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

Optimal. Leaf size=92 \[ \frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{5/2}}-\frac{2 b \sqrt{a+b x}}{d^2 \sqrt{c+d x}}-\frac{2 (a+b x)^{3/2}}{3 d (c+d x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0433227, antiderivative size = 92, 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 = {47, 63, 217, 206} \[ \frac{2 b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{d^{5/2}}-\frac{2 b \sqrt{a+b x}}{d^2 \sqrt{c+d x}}-\frac{2 (a+b x)^{3/2}}{3 d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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}}{(c+d x)^{5/2}} \, dx &=-\frac{2 (a+b x)^{3/2}}{3 d (c+d x)^{3/2}}+\frac{b \int \frac{\sqrt{a+b x}}{(c+d x)^{3/2}} \, dx}{d}\\ &=-\frac{2 (a+b x)^{3/2}}{3 d (c+d x)^{3/2}}-\frac{2 b \sqrt{a+b x}}{d^2 \sqrt{c+d x}}+\frac{b^2 \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{d^2}\\ &=-\frac{2 (a+b x)^{3/2}}{3 d (c+d x)^{3/2}}-\frac{2 b \sqrt{a+b x}}{d^2 \sqrt{c+d x}}+\frac{(2 b) \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 (a+b x)^{3/2}}{3 d (c+d x)^{3/2}}-\frac{2 b \sqrt{a+b x}}{d^2 \sqrt{c+d x}}+\frac{(2 b) \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 (a+b x)^{3/2}}{3 d (c+d x)^{3/2}}-\frac{2 b \sqrt{a+b x}}{d^2 \sqrt{c+d x}}+\frac{2 b^{3/2} \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 = 0.525808, size = 111, normalized size = 1.21 \[ \frac{6 (b c-a d)^{3/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{3/2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )-2 \sqrt{d} \sqrt{a+b x} (a d+3 b c+4 b d x)}{3 d^{5/2} (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{{\frac{3}{2}}} \left ( dx+c \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.96902, size = 734, normalized size = 7.98 \begin{align*} \left [\frac{3 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \sqrt{\frac{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^{2} x + b c d + a d^{2}\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{\frac{b}{d}} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \,{\left (4 \, b d x + 3 \, b c + a d\right )} \sqrt{b x + a} \sqrt{d x + c}}{6 \,{\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}}, -\frac{3 \,{\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \sqrt{-\frac{b}{d}} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{-\frac{b}{d}}}{2 \,{\left (b^{2} d x^{2} + a b c +{\left (b^{2} c + a b d\right )} x\right )}}\right ) + 2 \,{\left (4 \, b d x + 3 \, b c + a d\right )} \sqrt{b x + a} \sqrt{d x + c}}{3 \,{\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(b*d^2*x^2 + 2*b*c*d*x + b*c^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) - 4*(4*b*d*x + 3*b*c +
a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(d^4*x^2 + 2*c*d^3*x + c^2*d^2), -1/3*(3*(b*d^2*x^2 + 2*b*c*d*x + b*c^2)*sqr
t(-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*d*x + 3*b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c))/(d^4*x^2 + 2*c*d^3*x + c^2*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}}}{\left (c + d x\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.37068, size = 296, normalized size = 3.22 \begin{align*} \frac{\sqrt{b d} \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 )}{16 \,{\left (b^{5} c d^{4} - a b^{4} d^{5}\right )}} + \frac{\sqrt{b x + a}{\left (\frac{4 \,{\left (b^{5} c d^{2} - a b^{4} d^{3}\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^{6} c^{2} d - 2 \, a b^{5} c d^{2} + a^{2} b^{4} d^{3}\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)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

1/16*sqrt(b*d)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(b^5*c*d^4 - a*b^4*d^5
) + 1/48*sqrt(b*x + a)*(4*(b^5*c*d^2 - a*b^4*d^3)*(b*x + a)/(b^8*c^2*d^4 - 2*a*b^7*c*d^5 + a^2*b^6*d^6) + 3*(b
^6*c^2*d - 2*a*b^5*c*d^2 + a^2*b^4*d^3)/(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)