3.1433 \(\int \frac{1}{(a+b x)^4 (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=173 \[ -\frac{35 d^3}{8 \sqrt{c+d x} (b c-a d)^4}-\frac{35 d^2}{24 (a+b x) \sqrt{c+d x} (b c-a d)^3}+\frac{35 \sqrt{b} d^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{8 (b c-a d)^{9/2}}+\frac{7 d}{12 (a+b x)^2 \sqrt{c+d x} (b c-a d)^2}-\frac{1}{3 (a+b x)^3 \sqrt{c+d x} (b c-a d)} \]

[Out]

(-35*d^3)/(8*(b*c - a*d)^4*Sqrt[c + d*x]) - 1/(3*(b*c - a*d)*(a + b*x)^3*Sqrt[c + d*x]) + (7*d)/(12*(b*c - a*d
)^2*(a + b*x)^2*Sqrt[c + d*x]) - (35*d^2)/(24*(b*c - a*d)^3*(a + b*x)*Sqrt[c + d*x]) + (35*Sqrt[b]*d^3*ArcTanh
[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(8*(b*c - a*d)^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0660012, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {51, 63, 208} \[ -\frac{35 d^3}{8 \sqrt{c+d x} (b c-a d)^4}-\frac{35 d^2}{24 (a+b x) \sqrt{c+d x} (b c-a d)^3}+\frac{35 \sqrt{b} d^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{8 (b c-a d)^{9/2}}+\frac{7 d}{12 (a+b x)^2 \sqrt{c+d x} (b c-a d)^2}-\frac{1}{3 (a+b x)^3 \sqrt{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*d^3)/(8*(b*c - a*d)^4*Sqrt[c + d*x]) - 1/(3*(b*c - a*d)*(a + b*x)^3*Sqrt[c + d*x]) + (7*d)/(12*(b*c - a*d
)^2*(a + b*x)^2*Sqrt[c + d*x]) - (35*d^2)/(24*(b*c - a*d)^3*(a + b*x)*Sqrt[c + d*x]) + (35*Sqrt[b]*d^3*ArcTanh
[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(8*(b*c - a*d)^(9/2))

