3.688 \(\int \frac{(a+b x)^{5/2}}{x^4 (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=226 \[ -\frac{(a+b x)^{5/2} (b c-7 a d)}{12 a c^2 x^2 \sqrt{c+d x}}-\frac{5 (a+b x)^{3/2} (b c-7 a d) (b c-a d)}{24 a c^3 x \sqrt{c+d x}}+\frac{5 \sqrt{a+b x} (b c-7 a d) (b c-a d)^2}{8 a c^4 \sqrt{c+d x}}-\frac{5 (b c-7 a d) (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 \sqrt{a} c^{9/2}}-\frac{(a+b x)^{7/2}}{3 a c x^3 \sqrt{c+d x}} \]

[Out]

(5*(b*c - 7*a*d)*(b*c - a*d)^2*Sqrt[a + b*x])/(8*a*c^4*Sqrt[c + d*x]) - (5*(b*c - 7*a*d)*(b*c - a*d)*(a + b*x)
^(3/2))/(24*a*c^3*x*Sqrt[c + d*x]) - ((b*c - 7*a*d)*(a + b*x)^(5/2))/(12*a*c^2*x^2*Sqrt[c + d*x]) - (a + b*x)^
(7/2)/(3*a*c*x^3*Sqrt[c + d*x]) - (5*(b*c - 7*a*d)*(b*c - a*d)^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt
[c + d*x])])/(8*Sqrt[a]*c^(9/2))

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Rubi [A]  time = 0.111685, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {96, 94, 93, 208} \[ -\frac{(a+b x)^{5/2} (b c-7 a d)}{12 a c^2 x^2 \sqrt{c+d x}}-\frac{5 (a+b x)^{3/2} (b c-7 a d) (b c-a d)}{24 a c^3 x \sqrt{c+d x}}+\frac{5 \sqrt{a+b x} (b c-7 a d) (b c-a d)^2}{8 a c^4 \sqrt{c+d x}}-\frac{5 (b c-7 a d) (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 \sqrt{a} c^{9/2}}-\frac{(a+b x)^{7/2}}{3 a c x^3 \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/(x^4*(c + d*x)^(3/2)),x]

[Out]

(5*(b*c - 7*a*d)*(b*c - a*d)^2*Sqrt[a + b*x])/(8*a*c^4*Sqrt[c + d*x]) - (5*(b*c - 7*a*d)*(b*c - a*d)*(a + b*x)
^(3/2))/(24*a*c^3*x*Sqrt[c + d*x]) - ((b*c - 7*a*d)*(a + b*x)^(5/2))/(12*a*c^2*x^2*Sqrt[c + d*x]) - (a + b*x)^
(7/2)/(3*a*c*x^3*Sqrt[c + d*x]) - (5*(b*c - 7*a*d)*(b*c - a*d)^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt
[c + d*x])])/(8*Sqrt[a]*c^(9/2))

