3.720 \(\int \frac{(c+d x)^{5/2}}{x^4 \sqrt{a+b x}} \, dx\)

Optimal. Leaf size=157 \[ \frac{5 \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}{12 a^2 x^2}-\frac{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^2}{8 a^3 x}+\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 a^{7/2} \sqrt{c}}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 a x^3} \]

[Out]

(-5*(b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*a^3*x) + (5*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(12*a
^2*x^2) - (Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*a*x^3) + (5*(b*c - a*d)^3*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a
]*Sqrt[c + d*x])])/(8*a^(7/2)*Sqrt[c])

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Rubi [A]  time = 0.0751159, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {94, 93, 208} \[ \frac{5 \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}{12 a^2 x^2}-\frac{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^2}{8 a^3 x}+\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 a^{7/2} \sqrt{c}}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(x^4*Sqrt[a + b*x]),x]

[Out]

(-5*(b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*a^3*x) + (5*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(12*a
^2*x^2) - (Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*a*x^3) + (5*(b*c - a*d)^3*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a
]*Sqrt[c + d*x])])/(8*a^(7/2)*Sqrt[c])

Rule 94

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 + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

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{(c+d x)^{5/2}}{x^4 \sqrt{a+b x}} \, dx &=-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 a x^3}-\frac{(5 (b c-a d)) \int \frac{(c+d x)^{3/2}}{x^3 \sqrt{a+b x}} \, dx}{6 a}\\ &=\frac{5 (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 a^2 x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 a x^3}+\frac{\left (5 (b c-a d)^2\right ) \int \frac{\sqrt{c+d x}}{x^2 \sqrt{a+b x}} \, dx}{8 a^2}\\ &=-\frac{5 (b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 a^3 x}+\frac{5 (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 a^2 x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 a x^3}-\frac{\left (5 (b c-a d)^3\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{16 a^3}\\ &=-\frac{5 (b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 a^3 x}+\frac{5 (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 a^2 x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 a x^3}-\frac{\left (5 (b c-a d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{8 a^3}\\ &=-\frac{5 (b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 a^3 x}+\frac{5 (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 a^2 x^2}-\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 a x^3}+\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 a^{7/2} \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.133975, size = 142, normalized size = 0.9 \[ \frac{\frac{5 x (b c-a d) \left (3 x^2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )+\sqrt{a} \sqrt{c} \sqrt{a+b x} \sqrt{c+d x} (2 a c+5 a d x-3 b c x)\right )}{a^{5/2} \sqrt{c}}-8 \sqrt{a+b x} (c+d x)^{5/2}}{24 a x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/2)/(x^4*Sqrt[a + b*x]),x]

[Out]

(-8*Sqrt[a + b*x]*(c + d*x)^(5/2) + (5*(b*c - a*d)*x*(Sqrt[a]*Sqrt[c]*Sqrt[a + b*x]*Sqrt[c + d*x]*(2*a*c - 3*b
*c*x + 5*a*d*x) + 3*(b*c - a*d)^2*x^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(a^(5/2)*Sqrt
[c]))/(24*a*x^3)

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Maple [B]  time = 0.02, size = 405, normalized size = 2.6 \begin{align*} -{\frac{1}{48\,{a}^{3}{x}^{3}}\sqrt{bx+a}\sqrt{dx+c} \left ( 15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}{a}^{3}{d}^{3}-45\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}{a}^{2}bc{d}^{2}+45\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}a{b}^{2}{c}^{2}d-15\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{3}{b}^{3}{c}^{3}+66\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{x}^{2}{a}^{2}{d}^{2}-80\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{x}^{2}abcd+30\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{x}^{2}{b}^{2}{c}^{2}+52\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}x{a}^{2}cd-20\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}xab{c}^{2}+16\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}{a}^{2}{c}^{2} \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)/x^4/(b*x+a)^(1/2),x)

[Out]

-1/48*(d*x+c)^(1/2)*(b*x+a)^(1/2)/a^3*(15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^3*
a^3*d^3-45*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^3*a^2*b*c*d^2+45*ln((a*d*x+b*c*x+
2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+2*a*c)/x)*x^3*a*b^2*c^2*d-15*ln((a*d*x+b*c*x+2*(a*c)^(1/2)*((b*x+a)*(d*x
+c))^(1/2)+2*a*c)/x)*x^3*b^3*c^3+66*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x^2*a^2*d^2-80*((b*x+a)*(d*x+c))^(1/2)
*(a*c)^(1/2)*x^2*a*b*c*d+30*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x^2*b^2*c^2+52*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(
1/2)*x*a^2*c*d-20*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*x*a*b*c^2+16*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)*a^2*c^2
)/((b*x+a)*(d*x+c))^(1/2)/x^3/(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)^(5/2)/x^4/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.50628, size = 976, normalized size = 6.22 \begin{align*} \left [-\frac{15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{a c} x^{3} \log \left (\frac{8 \, a^{2} c^{2} +{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \,{\left (2 \, a c +{\left (b c + a d\right )} x\right )} \sqrt{a c} \sqrt{b x + a} \sqrt{d x + c} + 8 \,{\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \,{\left (8 \, a^{3} c^{3} +{\left (15 \, a b^{2} c^{3} - 40 \, a^{2} b c^{2} d + 33 \, a^{3} c d^{2}\right )} x^{2} - 2 \,{\left (5 \, a^{2} b c^{3} - 13 \, a^{3} c^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{96 \, a^{4} c x^{3}}, -\frac{15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-a c} x^{3} \arctan \left (\frac{{\left (2 \, a c +{\left (b c + a d\right )} x\right )} \sqrt{-a c} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (a b c d x^{2} + a^{2} c^{2} +{\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) + 2 \,{\left (8 \, a^{3} c^{3} +{\left (15 \, a b^{2} c^{3} - 40 \, a^{2} b c^{2} d + 33 \, a^{3} c d^{2}\right )} x^{2} - 2 \,{\left (5 \, a^{2} b c^{3} - 13 \, a^{3} c^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{48 \, a^{4} c x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^4/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(a*c)*x^3*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)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*
x)/x^2) + 4*(8*a^3*c^3 + (15*a*b^2*c^3 - 40*a^2*b*c^2*d + 33*a^3*c*d^2)*x^2 - 2*(5*a^2*b*c^3 - 13*a^3*c^2*d)*x
)*sqrt(b*x + a)*sqrt(d*x + c))/(a^4*c*x^3), -1/48*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt
(-a*c)*x^3*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)) + 2*(8*a^3*c^3 + (15*a*b^2*c^3 - 40*a^2*b*c^2*d + 33*a^3*c*d^2)*x^2 - 2*(5*a^2*b*c^3 -
13*a^3*c^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^4*c*x^3)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/2)/x**4/(b*x+a)**(1/2),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^4/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError