3.1423 \(\int \frac{1}{(a+b x)^5 \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=180 \[ \frac{35 d^3 \sqrt{c+d x}}{64 (a+b x) (b c-a d)^4}-\frac{35 d^2 \sqrt{c+d x}}{96 (a+b x)^2 (b c-a d)^3}-\frac{35 d^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{64 \sqrt{b} (b c-a d)^{9/2}}+\frac{7 d \sqrt{c+d x}}{24 (a+b x)^3 (b c-a d)^2}-\frac{\sqrt{c+d x}}{4 (a+b x)^4 (b c-a d)} \]

[Out]

-Sqrt[c + d*x]/(4*(b*c - a*d)*(a + b*x)^4) + (7*d*Sqrt[c + d*x])/(24*(b*c - a*d)^2*(a + b*x)^3) - (35*d^2*Sqrt
[c + d*x])/(96*(b*c - a*d)^3*(a + b*x)^2) + (35*d^3*Sqrt[c + d*x])/(64*(b*c - a*d)^4*(a + b*x)) - (35*d^4*ArcT
anh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(64*Sqrt[b]*(b*c - a*d)^(9/2))

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Rubi [A]  time = 0.0647658, antiderivative size = 180, 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} \[ \frac{35 d^3 \sqrt{c+d x}}{64 (a+b x) (b c-a d)^4}-\frac{35 d^2 \sqrt{c+d x}}{96 (a+b x)^2 (b c-a d)^3}-\frac{35 d^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{64 \sqrt{b} (b c-a d)^{9/2}}+\frac{7 d \sqrt{c+d x}}{24 (a+b x)^3 (b c-a d)^2}-\frac{\sqrt{c+d x}}{4 (a+b x)^4 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^5*Sqrt[c + d*x]),x]

[Out]

-Sqrt[c + d*x]/(4*(b*c - a*d)*(a + b*x)^4) + (7*d*Sqrt[c + d*x])/(24*(b*c - a*d)^2*(a + b*x)^3) - (35*d^2*Sqrt
[c + d*x])/(96*(b*c - a*d)^3*(a + b*x)^2) + (35*d^3*Sqrt[c + d*x])/(64*(b*c - a*d)^4*(a + b*x)) - (35*d^4*ArcT
anh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(64*Sqrt[b]*(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)^5 \sqrt{c+d x}} \, dx &=-\frac{\sqrt{c+d x}}{4 (b c-a d) (a+b x)^4}-\frac{(7 d) \int \frac{1}{(a+b x)^4 \sqrt{c+d x}} \, dx}{8 (b c-a d)}\\ &=-\frac{\sqrt{c+d x}}{4 (b c-a d) (a+b x)^4}+\frac{7 d \sqrt{c+d x}}{24 (b c-a d)^2 (a+b x)^3}+\frac{\left (35 d^2\right ) \int \frac{1}{(a+b x)^3 \sqrt{c+d x}} \, dx}{48 (b c-a d)^2}\\ &=-\frac{\sqrt{c+d x}}{4 (b c-a d) (a+b x)^4}+\frac{7 d \sqrt{c+d x}}{24 (b c-a d)^2 (a+b x)^3}-\frac{35 d^2 \sqrt{c+d x}}{96 (b c-a d)^3 (a+b x)^2}-\frac{\left (35 d^3\right ) \int \frac{1}{(a+b x)^2 \sqrt{c+d x}} \, dx}{64 (b c-a d)^3}\\ &=-\frac{\sqrt{c+d x}}{4 (b c-a d) (a+b x)^4}+\frac{7 d \sqrt{c+d x}}{24 (b c-a d)^2 (a+b x)^3}-\frac{35 d^2 \sqrt{c+d x}}{96 (b c-a d)^3 (a+b x)^2}+\frac{35 d^3 \sqrt{c+d x}}{64 (b c-a d)^4 (a+b x)}+\frac{\left (35 d^4\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{128 (b c-a d)^4}\\ &=-\frac{\sqrt{c+d x}}{4 (b c-a d) (a+b x)^4}+\frac{7 d \sqrt{c+d x}}{24 (b c-a d)^2 (a+b x)^3}-\frac{35 d^2 \sqrt{c+d x}}{96 (b c-a d)^3 (a+b x)^2}+\frac{35 d^3 \sqrt{c+d x}}{64 (b c-a d)^4 (a+b x)}+\frac{\left (35 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{64 (b c-a d)^4}\\ &=-\frac{\sqrt{c+d x}}{4 (b c-a d) (a+b x)^4}+\frac{7 d \sqrt{c+d x}}{24 (b c-a d)^2 (a+b x)^3}-\frac{35 d^2 \sqrt{c+d x}}{96 (b c-a d)^3 (a+b x)^2}+\frac{35 d^3 \sqrt{c+d x}}{64 (b c-a d)^4 (a+b x)}-\frac{35 d^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{64 \sqrt{b} (b c-a d)^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0117945, size = 50, normalized size = 0.28 \[ \frac{2 d^4 \sqrt{c+d x} \, _2F_1\left (\frac{1}{2},5;\frac{3}{2};-\frac{b (c+d x)}{a d-b c}\right )}{(a d-b c)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^4*Sqrt[c + d*x]*Hypergeometric2F1[1/2, 5, 3/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(-(b*c) + a*d)^5

