3 (b c-a d)^{5/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{b}}{4 d^{5/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.004, size = 308, normalized size = 2.7 \begin{align*}{\frac{1}{2\,d} \left ( bx+a \right ) ^{{\frac{3}{2}}}\sqrt{dx+c}}+{\frac{3\,a}{4\,d}\sqrt{bx+a}\sqrt{dx+c}}-{\frac{3\,bc}{4\,{d}^{2}}\sqrt{bx+a}\sqrt{dx+c}}+{\frac{3\,{a}^{2}}{8}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({ \left ({\frac{ad}{2}}+{\frac{bc}{2}}+bdx \right ){\frac{1}{\sqrt{bd}}}}+\sqrt{d{x}^{2}b+ \left ( ad+bc \right ) x+ac} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{bd}}}}-{\frac{3\,abc}{4\,d}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({ \left ({\frac{ad}{2}}+{\frac{bc}{2}}+bdx \right ){\frac{1}{\sqrt{bd}}}}+\sqrt{d{x}^{2}b+ \left ( ad+bc \right ) x+ac} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{bd}}}}+{\frac{3\,{b}^{2}{c}^{2}}{8\,{d}^{2}}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({ \left ({\frac{ad}{2}}+{\frac{bc}{2}}+bdx \right ){\frac{1}{\sqrt{bd}}}}+\sqrt{d{x}^{2}b+ \left ( ad+bc \right ) x+ac} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{bd}}}} \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)

[Out]

1/2*(b*x+a)^(3/2)*(d*x+c)^(1/2)/d+3/4/d*(b*x+a)^(1/2)*(d*x+c)^(1/2)*a-3/4/d^2*(b*x+a)^(1/2)*(d*x+c)^(1/2)*b*c+
3/8*((b*x+a)*(d*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(b*d)^(1/2)+(d*x^2*b+(a*d+b
*c)*x+a*c)^(1/2))/(b*d)^(1/2)*a^2-3/4/d*((b*x+a)*(d*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(1/2)*ln((1/2*a*d+1/2*b*
c+b*d*x)/(b*d)^(1/2)+(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*a*b*c+3/8/d^2*((b*x+a)*(d*x+c))^(1/2)/(b*x+a
)^(1/2)/(d*x+c)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(b*d)^(1/2)+(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*b^2*
c^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, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.5145, size = 711, normalized size = 6.29 \begin{align*} \left [\frac{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \,{\left (2 \, b d x + b c + a d\right )} \sqrt{b d} \sqrt{b x + a} \sqrt{d x + c} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \,{\left (2 \, b^{2} d^{2} x - 3 \, b^{2} c d + 5 \, a b d^{2}\right )} \sqrt{b x + a} \sqrt{d x + c}}{16 \, b d^{3}}, -\frac{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-b d} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{-b d} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (b^{2} d^{2} x^{2} + a b c d +{\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \,{\left (2 \, b^{2} d^{2} x - 3 \, b^{2} c d + 5 \, a b d^{2}\right )} \sqrt{b x + a} \sqrt{d x + c}}{8 \, b d^{3}}\right ] \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, algorithm="fricas")

[Out]

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

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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}}\, 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)

[Out]

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

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Giac [A]  time = 1.28775, size = 188, normalized size = 1.66 \begin{align*} \frac{{\left (\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 )}}{b d} - \frac{3 \,{\left (b c d - a d^{2}\right )}}{b d^{3}}\right )} - \frac{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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 |}\right )}{\sqrt{b d} d^{2}}\right )} b}{4 \,{\left | b \right |}} \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, algorithm="giac")

[Out]

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