3.1385 \(\int \frac{\sqrt{c+d x}}{(a+b x)^5} \, dx\)

Optimal. Leaf size=182 \[ \frac{5 d^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{64 b^{3/2} (b c-a d)^{7/2}}-\frac{5 d^3 \sqrt{c+d x}}{64 b (a+b x) (b c-a d)^3}+\frac{5 d^2 \sqrt{c+d x}}{96 b (a+b x)^2 (b c-a d)^2}-\frac{d \sqrt{c+d x}}{24 b (a+b x)^3 (b c-a d)}-\frac{\sqrt{c+d x}}{4 b (a+b x)^4} \]

[Out]

-Sqrt[c + d*x]/(4*b*(a + b*x)^4) - (d*Sqrt[c + d*x])/(24*b*(b*c - a*d)*(a + b*x)^3) + (5*d^2*Sqrt[c + d*x])/(9
6*b*(b*c - a*d)^2*(a + b*x)^2) - (5*d^3*Sqrt[c + d*x])/(64*b*(b*c - a*d)^3*(a + b*x)) + (5*d^4*ArcTanh[(Sqrt[b
]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(64*b^(3/2)*(b*c - a*d)^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.122998, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {47, 51, 63, 208} \[ \frac{5 d^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{64 b^{3/2} (b c-a d)^{7/2}}-\frac{5 d^3 \sqrt{c+d x}}{64 b (a+b x) (b c-a d)^3}+\frac{5 d^2 \sqrt{c+d x}}{96 b (a+b x)^2 (b c-a d)^2}-\frac{d \sqrt{c+d x}}{24 b (a+b x)^3 (b c-a d)}-\frac{\sqrt{c+d x}}{4 b (a+b x)^4} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]/(a + b*x)^5,x]

[Out]

-Sqrt[c + d*x]/(4*b*(a + b*x)^4) - (d*Sqrt[c + d*x])/(24*b*(b*c - a*d)*(a + b*x)^3) + (5*d^2*Sqrt[c + d*x])/(9
6*b*(b*c - a*d)^2*(a + b*x)^2) - (5*d^3*Sqrt[c + d*x])/(64*b*(b*c - a*d)^3*(a + b*x)) + (5*d^4*ArcTanh[(Sqrt[b
]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(64*b^(3/2)*(b*c - a*d)^(7/2))

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [C]  time = 0.0142361, size = 52, normalized size = 0.29 \[ \frac{2 d^4 (c+d x)^{3/2} \, _2F_1\left (\frac{3}{2},5;\frac{5}{2};-\frac{b (c+d x)}{a d-b c}\right )}{3 (a d-b c)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 248, normalized size = 1.4 \begin{align*}{\frac{5\,{d}^{4}{b}^{2}}{64\, \left ( bdx+ad \right ) ^{4} \left ({a}^{3}{d}^{3}-3\,{a}^{2}bc{d}^{2}+3\,a{b}^{2}{c}^{2}d-{b}^{3}{c}^{3} \right ) } \left ( dx+c \right ) ^{{\frac{7}{2}}}}+{\frac{55\,{d}^{4}b}{192\, \left ( bdx+ad \right ) ^{4} \left ({a}^{2}{d}^{2}-2\,abcd+{b}^{2}{c}^{2} \right ) } \left ( dx+c \right ) ^{{\frac{5}{2}}}}+{\frac{73\,{d}^{4}}{192\, \left ( bdx+ad \right ) ^{4} \left ( ad-bc \right ) } \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{d}^{4}}{64\, \left ( bdx+ad \right ) ^{4}b}\sqrt{dx+c}}+{\frac{5\,{d}^{4}}{64\,b \left ({a}^{3}{d}^{3}-3\,{a}^{2}bc{d}^{2}+3\,a{b}^{2}{c}^{2}d-{b}^{3}{c}^{3} \right ) }\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((d*x+c)^(1/2)/(b*x+a)^5,x)

[Out]

