3.1443 \(\int \frac{1}{(a+b x)^4 (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=200 \[ \frac{105 b^{3/2} d^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{8 (b c-a d)^{11/2}}-\frac{105 b d^3}{8 \sqrt{c+d x} (b c-a d)^5}-\frac{35 d^3}{8 (c+d x)^{3/2} (b c-a d)^4}-\frac{21 d^2}{8 (a+b x) (c+d x)^{3/2} (b c-a d)^3}+\frac{3 d}{4 (a+b x)^2 (c+d x)^{3/2} (b c-a d)^2}-\frac{1}{3 (a+b x)^3 (c+d x)^{3/2} (b c-a d)} \]

[Out]

(-35*d^3)/(8*(b*c - a*d)^4*(c + d*x)^(3/2)) - 1/(3*(b*c - a*d)*(a + b*x)^3*(c + d*x)^(3/2)) + (3*d)/(4*(b*c -
a*d)^2*(a + b*x)^2*(c + d*x)^(3/2)) - (21*d^2)/(8*(b*c - a*d)^3*(a + b*x)*(c + d*x)^(3/2)) - (105*b*d^3)/(8*(b
*c - a*d)^5*Sqrt[c + d*x]) + (105*b^(3/2)*d^3*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(8*(b*c - a*d)
^(11/2))

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Rubi [A]  time = 0.134447, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {51, 63, 208} \[ \frac{105 b^{3/2} d^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{8 (b c-a d)^{11/2}}-\frac{105 b d^3}{8 \sqrt{c+d x} (b c-a d)^5}-\frac{35 d^3}{8 (c+d x)^{3/2} (b c-a d)^4}-\frac{21 d^2}{8 (a+b x) (c+d x)^{3/2} (b c-a d)^3}+\frac{3 d}{4 (a+b x)^2 (c+d x)^{3/2} (b c-a d)^2}-\frac{1}{3 (a+b x)^3 (c+d x)^{3/2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*d^3)/(8*(b*c - a*d)^4*(c + d*x)^(3/2)) - 1/(3*(b*c - a*d)*(a + b*x)^3*(c + d*x)^(3/2)) + (3*d)/(4*(b*c -
a*d)^2*(a + b*x)^2*(c + d*x)^(3/2)) - (21*d^2)/(8*(b*c - a*d)^3*(a + b*x)*(c + d*x)^(3/2)) - (105*b*d^3)/(8*(b
*c - a*d)^5*Sqrt[c + d*x]) + (105*b^(3/2)*d^3*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(8*(b*c - a*d)
^(11/2))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 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{1}{(a+b x)^4 (c+d x)^{5/2}} \, dx &=-\frac{1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}-\frac{(3 d) \int \frac{1}{(a+b x)^3 (c+d x)^{5/2}} \, dx}{2 (b c-a d)}\\ &=-\frac{1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac{3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}+\frac{\left (21 d^2\right ) \int \frac{1}{(a+b x)^2 (c+d x)^{5/2}} \, dx}{8 (b c-a d)^2}\\ &=-\frac{1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac{3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}-\frac{21 d^2}{8 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}-\frac{\left (105 d^3\right ) \int \frac{1}{(a+b x) (c+d x)^{5/2}} \, dx}{16 (b c-a d)^3}\\ &=-\frac{35 d^3}{8 (b c-a d)^4 (c+d x)^{3/2}}-\frac{1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac{3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}-\frac{21 d^2}{8 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}-\frac{\left (105 b d^3\right ) \int \frac{1}{(a+b x) (c+d x)^{3/2}} \, dx}{16 (b c-a d)^4}\\ &=-\frac{35 d^3}{8 (b c-a d)^4 (c+d x)^{3/2}}-\frac{1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac{3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}-\frac{21 d^2}{8 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}-\frac{105 b d^3}{8 (b c-a d)^5 \sqrt{c+d x}}-\frac{\left (105 b^2 d^3\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{16 (b c-a d)^5}\\ &=-\frac{35 d^3}{8 (b c-a d)^4 (c+d x)^{3/2}}-\frac{1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac{3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}-\frac{21 d^2}{8 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}-\frac{105 b d^3}{8 (b c-a d)^5 \sqrt{c+d x}}-\frac{\left (105 b^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{8 (b c-a d)^5}\\ &=-\frac{35 d^3}{8 (b c-a d)^4 (c+d x)^{3/2}}-\frac{1}{3 (b c-a d) (a+b x)^3 (c+d x)^{3/2}}+\frac{3 d}{4 (b c-a d)^2 (a+b x)^2 (c+d x)^{3/2}}-\frac{21 d^2}{8 (b c-a d)^3 (a+b x) (c+d x)^{3/2}}-\frac{105 b d^3}{8 (b c-a d)^5 \sqrt{c+d x}}+\frac{105 b^{3/2} d^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{8 (b c-a d)^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0213188, size = 52, normalized size = 0.26 \[ -\frac{2 d^3 \, _2F_1\left (-\frac{3}{2},4;-\frac{1}{2};-\frac{b (c+d x)}{a d-b c}\right )}{3 (c+d x)^{3/2} (a d-b c)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^3*Hypergeometric2F1[-3/2, 4, -1/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(3*(-(b*c) + a*d)^4*(c + d*x)^(3/2)
)