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Maple [A]  time = 0.006, size = 179, normalized size = 1. \begin{align*}{\frac{{d}^{4}}{ \left ( 4\,ad-4\,bc \right ) \left ( bdx+ad \right ) ^{4}}\sqrt{dx+c}}+{\frac{7\,{d}^{4}}{24\, \left ( ad-bc \right ) ^{2} \left ( bdx+ad \right ) ^{3}}\sqrt{dx+c}}+{\frac{35\,{d}^{4}}{96\, \left ( ad-bc \right ) ^{3} \left ( bdx+ad \right ) ^{2}}\sqrt{dx+c}}+{\frac{35\,{d}^{4}}{64\, \left ( ad-bc \right ) ^{4} \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{35\,{d}^{4}}{64\, \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)^5/(d*x+c)^(1/2),x)

[Out]

1/4*d^4*(d*x+c)^(1/2)/(a*d-b*c)/(b*d*x+a*d)^4+7/24*d^4/(a*d-b*c)^2*(d*x+c)^(1/2)/(b*d*x+a*d)^3+35/96*d^4/(a*d-
b*c)^3*(d*x+c)^(1/2)/(b*d*x+a*d)^2+35/64*d^4/(a*d-b*c)^4*(d*x+c)^(1/2)/(b*d*x+a*d)+35/64*d^4/(a*d-b*c)^4/((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)^5/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.39643, size = 2709, normalized size = 15.05 \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)^5/(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/384*(105*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)*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*(48*b^5*c^4 - 248*a*b^4*c^3*d +
526*a^2*b^3*c^2*d^2 - 605*a^3*b^2*c*d^3 + 279*a^4*b*d^4 - 105*(b^5*c*d^3 - a*b^4*d^4)*x^3 + 35*(2*b^5*c^2*d^2
- 13*a*b^4*c*d^3 + 11*a^2*b^3*d^4)*x^2 - 7*(8*b^5*c^3*d - 44*a*b^4*c^2*d^2 + 109*a^2*b^3*c*d^3 - 73*a^3*b^2*d
^4)*x)*sqrt(d*x + c))/(a^4*b^6*c^5 - 5*a^5*b^5*c^4*d + 10*a^6*b^4*c^3*d^2 - 10*a^7*b^3*c^2*d^3 + 5*a^8*b^2*c*d
^4 - a^9*b*d^5 + (b^10*c^5 - 5*a*b^9*c^4*d + 10*a^2*b^8*c^3*d^2 - 10*a^3*b^7*c^2*d^3 + 5*a^4*b^6*c*d^4 - a^5*b
^5*d^5)*x^4 + 4*(a*b^9*c^5 - 5*a^2*b^8*c^4*d + 10*a^3*b^7*c^3*d^2 - 10*a^4*b^6*c^2*d^3 + 5*a^5*b^5*c*d^4 - a^6
*b^4*d^5)*x^3 + 6*(a^2*b^8*c^5 - 5*a^3*b^7*c^4*d + 10*a^4*b^6*c^3*d^2 - 10*a^5*b^5*c^2*d^3 + 5*a^6*b^4*c*d^4 -
a^7*b^3*d^5)*x^2 + 4*(a^3*b^7*c^5 - 5*a^4*b^6*c^4*d + 10*a^5*b^5*c^3*d^2 - 10*a^6*b^4*c^2*d^3 + 5*a^7*b^3*c*d
^4 - a^8*b^2*d^5)*x), 1/192*(105*(b^4*d^4*x^4 + 4*a*b^3*d^4*x^3 + 6*a^2*b^2*d^4*x^2 + 4*a^3*b*d^4*x + a^4*d^4)
*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - (48*b^5*c^4 - 248*a*b^4*c^3*d