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)^{3/2}} \, dx &=-\frac{1}{3 (b c-a d) (a+b x)^3 \sqrt{c+d x}}-\frac{(7 d) \int \frac{1}{(a+b x)^3 (c+d x)^{3/2}} \, dx}{6 (b c-a d)}\\ &=-\frac{1}{3 (b c-a d) (a+b x)^3 \sqrt{c+d x}}+\frac{7 d}{12 (b c-a d)^2 (a+b x)^2 \sqrt{c+d x}}+\frac{\left (35 d^2\right ) \int \frac{1}{(a+b x)^2 (c+d x)^{3/2}} \, dx}{24 (b c-a d)^2}\\ &=-\frac{1}{3 (b c-a d) (a+b x)^3 \sqrt{c+d x}}+\frac{7 d}{12 (b c-a d)^2 (a+b x)^2 \sqrt{c+d x}}-\frac{35 d^2}{24 (b c-a d)^3 (a+b x) \sqrt{c+d x}}-\frac{\left (35 d^3\right ) \int \frac{1}{(a+b x) (c+d x)^{3/2}} \, dx}{16 (b c-a d)^3}\\ &=-\frac{35 d^3}{8 (b c-a d)^4 \sqrt{c+d x}}-\frac{1}{3 (b c-a d) (a+b x)^3 \sqrt{c+d x}}+\frac{7 d}{12 (b c-a d)^2 (a+b x)^2 \sqrt{c+d x}}-\frac{35 d^2}{24 (b c-a d)^3 (a+b x) \sqrt{c+d x}}-\frac{\left (35 b d^3\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{16 (b c-a d)^4}\\ &=-\frac{35 d^3}{8 (b c-a d)^4 \sqrt{c+d x}}-\frac{1}{3 (b c-a d) (a+b x)^3 \sqrt{c+d x}}+\frac{7 d}{12 (b c-a d)^2 (a+b x)^2 \sqrt{c+d x}}-\frac{35 d^2}{24 (b c-a d)^3 (a+b x) \sqrt{c+d x}}-\frac{\left (35 b 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)^4}\\ &=-\frac{35 d^3}{8 (b c-a d)^4 \sqrt{c+d x}}-\frac{1}{3 (b c-a d) (a+b x)^3 \sqrt{c+d x}}+\frac{7 d}{12 (b c-a d)^2 (a+b x)^2 \sqrt{c+d x}}-\frac{35 d^2}{24 (b c-a d)^3 (a+b x) \sqrt{c+d x}}+\frac{35 \sqrt{b} d^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{8 (b c-a d)^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0156973, size = 50, normalized size = 0.29 \[ -\frac{2 d^3 \, _2F_1\left (-\frac{1}{2},4;\frac{1}{2};-\frac{b (c+d x)}{a d-b c}\right )}{\sqrt{c+d x} (a d-b c)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.018, size = 292, normalized size = 1.7 \begin{align*} -2\,{\frac{{d}^{3}}{ \left ( ad-bc \right ) ^{4}\sqrt{dx+c}}}-{\frac{19\,{d}^{3}{b}^{3}}{8\, \left ( ad-bc \right ) ^{4} \left ( bdx+ad \right ) ^{3}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}-{\frac{17\,{d}^{4}{b}^{2}a}{3\, \left ( ad-bc \right ) ^{4} \left ( bdx+ad \right ) ^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{17\,{d}^{3}{b}^{3}c}{3\, \left ( ad-bc \right ) ^{4} \left ( bdx+ad \right ) ^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{29\,{d}^{5}b{a}^{2}}{8\, \left ( ad-bc \right ) ^{4} \left ( bdx+ad \right ) ^{3}}\sqrt{dx+c}}+{\frac{29\,{d}^{4}{b}^{2}ac}{4\, \left ( ad-bc \right ) ^{4} \left ( bdx+ad \right ) ^{3}}\sqrt{dx+c}}-{\frac{29\,{d}^{3}{b}^{3}{c}^{2}}{8\, \left ( ad-bc \right ) ^{4} \left ( bdx+ad \right ) ^{3}}\sqrt{dx+c}}-{\frac{35\,{d}^{3}b}{8\, \left ( ad-bc \right ) ^{4}}\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)^(3/2),x)

[Out]

-2*d^3/(a*d-b*c)^4/(d*x+c)^(1/2)-19/8*d^3/(a*d-b*c)^4*b^3/(b*d*x+a*d)^3*(d*x+c)^(5/2)-17/3*d^4/(a*d-b*c)^4*b^2
/(b*d*x+a*d)^3*(d*x+c)^(3/2)*a+17/3*d^3/(a*d-b*c)^4*b^3/(b*d*x+a*d)^3*(d*x+c)^(3/2)*c-29/8*d^5/(a*d-b*c)^4*b/(
b*d*x+a*d)^3*(d*x+c)^(1/2)*a^2+29/4*d^4/(a*d-b*c)^4*b^2/(b*d*x+a*d)^3*(d*x+c)^(1/2)*a*c-29/8*d^3/(a*d-b*c)^4*b
^3/(b*d*x+a*d)^3*(d*x+c)^(1/2)*c^2-35/8*d^3/(a*d-b*c)^4*b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c
)*b)^(1/2))

________________________________________________________________________________________

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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.37498, size = 2441, normalized size = 14.11 \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)^(3/2),x, algorithm="fricas")

[Out]

