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

Optimal. Leaf size=174 \[ \frac{\sqrt{a+b x} \left (c \left (3 a^2 d^2-22 a b c d+15 b^2 c^2\right )+d x (5 b c-3 a d) (b c-a d)\right )}{3 b d^3 \sqrt{c+d x} (b c-a d)^2}-\frac{(a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{3/2} d^{7/2}}-\frac{2 c x^2 \sqrt{a+b x}}{3 d (c+d x)^{3/2} (b c-a d)} \]

[Out]

(-2*c*x^2*Sqrt[a + b*x])/(3*d*(b*c - a*d)*(c + d*x)^(3/2)) + (Sqrt[a + b*x]*(c*(15*b^2*c^2 - 22*a*b*c*d + 3*a^
2*d^2) + d*(5*b*c - 3*a*d)*(b*c - a*d)*x))/(3*b*d^3*(b*c - a*d)^2*Sqrt[c + d*x]) - ((5*b*c + a*d)*ArcTanh[(Sqr
t[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(b^(3/2)*d^(7/2))

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Rubi [A]  time = 0.145001, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {98, 143, 63, 217, 206} \[ \frac{\sqrt{a+b x} \left (c \left (3 a^2 d^2-22 a b c d+15 b^2 c^2\right )+d x (5 b c-3 a d) (b c-a d)\right )}{3 b d^3 \sqrt{c+d x} (b c-a d)^2}-\frac{(a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{3/2} d^{7/2}}-\frac{2 c x^2 \sqrt{a+b x}}{3 d (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*x^2*Sqrt[a + b*x])/(3*d*(b*c - a*d)*(c + d*x)^(3/2)) + (Sqrt[a + b*x]*(c*(15*b^2*c^2 - 22*a*b*c*d + 3*a^
2*d^2) + d*(5*b*c - 3*a*d)*(b*c - a*d)*x))/(3*b*d^3*(b*c - a*d)^2*Sqrt[c + d*x]) - ((5*b*c + a*d)*ArcTanh[(Sqr
t[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(b^(3/2)*d^(7/2))

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 143

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^2*d*e*g - a^2*d*f*h*m - a*b*(d*(f*g + e*h) - c*f*h*(m + 1)) + b*f*h*(b*c - a*d)*(m + 1)*x)*(a + b*x
)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)), x] + Dist[(a*d*f*h*m + b*(d*(f*g + e*h) - c*f*h*(m +
2)))/(b^2*d), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[m
+ n + 2, 0] && NeQ[m, -1] &&  !(SumSimplerQ[n, 1] &&  !SumSimplerQ[m, 1])

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{x^3}{\sqrt{a+b x} (c+d x)^{5/2}} \, dx &=-\frac{2 c x^2 \sqrt{a+b x}}{3 d (b c-a d) (c+d x)^{3/2}}+\frac{2 \int \frac{x \left (2 a c+\frac{1}{2} (5 b c-3 a d) x\right )}{\sqrt{a+b x} (c+d x)^{3/2}} \, dx}{3 d (b c-a d)}\\ &=-\frac{2 c x^2 \sqrt{a+b x}}{3 d (b c-a d) (c+d x)^{3/2}}+\frac{\sqrt{a+b x} \left (c \left (15 b^2 c^2-22 a b c d+3 a^2 d^2\right )+d (5 b c-3 a d) (b c-a d) x\right )}{3 b d^3 (b c-a d)^2 \sqrt{c+d x}}-\frac{(5 b c+a d) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 b d^3}\\ &=-\frac{2 c x^2 \sqrt{a+b x}}{3 d (b c-a d) (c+d x)^{3/2}}+\frac{\sqrt{a+b x} \left (c \left (15 b^2 c^2-22 a b c d+3 a^2 d^2\right )+d (5 b c-3 a d) (b c-a d) x\right )}{3 b d^3 (b c-a d)^2 \sqrt{c+d x}}-\frac{(5 b c+a d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{b^2 d^3}\\ &=-\frac{2 c x^2 \sqrt{a+b x}}{3 d (b c-a d) (c+d x)^{3/2}}+\frac{\sqrt{a+b x} \left (c \left (15 b^2 c^2-22 a b c d+3 a^2 d^2\right )+d (5 b c-3 a d) (b c-a d) x\right )}{3 b d^3 (b c-a d)^2 \sqrt{c+d x}}-\frac{(5 b c+a d) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{b^2 d^3}\\ &=-\frac{2 c x^2 \sqrt{a+b x}}{3 d (b c-a d) (c+d x)^{3/2}}+\frac{\sqrt{a+b x} \left (c \left (15 b^2 c^2-22 a b c d+3 a^2 d^2\right )+d (5 b c-3 a d) (b c-a d) x\right )}{3 b d^3 (b c-a d)^2 \sqrt{c+d x}}-\frac{(5 b c+a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{3/2} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.382548, size = 182, normalized size = 1.05 \[ \frac{\frac{c \sqrt{a+b x} \left (3 a^2 d^2 (c+2 d x)-2 a b c d (11 c+15 d x)+5 b^2 c^2 (3 c+4 d x)\right )}{d^2 (b c-a d)^2}-\frac{3 (b c-a d)^{3/2} (a d+5 b c) \left (\frac{b (c+d x)}{b c-a d}\right )^{3/2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{b^2 d^{5/2}}+3 x^2 \sqrt{a+b x}}{3 b d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x^2*Sqrt[a + b*x] + (c*Sqrt[a + b*x]*(3*a^2*d^2*(c + 2*d*x) + 5*b^2*c^2*(3*c + 4*d*x) - 2*a*b*c*d*(11*c + 1
5*d*x)))/(d^2*(b*c - a*d)^2) - (3*(b*c - a*d)^(3/2)*(5*b*c + a*d)*((b*(c + d*x))/(b*c - a*d))^(3/2)*ArcSinh[(S
qrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(b^2*d^(5/2)))/(3*b*d*(c + d*x)^(3/2))