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Maple [A]  time = 0.019, size = 319, normalized size = 1.6 \begin{align*} -{\frac{2\,{d}^{3}}{3\, \left ( ad-bc \right ) ^{4}} \left ( dx+c \right ) ^{-{\frac{3}{2}}}}+8\,{\frac{{d}^{3}b}{ \left ( ad-bc \right ) ^{5}\sqrt{dx+c}}}+{\frac{41\,{d}^{3}{b}^{4}}{8\, \left ( ad-bc \right ) ^{5} \left ( bdx+ad \right ) ^{3}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}+{\frac{35\,{d}^{4}{b}^{3}a}{3\, \left ( ad-bc \right ) ^{5} \left ( bdx+ad \right ) ^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{35\,{d}^{3}{b}^{4}c}{3\, \left ( ad-bc \right ) ^{5} \left ( bdx+ad \right ) ^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{55\,{d}^{5}{b}^{2}{a}^{2}}{8\, \left ( ad-bc \right ) ^{5} \left ( bdx+ad \right ) ^{3}}\sqrt{dx+c}}-{\frac{55\,{d}^{4}{b}^{3}ac}{4\, \left ( ad-bc \right ) ^{5} \left ( bdx+ad \right ) ^{3}}\sqrt{dx+c}}+{\frac{55\,{d}^{3}{b}^{4}{c}^{2}}{8\, \left ( ad-bc \right ) ^{5} \left ( bdx+ad \right ) ^{3}}\sqrt{dx+c}}+{\frac{105\,{d}^{3}{b}^{2}}{8\, \left ( ad-bc \right ) ^{5}}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*d^3/(a*d-b*c)^4/(d*x+c)^(3/2)+8*d^3/(a*d-b*c)^5*b/(d*x+c)^(1/2)+41/8*d^3/(a*d-b*c)^5*b^4/(b*d*x+a*d)^3*(d
*x+c)^(5/2)+35/3*d^4/(a*d-b*c)^5*b^3/(b*d*x+a*d)^3*(d*x+c)^(3/2)*a-35/3*d^3/(a*d-b*c)^5*b^4/(b*d*x+a*d)^3*(d*x
+c)^(3/2)*c+55/8*d^5/(a*d-b*c)^5*b^2/(b*d*x+a*d)^3*(d*x+c)^(1/2)*a^2-55/4*d^4/(a*d-b*c)^5*b^3/(b*d*x+a*d)^3*(d
*x+c)^(1/2)*a*c+55/8*d^3/(a*d-b*c)^5*b^4/(b*d*x+a*d)^3*(d*x+c)^(1/2)*c^2+105/8*d^3/(a*d-b*c)^5*b^2/((a*d-b*c)*
b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(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(1/(b*x+a)^4/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.55919, size = 3717, normalized size = 18.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(315*(b^4*d^5*x^5 + a^3*b*c^2*d^3 + (2*b^4*c*d^4 + 3*a*b^3*d^5)*x^4 + (b^4*c^2*d^3 + 6*a*b^3*c*d^4 + 3*
a^2*b^2*d^5)*x^3 + (3*a*b^3*c^2*d^3 + 6*a^2*b^2*c*d^4 + a^3*b*d^5)*x^2 + (3*a^2*b^2*c^2*d^3 + 2*a^3*b*c*d^4)*x
)*sqrt(b/(b*c - a*d))*log((b*d*x + 2*b*c - a*d - 2*(b*c - a*d)*sqrt(d*x + c)*sqrt(b/(b*c - a*d)))/(b*x + a)) +
2*(315*b^4*d^4*x^4 + 8*b^4*c^4 - 50*a*b^3*c^3*d + 165*a^2*b^2*c^2*d^2 + 208*a^3*b*c*d^3 - 16*a^4*d^4 + 420*(b