5/64*d^4/(b*d*x+a*d)^4*b^2/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)*(d*x+c)^(7/2)+55/192*d^4/(b*d*x+a*d)^
4*b/(a^2*d^2-2*a*b*c*d+b^2*c^2)*(d*x+c)^(5/2)+73/192*d^4/(b*d*x+a*d)^4/(a*d-b*c)*(d*x+c)^(3/2)-5/64*d^4/(b*d*x
+a*d)^4/b*(d*x+c)^(1/2)+5/64*d^4/b/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/((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((d*x+c)^(1/2)/(b*x+a)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.32994, size = 2404, normalized size = 13.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(15*(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 - 184*a*b^4*c^3*d +
254*a^2*b^3*c^2*d^2 - 133*a^3*b^2*c*d^3 + 15*a^4*b*d^4 + 15*(b^5*c*d^3 - a*b^4*d^4)*x^3 - 5*(2*b^5*c^2*d^2 -
13*a*b^4*c*d^3 + 11*a^2*b^3*d^4)*x^2 + (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^4 - 4*a^5*b^5*c^3*d + 6*a^6*b^4*c^2*d^2 - 4*a^7*b^3*c*d^3 + a^8*b^2*d^4 + (b^10*c^
4 - 4*a*b^9*c^3*d + 6*a^2*b^8*c^2*d^2 - 4*a^3*b^7*c*d^3 + a^4*b^6*d^4)*x^4 + 4*(a*b^9*c^4 - 4*a^2*b^8*c^3*d +
6*a^3*b^7*c^2*d^2 - 4*a^4*b^6*c*d^3 + a^5*b^5*d^4)*x^3 + 6*(a^2*b^8*c^4 - 4*a^3*b^7*c^3*d + 6*a^4*b^6*c^2*d^2
- 4*a^5*b^5*c*d^3 + a^6*b^4*d^4)*x^2 + 4*(a^3*b^7*c^4 - 4*a^4*b^6*c^3*d + 6*a^5*b^5*c^2*d^2 - 4*a^6*b^4*c*d^3
+ a^7*b^3*d^4)*x), -1/192*(15*(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)*sq
rt(-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 - 184*a*b^4*c^3*d +
254*a^2*b^3*c^2*d^2 - 133*a^3*b^2*c*d^3 + 15*a^4*b*d^4 + 15*(b^5*c*d^3 - a*b^4*d^4)*x^3 - 5*(2*b^5*c^2*d^2 - 1
3*a*b^4*c*d^3 + 11*a^2*b^3*d^4)*x^2 + (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^4 - 4*a^5*b^5*c^3*d + 6*a^6*b^4*c^2*d^2 - 4*a^7*b^3*c*d^3 + a^8*b^2*d^4 + (b^10*c^4
- 4*a*b^9*c^3*d + 6*a^2*b^8*c^2*d^2 - 4*a^3*b^7*c*d^3 + a^4*b^6*d^4)*x^4 + 4*(a*b^9*c^4 - 4*a^2*b^8*c^3*d + 6
*a^3*b^7*c^2*d^2 - 4*a^4*b^6*c*d^3 + a^5*b^5*d^4)*x^3 + 6*(a^2*b^8*c^4 - 4*a^3*b^7*c^3*d + 6*a^4*b^6*c^2*d^2 -
4*a^5*b^5*c*d^3 + a^6*b^4*d^4)*x^2 + 4*(a^3*b^7*c^4 - 4*a^4*b^6*c^3*d + 6*a^5*b^5*c^2*d^2 - 4*a^6*b^4*c*d^3 +
a^7*b^3*d^4)*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((d*x+c)**(1/2)/(b*x+a)**5,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.10069, size = 420, normalized size = 2.31 \begin{align*} -\frac{5 \, d^{4} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{64 \,{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} \sqrt{-b^{2} c + a b d}} - \frac{15 \,{\left (d x + c\right )}^{\frac{7}{2}} b^{3} d^{4} - 55 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{3} c d^{4} + 73 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{3} c^{2} d^{4} + 15 \, \sqrt{d x + c} b^{3} c^{3} d^{4} + 55 \,{\left (d x + c\right )}^{\frac{5}{2}} a b^{2} d^{5} - 146 \,{\left (d x + c\right )}^{\frac{3}{2}} a b^{2} c d^{5} - 45 \, \sqrt{d x + c} a b^{2} c^{2} d^{5} + 73 \,{\left (d x + c\right )}^{\frac{3}{2}} a^{2} b d^{6} + 45 \, \sqrt{d x + c} a^{2} b c d^{6} - 15 \, \sqrt{d x + c} a^{3} d^{7}}{192 \,{\left (b^{4} c^{3} - 3 \, a b^{3} c^{2} d + 3 \, a^{2} b^{2} c d^{2} - a^{3} b d^{3}\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((d*x+c)^(1/2)/(b*x+a)^5,x, algorithm="giac")

[Out]

-5/64*d^4*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3
)*sqrt(-b^2*c + a*b*d)) - 1/192*(15*(d*x + c)^(7/2)*b^3*d^4 - 55*(d*x + c)^(5/2)*b^3*c*d^4 + 73*(d*x + c)^(3/2
)*b^3*c^2*d^4 + 15*sqrt(d*x + c)*b^3*c^3*d^4 + 55*(d*x + c)^(5/2)*a*b^2*d^5 - 146*(d*x + c)^(3/2)*a*b^2*c*d^5
- 45*sqrt(d*x + c)*a*b^2*c^2*d^5 + 73*(d*x + c)^(3/2)*a^2*b*d^6 + 45*sqrt(d*x + c)*a^2*b*c*d^6 - 15*sqrt(d*x +
c)*a^3*d^7)/((b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*((d*x + c)*b - b*c + a*d)^4)