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Maple [B]  time = 0.024, size = 928, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x)

[Out]

-1/6*(b*x+a)^(1/2)*(3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a^3*d^5+
9*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a^2*b*c*d^4-27*ln(1/2*(2*b*d
*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*a*b^2*c^2*d^3+15*ln(1/2*(2*b*d*x+2*((b*x+a)
*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x^2*b^3*c^3*d^2+6*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*
(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*a^3*c*d^4+18*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c
)/(b*d)^(1/2))*x*a^2*b*c^2*d^3-54*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*
x*a*b^2*c^3*d^2+30*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*x*b^3*c^4*d-6*(
b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a^2*d^4+12*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*a*b*c*d^3-6*(b*d)^(1
/2)*((b*x+a)*(d*x+c))^(1/2)*x^2*b^2*c^2*d^2+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(
b*d)^(1/2))*a^3*c^2*d^3+9*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b*c^
3*d^2-27*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^2*c^4*d+15*ln(1/2*(2*
b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^3*c^5-12*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/
2)*x*a^2*c*d^3+60*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*a*b*c^2*d^2-40*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*x*b
^2*c^3*d-6*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*c^2*d^2+44*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c^3*d-30
*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c^4)/b/(b*d)^(1/2)/(a*d-b*c)^2/((b*x+a)*(d*x+c))^(1/2)/d^3/(d*x+c)^(3
/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(x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.69322, size = 1871, normalized size = 10.75 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/12*(3*(5*b^3*c^5 - 9*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 + a^3*c^2*d^3 + (5*b^3*c^3*d^2 - 9*a*b^2*c^2*d^3 + 3*a^2
*b*c*d^4 + a^3*d^5)*x^2 + 2*(5*b^3*c^4*d - 9*a*b^2*c^3*d^2 + 3*a^2*b*c^2*d^3 + a^3*c*d^4)*x)*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*(15*b^3*c^4*d - 22*a*b^2*c^3*d^2 + 3*a^2*b*c^2*d^3 + 3*(b^3*c^2*d^3 - 2*a*b^2*c*d^
4 + a^2*b*d^5)*x^2 + 2*(10*b^3*c^3*d^2 - 15*a*b^2*c^2*d^3 + 3*a^2*b*c*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^
4*c^4*d^4 - 2*a*b^3*c^3*d^5 + a^2*b^2*c^2*d^6 + (b^4*c^2*d^6 - 2*a*b^3*c*d^7 + a^2*b^2*d^8)*x^2 + 2*(b^4*c^3*d
^5 - 2*a*b^3*c^2*d^6 + a^2*b^2*c*d^7)*x), 1/6*(3*(5*b^3*c^5 - 9*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 + a^3*c^2*d^3 +
(5*b^3*c^3*d^2 - 9*a*b^2*c^2*d^3 + 3*a^2*b*c*d^4 + a^3*d^5)*x^2 + 2*(5*b^3*c^4*d - 9*a*b^2*c^3*d^2 + 3*a^2*b*c
^2*d^3 + a^3*c*d^4)*x)*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*(15*b^3*c^4*d - 22*a*b^2*c^3*d^2 + 3*a^2*b*c^2*d^3 + 3*(b^3*c
^2*d^3 - 2*a*b^2*c*d^4 + a^2*b*d^5)*x^2 + 2*(10*b^3*c^3*d^2 - 15*a*b^2*c^2*d^3 + 3*a^2*b*c*d^4)*x)*sqrt(b*x +
a)*sqrt(d*x + c))/(b^4*c^4*d^4 - 2*a*b^3*c^3*d^5 + a^2*b^2*c^2*d^6 + (b^4*c^2*d^6 - 2*a*b^3*c*d^7 + a^2*b^2*d^
8)*x^2 + 2*(b^4*c^3*d^5 - 2*a*b^3*c^2*d^6 + a^2*b^2*c*d^7)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\sqrt{a + b x} \left (c + d x\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(d*x+c)**(5/2)/(b*x+a)**(1/2),x)