^4*c*d^3 + 2*a*b^3*d^4)*x^3 + 63*(b^4*c^2*d^2 + 18*a*b^3*c*d^3 + 11*a^2*b^2*d^4)*x^2 - 18*(b^4*c^3*d - 10*a*b^
3*c^2*d^2 - 53*a^2*b^2*c*d^3 - 8*a^3*b*d^4)*x)*sqrt(d*x + c))/(a^3*b^5*c^7 - 5*a^4*b^4*c^6*d + 10*a^5*b^3*c^5*
d^2 - 10*a^6*b^2*c^4*d^3 + 5*a^7*b*c^3*d^4 - a^8*c^2*d^5 + (b^8*c^5*d^2 - 5*a*b^7*c^4*d^3 + 10*a^2*b^6*c^3*d^4
- 10*a^3*b^5*c^2*d^5 + 5*a^4*b^4*c*d^6 - a^5*b^3*d^7)*x^5 + (2*b^8*c^6*d - 7*a*b^7*c^5*d^2 + 5*a^2*b^6*c^4*d^
3 + 10*a^3*b^5*c^3*d^4 - 20*a^4*b^4*c^2*d^5 + 13*a^5*b^3*c*d^6 - 3*a^6*b^2*d^7)*x^4 + (b^8*c^7 + a*b^7*c^6*d -
17*a^2*b^6*c^5*d^2 + 35*a^3*b^5*c^4*d^3 - 25*a^4*b^4*c^3*d^4 - a^5*b^3*c^2*d^5 + 9*a^6*b^2*c*d^6 - 3*a^7*b*d^
7)*x^3 + (3*a*b^7*c^7 - 9*a^2*b^6*c^6*d + a^3*b^5*c^5*d^2 + 25*a^4*b^4*c^4*d^3 - 35*a^5*b^3*c^3*d^4 + 17*a^6*b
^2*c^2*d^5 - a^7*b*c*d^6 - a^8*d^7)*x^2 + (3*a^2*b^6*c^7 - 13*a^3*b^5*c^6*d + 20*a^4*b^4*c^5*d^2 - 10*a^5*b^3*
c^4*d^3 - 5*a^6*b^2*c^3*d^4 + 7*a^7*b*c^2*d^5 - 2*a^8*c*d^6)*x), 1/24*(315*(b^4*d^5*x^5 + a^3*b*c^2*d^3 + (2*b
^4*c*d^4 + 3*a*b^3*d^5)*x^4 + (b^4*c^2*d^3 + 6*a*b^3*c*d^4 + 3*a^2*b^2*d^5)*x^3 + (3*a*b^3*c^2*d^3 + 6*a^2*b^2
*c*d^4 + a^3*b*d^5)*x^2 + (3*a^2*b^2*c^2*d^3 + 2*a^3*b*c*d^4)*x)*sqrt(-b/(b*c - a*d))*arctan(-(b*c - a*d)*sqrt
(d*x + c)*sqrt(-b/(b*c - a*d))/(b*d*x + b*c)) - (315*b^4*d^4*x^4 + 8*b^4*c^4 - 50*a*b^3*c^3*d + 165*a^2*b^2*c^
2*d^2 + 208*a^3*b*c*d^3 - 16*a^4*d^4 + 420*(b^4*c*d^3 + 2*a*b^3*d^4)*x^3 + 63*(b^4*c^2*d^2 + 18*a*b^3*c*d^3 +
11*a^2*b^2*d^4)*x^2 - 18*(b^4*c^3*d - 10*a*b^3*c^2*d^2 - 53*a^2*b^2*c*d^3 - 8*a^3*b*d^4)*x)*sqrt(d*x + c))/(a^
3*b^5*c^7 - 5*a^4*b^4*c^6*d + 10*a^5*b^3*c^5*d^2 - 10*a^6*b^2*c^4*d^3 + 5*a^7*b*c^3*d^4 - a^8*c^2*d^5 + (b^8*c
^5*d^2 - 5*a*b^7*c^4*d^3 + 10*a^2*b^6*c^3*d^4 - 10*a^3*b^5*c^2*d^5 + 5*a^4*b^4*c*d^6 - a^5*b^3*d^7)*x^5 + (2*b
^8*c^6*d - 7*a*b^7*c^5*d^2 + 5*a^2*b^6*c^4*d^3 + 10*a^3*b^5*c^3*d^4 - 20*a^4*b^4*c^2*d^5 + 13*a^5*b^3*c*d^6 -
3*a^6*b^2*d^7)*x^4 + (b^8*c^7 + a*b^7*c^6*d - 17*a^2*b^6*c^5*d^2 + 35*a^3*b^5*c^4*d^3 - 25*a^4*b^4*c^3*d^4 - a
^5*b^3*c^2*d^5 + 9*a^6*b^2*c*d^6 - 3*a^7*b*d^7)*x^3 + (3*a*b^7*c^7 - 9*a^2*b^6*c^6*d + a^3*b^5*c^5*d^2 + 25*a^
4*b^4*c^4*d^3 - 35*a^5*b^3*c^3*d^4 + 17*a^6*b^2*c^2*d^5 - a^7*b*c*d^6 - a^8*d^7)*x^2 + (3*a^2*b^6*c^7 - 13*a^3
*b^5*c^6*d + 20*a^4*b^4*c^5*d^2 - 10*a^5*b^3*c^4*d^3 - 5*a^6*b^2*c^3*d^4 + 7*a^7*b*c^2*d^5 - 2*a^8*c*d^6)*x)]

