3.1422 \(\int \frac{1}{(a+b x)^4 \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=147 \[ -\frac{5 d^2 \sqrt{c+d x}}{8 (a+b x) (b c-a d)^3}+\frac{5 d^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{8 \sqrt{b} (b c-a d)^{7/2}}+\frac{5 d \sqrt{c+d x}}{12 (a+b x)^2 (b c-a d)^2}-\frac{\sqrt{c+d x}}{3 (a+b x)^3 (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0504401, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {51, 63, 208} \[ -\frac{5 d^2 \sqrt{c+d x}}{8 (a+b x) (b c-a d)^3}+\frac{5 d^3 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{8 \sqrt{b} (b c-a d)^{7/2}}+\frac{5 d \sqrt{c+d x}}{12 (a+b x)^2 (b c-a d)^2}-\frac{\sqrt{c+d x}}{3 (a+b x)^3 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^4*Sqrt[c + d*x]),x]

[Out]

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

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

Mathematica [C]  time = 0.0115655, size = 50, normalized size = 0.34 \[ \frac{2 d^3 \sqrt{c+d x} \, _2F_1\left (\frac{1}{2},4;\frac{3}{2};-\frac{b (c+d x)}{a d-b c}\right )}{(a d-b c)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 147, normalized size = 1. \begin{align*}{\frac{{d}^{3}}{ \left ( 3\,ad-3\,bc \right ) \left ( bdx+ad \right ) ^{3}}\sqrt{dx+c}}+{\frac{5\,{d}^{3}}{12\, \left ( ad-bc \right ) ^{2} \left ( bdx+ad \right ) ^{2}}\sqrt{dx+c}}+{\frac{5\,{d}^{3}}{8\, \left ( ad-bc \right ) ^{3} \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{5\,{d}^{3}}{8\, \left ( ad-bc \right ) ^{3}}\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)^(1/2),x)

[Out]

1/3*d^3*(d*x+c)^(1/2)/(a*d-b*c)/(b*d*x+a*d)^3+5/12*d^3/(a*d-b*c)^2*(d*x+c)^(1/2)/(b*d*x+a*d)^2+5/8*d^3/(a*d-b*
c)^3*(d*x+c)^(1/2)/(b*d*x+a*d)+5/8*d^3/(a*d-b*c)^3/((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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.26014, size = 1805, normalized size = 12.28 \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)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(15*(b^3*d^3*x^3 + 3*a*b^2*d^3*x^2 + 3*a^2*b*d^3*x + a^3*d^3)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b*c -
a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) + 2*(8*b^4*c^3 - 34*a*b^3*c^2*d + 59*a^2*b^2*c*d^2 - 33*
a^3*b*d^3 + 15*(b^4*c*d^2 - a*b^3*d^3)*x^2 - 10*(b^4*c^2*d - 5*a*b^3*c*d^2 + 4*a^2*b^2*d^3)*x)*sqrt(d*x + c))/
(a^3*b^5*c^4 - 4*a^4*b^4*c^3*d + 6*a^5*b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^4 + (b^8*c^4 - 4*a*b^7*c^3*d +
6*a^2*b^6*c^2*d^2 - 4*a^3*b^5*c*d^3 + a^4*b^4*d^4)*x^3 + 3*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 -
4*a^4*b^4*c*d^3 + a^5*b^3*d^4)*x^2 + 3*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4*a^5*b^3*c*d^3 +
a^6*b^2*d^4)*x), -1/24*(15*(b^3*d^3*x^3 + 3*a*b^2*d^3*x^2 + 3*a^2*b*d^3*x + a^3*d^3)*sqrt(-b^2*c + a*b*d)*arct
an(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) + (8*b^4*c^3 - 34*a*b^3*c^2*d + 59*a^2*b^2*c*d^2 - 33*a^3
*b*d^3 + 15*(b^4*c*d^2 - a*b^3*d^3)*x^2 - 10*(b^4*c^2*d - 5*a*b^3*c*d^2 + 4*a^2*b^2*d^3)*x)*sqrt(d*x + c))/(a^
3*b^5*c^4 - 4*a^4*b^4*c^3*d + 6*a^5*b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^4 + (b^8*c^4 - 4*a*b^7*c^3*d + 6*a
^2*b^6*c^2*d^2 - 4*a^3*b^5*c*d^3 + a^4*b^4*d^4)*x^3 + 3*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 - 4*a
^4*b^4*c*d^3 + a^5*b^3*d^4)*x^2 + 3*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4*a^5*b^3*c*d^3 + a^6
*b^2*d^4)*x)]

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Sympy [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)**(1/2),x)

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.07195, size = 312, normalized size = 2.12 \begin{align*} -\frac{5 \, d^{3} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{8 \,{\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{-b^{2} c + a b d}} - \frac{15 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{2} d^{3} - 40 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{2} c d^{3} + 33 \, \sqrt{d x + c} b^{2} c^{2} d^{3} + 40 \,{\left (d x + c\right )}^{\frac{3}{2}} a b d^{4} - 66 \, \sqrt{d x + c} a b c d^{4} + 33 \, \sqrt{d x + c} a^{2} d^{5}}{24 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\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)^(1/2),x, algorithm="giac")

[Out]

-5/8*d^3*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqr
t(-b^2*c + a*b*d)) - 1/24*(15*(d*x + c)^(5/2)*b^2*d^3 - 40*(d*x + c)^(3/2)*b^2*c*d^3 + 33*sqrt(d*x + c)*b^2*c^
2*d^3 + 40*(d*x + c)^(3/2)*a*b*d^4 - 66*sqrt(d*x + c)*a*b*c*d^4 + 33*sqrt(d*x + c)*a^2*d^5)/((b^3*c^3 - 3*a*b^
2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*((d*x + c)*b - b*c + a*d)^3)