[Out]

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

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Giac [B]  time = 3.21873, size = 504, normalized size = 2.9 \begin{align*} \frac{{\left ({\left (b x + a\right )}{\left (\frac{3 \,{\left (b^{6} c^{2} d^{4}{\left | b \right |} - 2 \, a b^{5} c d^{5}{\left | b \right |} + a^{2} b^{4} d^{6}{\left | b \right |}\right )}{\left (b x + a\right )}}{b^{7} c^{2} d^{5} - 2 \, a b^{6} c d^{6} + a^{2} b^{5} d^{7}} + \frac{2 \,{\left (10 \, b^{7} c^{3} d^{3}{\left | b \right |} - 18 \, a b^{6} c^{2} d^{4}{\left | b \right |} + 9 \, a^{2} b^{5} c d^{5}{\left | b \right |} - 3 \, a^{3} b^{4} d^{6}{\left | b \right |}\right )}}{b^{7} c^{2} d^{5} - 2 \, a b^{6} c d^{6} + a^{2} b^{5} d^{7}}\right )} + \frac{3 \,{\left (5 \, b^{8} c^{4} d^{2}{\left | b \right |} - 14 \, a b^{7} c^{3} d^{3}{\left | b \right |} + 12 \, a^{2} b^{6} c^{2} d^{4}{\left | b \right |} - 4 \, a^{3} b^{5} c d^{5}{\left | b \right |} + a^{4} b^{4} d^{6}{\left | b \right |}\right )}}{b^{7} c^{2} d^{5} - 2 \, a b^{6} c d^{6} + a^{2} b^{5} d^{7}}\right )} \sqrt{b x + a}}{3 \,{\left (b^{2} c +{\left (b x + a\right )} b d - a b d\right )}^{\frac{3}{2}}} + \frac{{\left (5 \, b c{\left | b \right |} + a d{\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 |}\right )}{\sqrt{b d} b^{2} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(d*x+c)^(5/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/3*((b*x + a)*(3*(b^6*c^2*d^4*abs(b) - 2*a*b^5*c*d^5*abs(b) + a^2*b^4*d^6*abs(b))*(b*x + a)/(b^7*c^2*d^5 - 2*
a*b^6*c*d^6 + a^2*b^5*d^7) + 2*(10*b^7*c^3*d^3*abs(b) - 18*a*b^6*c^2*d^4*abs(b) + 9*a^2*b^5*c*d^5*abs(b) - 3*a
^3*b^4*d^6*abs(b))/(b^7*c^2*d^5 - 2*a*b^6*c*d^6 + a^2*b^5*d^7)) + 3*(5*b^8*c^4*d^2*abs(b) - 14*a*b^7*c^3*d^3*a
bs(b) + 12*a^2*b^6*c^2*d^4*abs(b) - 4*a^3*b^5*c*d^5*abs(b) + a^4*b^4*d^6*abs(b))/(b^7*c^2*d^5 - 2*a*b^6*c*d^6
+ a^2*b^5*d^7))*sqrt(b*x + a)/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) + (5*b*c*abs(b) + a*d*abs(b))*log(abs(-sqr
t(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^3)