[1/48*(105*(b^3*d^4*x^4 + a^3*c*d^3 + (b^3*c*d^3 + 3*a*b^2*d^4)*x^3 + 3*(a*b^2*c*d^3 + a^2*b*d^4)*x^2 + (3*a^2
*b*c*d^3 + a^3*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*(105*b^3*d^3*x^3 + 8*b^3*c^3 - 38*a*b^2*c^2*d + 87*a^2*b*c*d^2 + 48*a^3*d^3 + 35*(b^3*
c*d^2 + 8*a*b^2*d^3)*x^2 - 7*(2*b^3*c^2*d - 14*a*b^2*c*d^2 - 33*a^2*b*d^3)*x)*sqrt(d*x + c))/(a^3*b^4*c^5 - 4*
a^4*b^3*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d - 4*a*b^6*c^3*d^2 + 6*a^2*b^5*c^2
*d^3 - 4*a^3*b^4*c*d^4 + a^4*b^3*d^5)*x^4 + (b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b^4*c^2*d^3 -
11*a^4*b^3*c*d^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3*b^4*c^3*d^2 + 2*a^4*b^3*c^2*d^3
- 3*a^5*b^2*c*d^4 + a^6*b*d^5)*x^2 + (3*a^2*b^5*c^5 - 11*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*a^5*b^2*c^2*d
^3 - a^6*b*c*d^4 + a^7*d^5)*x), 1/24*(105*(b^3*d^4*x^4 + a^3*c*d^3 + (b^3*c*d^3 + 3*a*b^2*d^4)*x^3 + 3*(a*b^2*
c*d^3 + a^2*b*d^4)*x^2 + (3*a^2*b*c*d^3 + a^3*d^4)*x)*sqrt(-b/(b*c - a*d))*arctan(-(b*c - a*d)*sqrt(d*x + c)*s
qrt(-b/(b*c - a*d))/(b*d*x + b*c)) - (105*b^3*d^3*x^3 + 8*b^3*c^3 - 38*a*b^2*c^2*d + 87*a^2*b*c*d^2 + 48*a^3*d
^3 + 35*(b^3*c*d^2 + 8*a*b^2*d^3)*x^2 - 7*(2*b^3*c^2*d - 14*a*b^2*c*d^2 - 33*a^2*b*d^3)*x)*sqrt(d*x + c))/(a^3
*b^4*c^5 - 4*a^4*b^3*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d - 4*a*b^6*c^3*d^2 +
6*a^2*b^5*c^2*d^3 - 4*a^3*b^4*c*d^4 + a^4*b^3*d^5)*x^4 + (b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b
^4*c^2*d^3 - 11*a^4*b^3*c*d^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3*b^4*c^3*d^2 + 2*a^
4*b^3*c^2*d^3 - 3*a^5*b^2*c*d^4 + a^6*b*d^5)*x^2 + (3*a^2*b^5*c^5 - 11*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*
a^5*b^2*c^2*d^3 - a^6*b*c*d^4 + a^7*d^5)*x)]

________________________________________________________________________________________

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)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.07874, size = 440, normalized size = 2.54 \begin{align*} -\frac{35 \, b d^{3} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{8 \,{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt{-b^{2} c + a b d}} - \frac{2 \, d^{3}}{{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt{d x + c}} - \frac{57 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{3} d^{3} - 136 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{3} c d^{3} + 87 \, \sqrt{d x + c} b^{3} c^{2} d^{3} + 136 \,{\left (d x + c\right )}^{\frac{3}{2}} a b^{2} d^{4} - 174 \, \sqrt{d x + c} a b^{2} c d^{4} + 87 \, \sqrt{d x + c} a^{2} b d^{5}}{24 \,{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )}{\left ({\left (d x + c\right )} b - b 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)^(3/2),x, algorithm="giac")

[Out]

-35/8*b*d^3*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3
*b*c*d^3 + a^4*d^4)*sqrt(-b^2*c + a*b*d)) - 2*d^3/((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^
3 + a^4*d^4)*sqrt(d*x + c)) - 1/24*(57*(d*x + c)^(5/2)*b^3*d^3 - 136*(d*x + c)^(3/2)*b^3*c*d^3 + 87*sqrt(d*x +
c)*b^3*c^2*d^3 + 136*(d*x + c)^(3/2)*a*b^2*d^4 - 174*sqrt(d*x + c)*a*b^2*c*d^4 + 87*sqrt(d*x + c)*a^2*b*d^5)/
((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*((d*x + c)*b - b*c + a*d)^3)