+ 526*a^2*b^3*c^2*d^2 - 605*a^3*b^2*c*d^3 + 279*a^4*b*d^4 - 105*(b^5*c*d^3 - a*b^4*d^4)*x^3 + 35*(2*b^5*c^2*d
^2 - 13*a*b^4*c*d^3 + 11*a^2*b^3*d^4)*x^2 - 7*(8*b^5*c^3*d - 44*a*b^4*c^2*d^2 + 109*a^2*b^3*c*d^3 - 73*a^3*b^2
*d^4)*x)*sqrt(d*x + c))/(a^4*b^6*c^5 - 5*a^5*b^5*c^4*d + 10*a^6*b^4*c^3*d^2 - 10*a^7*b^3*c^2*d^3 + 5*a^8*b^2*c
*d^4 - a^9*b*d^5 + (b^10*c^5 - 5*a*b^9*c^4*d + 10*a^2*b^8*c^3*d^2 - 10*a^3*b^7*c^2*d^3 + 5*a^4*b^6*c*d^4 - a^5
*b^5*d^5)*x^4 + 4*(a*b^9*c^5 - 5*a^2*b^8*c^4*d + 10*a^3*b^7*c^3*d^2 - 10*a^4*b^6*c^2*d^3 + 5*a^5*b^5*c*d^4 - a
^6*b^4*d^5)*x^3 + 6*(a^2*b^8*c^5 - 5*a^3*b^7*c^4*d + 10*a^4*b^6*c^3*d^2 - 10*a^5*b^5*c^2*d^3 + 5*a^6*b^4*c*d^4
- a^7*b^3*d^5)*x^2 + 4*(a^3*b^7*c^5 - 5*a^4*b^6*c^4*d + 10*a^5*b^5*c^3*d^2 - 10*a^6*b^4*c^2*d^3 + 5*a^7*b^3*c
*d^4 - a^8*b^2*d^5)*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)**5/(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [B]  time = 1.07056, size = 447, normalized size = 2.48 \begin{align*} \frac{35 \, d^{4} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{64 \,{\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{105 \,{\left (d x + c\right )}^{\frac{7}{2}} b^{3} d^{4} - 385 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{3} c d^{4} + 511 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{3} c^{2} d^{4} - 279 \, \sqrt{d x + c} b^{3} c^{3} d^{4} + 385 \,{\left (d x + c\right )}^{\frac{5}{2}} a b^{2} d^{5} - 1022 \,{\left (d x + c\right )}^{\frac{3}{2}} a b^{2} c d^{5} + 837 \, \sqrt{d x + c} a b^{2} c^{2} d^{5} + 511 \,{\left (d x + c\right )}^{\frac{3}{2}} a^{2} b d^{6} - 837 \, \sqrt{d x + c} a^{2} b c d^{6} + 279 \, \sqrt{d x + c} a^{3} d^{7}}{192 \,{\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 )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/64*d^4*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)) + 1/192*(105*(d*x + c)^(7/2)*b^3*d^4 - 385*(d*x + c)^(5/2)*b^3*c*d^4 +
511*(d*x + c)^(3/2)*b^3*c^2*d^4 - 279*sqrt(d*x + c)*b^3*c^3*d^4 + 385*(d*x + c)^(5/2)*a*b^2*d^5 - 1022*(d*x +
c)^(3/2)*a*b^2*c*d^5 + 837*sqrt(d*x + c)*a*b^2*c^2*d^5 + 511*(d*x + c)^(3/2)*a^2*b*d^6 - 837*sqrt(d*x + c)*a^
2*b*c*d^6 + 279*sqrt(d*x + c)*a^3*d^7)/((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)^4)