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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{(a+b x)^{5/2}}{x^4 (c+d x)^{3/2}} \, dx &=-\frac{(a+b x)^{7/2}}{3 a c x^3 \sqrt{c+d x}}-\frac{\left (-\frac{b c}{2}+\frac{7 a d}{2}\right ) \int \frac{(a+b x)^{5/2}}{x^3 (c+d x)^{3/2}} \, dx}{3 a c}\\ &=-\frac{(b c-7 a d) (a+b x)^{5/2}}{12 a c^2 x^2 \sqrt{c+d x}}-\frac{(a+b x)^{7/2}}{3 a c x^3 \sqrt{c+d x}}+\frac{(5 (b c-7 a d) (b c-a d)) \int \frac{(a+b x)^{3/2}}{x^2 (c+d x)^{3/2}} \, dx}{24 a c^2}\\ &=-\frac{5 (b c-7 a d) (b c-a d) (a+b x)^{3/2}}{24 a c^3 x \sqrt{c+d x}}-\frac{(b c-7 a d) (a+b x)^{5/2}}{12 a c^2 x^2 \sqrt{c+d x}}-\frac{(a+b x)^{7/2}}{3 a c x^3 \sqrt{c+d x}}+\frac{\left (5 (b c-7 a d) (b c-a d)^2\right ) \int \frac{\sqrt{a+b x}}{x (c+d x)^{3/2}} \, dx}{16 a c^3}\\ &=\frac{5 (b c-7 a d) (b c-a d)^2 \sqrt{a+b x}}{8 a c^4 \sqrt{c+d x}}-\frac{5 (b c-7 a d) (b c-a d) (a+b x)^{3/2}}{24 a c^3 x \sqrt{c+d x}}-\frac{(b c-7 a d) (a+b x)^{5/2}}{12 a c^2 x^2 \sqrt{c+d x}}-\frac{(a+b x)^{7/2}}{3 a c x^3 \sqrt{c+d x}}+\frac{\left (5 (b c-7 a d) (b c-a d)^2\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{16 c^4}\\ &=\frac{5 (b c-7 a d) (b c-a d)^2 \sqrt{a+b x}}{8 a c^4 \sqrt{c+d x}}-\frac{5 (b c-7 a d) (b c-a d) (a+b x)^{3/2}}{24 a c^3 x \sqrt{c+d x}}-\frac{(b c-7 a d) (a+b x)^{5/2}}{12 a c^2 x^2 \sqrt{c+d x}}-\frac{(a+b x)^{7/2}}{3 a c x^3 \sqrt{c+d x}}+\frac{\left (5 (b c-7 a d) (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{8 c^4}\\ &=\frac{5 (b c-7 a d) (b c-a d)^2 \sqrt{a+b x}}{8 a c^4 \sqrt{c+d x}}-\frac{5 (b c-7 a d) (b c-a d) (a+b x)^{3/2}}{24 a c^3 x \sqrt{c+d x}}-\frac{(b c-7 a d) (a+b x)^{5/2}}{12 a c^2 x^2 \sqrt{c+d x}}-\frac{(a+b x)^{7/2}}{3 a c x^3 \sqrt{c+d x}}-\frac{5 (b c-7 a d) (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{8 \sqrt{a} c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.285345, size = 167, normalized size = 0.74 \[ \frac{-\frac{1}{2} x (b c-7 a d) \left (2 c^{5/2} (a+b x)^{5/2}+5 x (b c-a d) \left (\sqrt{c} \sqrt{a+b x} (a (c+3 d x)-2 b c x)+3 \sqrt{a} x \sqrt{c+d x} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )\right )\right )-4 c^{7/2} (a+b x)^{7/2}}{12 a c^{9/2} x^3 \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(5/2)/(x^4*(c + d*x)^(3/2)),x]

[Out]

(-4*c^(7/2)*(a + b*x)^(7/2) - ((b*c - 7*a*d)*x*(2*c^(5/2)*(a + b*x)^(5/2) + 5*(b*c - a*d)*x*(Sqrt[c]*Sqrt[a +
b*x]*(-2*b*c*x + a*(c + 3*d*x)) + 3*Sqrt[a]*(b*c - a*d)*x*Sqrt[c + d*x]*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[
a]*Sqrt[c + d*x])])))/2)/(12*a*c^(9/2)*x^3*Sqrt[c + d*x])

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Maple [B]  time = 0.027, size = 704, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)/x^4/(d*x+c)^(3/2),x)

[Out]