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

[Out]

Timed out

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Giac [B]  time = 1.06702, size = 583, normalized size = 2.92 \begin{align*} -\frac{105 \, b^{2} d^{3} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{8 \,{\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} \sqrt{-b^{2} c + a b d}} - \frac{315 \,{\left (d x + c\right )}^{4} b^{4} d^{3} - 840 \,{\left (d x + c\right )}^{3} b^{4} c d^{3} + 693 \,{\left (d x + c\right )}^{2} b^{4} c^{2} d^{3} - 144 \,{\left (d x + c\right )} b^{4} c^{3} d^{3} - 16 \, b^{4} c^{4} d^{3} + 840 \,{\left (d x + c\right )}^{3} a b^{3} d^{4} - 1386 \,{\left (d x + c\right )}^{2} a b^{3} c d^{4} + 432 \,{\left (d x + c\right )} a b^{3} c^{2} d^{4} + 64 \, a b^{3} c^{3} d^{4} + 693 \,{\left (d x + c\right )}^{2} a^{2} b^{2} d^{5} - 432 \,{\left (d x + c\right )} a^{2} b^{2} c d^{5} - 96 \, a^{2} b^{2} c^{2} d^{5} + 144 \,{\left (d x + c\right )} a^{3} b d^{6} + 64 \, a^{3} b c d^{6} - 16 \, a^{4} d^{7}}{24 \,{\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )}{\left ({\left (d x + c\right )}^{\frac{3}{2}} b - \sqrt{d x + c} b c + \sqrt{d x + c} a d\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-105/8*b^2*d^3*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 1
0*a^3*b^2*c^2*d^3 + 5*a^4*b*c*d^4 - a^5*d^5)*sqrt(-b^2*c + a*b*d)) - 1/24*(315*(d*x + c)^4*b^4*d^3 - 840*(d*x
+ c)^3*b^4*c*d^3 + 693*(d*x + c)^2*b^4*c^2*d^3 - 144*(d*x + c)*b^4*c^3*d^3 - 16*b^4*c^4*d^3 + 840*(d*x + c)^3*
a*b^3*d^4 - 1386*(d*x + c)^2*a*b^3*c*d^4 + 432*(d*x + c)*a*b^3*c^2*d^4 + 64*a*b^3*c^3*d^4 + 693*(d*x + c)^2*a^
2*b^2*d^5 - 432*(d*x + c)*a^2*b^2*c*d^5 - 96*a^2*b^2*c^2*d^5 + 144*(d*x + c)*a^3*b*d^6 + 64*a^3*b*c*d^6 - 16*a
^4*d^7)/((b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 10*a^3*b^2*c^2*d^3 + 5*a^4*b*c*d^4 - a^5*d^5)*((d*x +
c)^(3/2)*b - sqrt(d*x + c)*b*c + sqrt(d*x + c)*a*d)^3)