1/48*(b*x+a)^(1/2)*(105*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^4*a^3*d^4-225*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^4*a^2*b*c*d^3+135*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^4*a*b^2*c^2*d^2-15*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^4*b^3*c^3*d+105*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*c*d^3-225*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^2*d^2+135*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^3*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^4-210*x^3*a^2*d^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)+380*x^3*a*b*c*d^2*((b*x+a)*(d*x+c))^
(1/2)*(a*c)^(1/2)-162*x^3*b^2*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-70*x^2*a^2*c*d^2*((b*x+a)*(d*x+c))^(1/
2)*(a*c)^(1/2)+136*x^2*a*b*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-66*x^2*b^2*c^3*((b*x+a)*(d*x+c))^(1/2)*(a
*c)^(1/2)+28*x*a^2*c^2*d*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-52*x*a*b*c^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2)-
16*a^2*c^3*((b*x+a)*(d*x+c))^(1/2)*(a*c)^(1/2))/c^4/((b*x+a)*(d*x+c))^(1/2)/(a*c)^(1/2)/x^3/(d*x+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)^(5/2)/x^4/(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 28.8942, size = 1377, normalized size = 6.09 \begin{align*} \left [-\frac{15 \,{\left ({\left (b^{3} c^{3} d - 9 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} - 7 \, a^{3} d^{4}\right )} x^{4} +{\left (b^{3} c^{4} - 9 \, a b^{2} c^{3} d + 15 \, a^{2} b c^{2} d^{2} - 7 \, a^{3} c d^{3}\right )} x^{3}\right )} \sqrt{a c} \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^{4} +{\left (81 \, a b^{2} c^{3} d - 190 \, a^{2} b c^{2} d^{2} + 105 \, a^{3} c d^{3}\right )} x^{3} +{\left (33 \, a b^{2} c^{4} - 68 \, a^{2} b c^{3} d + 35 \, a^{3} c^{2} d^{2}\right )} x^{2} + 2 \,{\left (13 \, a^{2} b c^{4} - 7 \, a^{3} c^{3} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{96 \,{\left (a c^{5} d x^{4} + a c^{6} x^{3}\right )}}, \frac{15 \,{\left ({\left (b^{3} c^{3} d - 9 \, a b^{2} c^{2} d^{2} + 15 \, a^{2} b c d^{3} - 7 \, a^{3} d^{4}\right )} x^{4} +{\left (b^{3} c^{4} - 9 \, a b^{2} c^{3} d + 15 \, a^{2} b c^{2} d^{2} - 7 \, a^{3} c d^{3}\right )} x^{3}\right )} \sqrt{-a c} \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^{4} +{\left (81 \, a b^{2} c^{3} d - 190 \, a^{2} b c^{2} d^{2} + 105 \, a^{3} c d^{3}\right )} x^{3} +{\left (33 \, a b^{2} c^{4} - 68 \, a^{2} b c^{3} d + 35 \, a^{3} c^{2} d^{2}\right )} x^{2} + 2 \,{\left (13 \, a^{2} b c^{4} - 7 \, a^{3} c^{3} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{48 \,{\left (a c^{5} d x^{4} + a c^{6} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/x^4/(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(15*((b^3*c^3*d - 9*a*b^2*c^2*d^2 + 15*a^2*b*c*d^3 - 7*a^3*d^4)*x^4 + (b^3*c^4 - 9*a*b^2*c^3*d + 15*a^2
*b*c^2*d^2 - 7*a^3*c*d^3)*x^3)*sqrt(a*c)*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^4 + (81*a*b^2*c
^3*d - 190*a^2*b*c^2*d^2 + 105*a^3*c*d^3)*x^3 + (33*a*b^2*c^4 - 68*a^2*b*c^3*d + 35*a^3*c^2*d^2)*x^2 + 2*(13*a
^2*b*c^4 - 7*a^3*c^3*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^5*d*x^4 + a*c^6*x^3), 1/48*(15*((b^3*c^3*d - 9*a*
b^2*c^2*d^2 + 15*a^2*b*c*d^3 - 7*a^3*d^4)*x^4 + (b^3*c^4 - 9*a*b^2*c^3*d + 15*a^2*b*c^2*d^2 - 7*a^3*c*d^3)*x^3
)*sqrt(-a*c)*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^4 + (81*a*b^2*c^3*d - 190*a^2*b*c^2*d^2 + 105*a^3*c*d^3)*x^3 + (33*a*b^
2*c^4 - 68*a^2*b*c^3*d + 35*a^3*c^2*d^2)*x^2 + 2*(13*a^2*b*c^4 - 7*a^3*c^3*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/
(a*c^5*d*x^4 + a*c^6*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((b*x+a)**(5/2)/x**4/(d*x+c)**(3/2),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((b*x+a)^(5/2)/x^4/(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError