3.1398 \(\int \frac{(c+d x)^{3/2}}{(a+b x)^6} \, dx\)

Optimal. Leaf size=208 \[ -\frac{3 d^4 \sqrt{c+d x}}{128 b^2 (a+b x) (b c-a d)^3}+\frac{d^3 \sqrt{c+d x}}{64 b^2 (a+b x)^2 (b c-a d)^2}-\frac{d^2 \sqrt{c+d x}}{80 b^2 (a+b x)^3 (b c-a d)}+\frac{3 d^5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{128 b^{5/2} (b c-a d)^{7/2}}-\frac{3 d \sqrt{c+d x}}{40 b^2 (a+b x)^4}-\frac{(c+d x)^{3/2}}{5 b (a+b x)^5} \]

[Out]

(-3*d*Sqrt[c + d*x])/(40*b^2*(a + b*x)^4) - (d^2*Sqrt[c + d*x])/(80*b^2*(b*c - a*d)*(a + b*x)^3) + (d^3*Sqrt[c
+ d*x])/(64*b^2*(b*c - a*d)^2*(a + b*x)^2) - (3*d^4*Sqrt[c + d*x])/(128*b^2*(b*c - a*d)^3*(a + b*x)) - (c + d
*x)^(3/2)/(5*b*(a + b*x)^5) + (3*d^5*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(128*b^(5/2)*(b*c - a*d
)^(7/2))

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Rubi [A]  time = 0.093295, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {47, 51, 63, 208} \[ -\frac{3 d^4 \sqrt{c+d x}}{128 b^2 (a+b x) (b c-a d)^3}+\frac{d^3 \sqrt{c+d x}}{64 b^2 (a+b x)^2 (b c-a d)^2}-\frac{d^2 \sqrt{c+d x}}{80 b^2 (a+b x)^3 (b c-a d)}+\frac{3 d^5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{128 b^{5/2} (b c-a d)^{7/2}}-\frac{3 d \sqrt{c+d x}}{40 b^2 (a+b x)^4}-\frac{(c+d x)^{3/2}}{5 b (a+b x)^5} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)/(a + b*x)^6,x]

[Out]

(-3*d*Sqrt[c + d*x])/(40*b^2*(a + b*x)^4) - (d^2*Sqrt[c + d*x])/(80*b^2*(b*c - a*d)*(a + b*x)^3) + (d^3*Sqrt[c
+ d*x])/(64*b^2*(b*c - a*d)^2*(a + b*x)^2) - (3*d^4*Sqrt[c + d*x])/(128*b^2*(b*c - a*d)^3*(a + b*x)) - (c + d
*x)^(3/2)/(5*b*(a + b*x)^5) + (3*d^5*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(128*b^(5/2)*(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{(c+d x)^{3/2}}{(a+b x)^6} \, dx &=-\frac{(c+d x)^{3/2}}{5 b (a+b x)^5}+\frac{(3 d) \int \frac{\sqrt{c+d x}}{(a+b x)^5} \, dx}{10 b}\\ &=-\frac{3 d \sqrt{c+d x}}{40 b^2 (a+b x)^4}-\frac{(c+d x)^{3/2}}{5 b (a+b x)^5}+\frac{\left (3 d^2\right ) \int \frac{1}{(a+b x)^4 \sqrt{c+d x}} \, dx}{80 b^2}\\ &=-\frac{3 d \sqrt{c+d x}}{40 b^2 (a+b x)^4}-\frac{d^2 \sqrt{c+d x}}{80 b^2 (b c-a d) (a+b x)^3}-\frac{(c+d x)^{3/2}}{5 b (a+b x)^5}-\frac{d^3 \int \frac{1}{(a+b x)^3 \sqrt{c+d x}} \, dx}{32 b^2 (b c-a d)}\\ &=-\frac{3 d \sqrt{c+d x}}{40 b^2 (a+b x)^4}-\frac{d^2 \sqrt{c+d x}}{80 b^2 (b c-a d) (a+b x)^3}+\frac{d^3 \sqrt{c+d x}}{64 b^2 (b c-a d)^2 (a+b x)^2}-\frac{(c+d x)^{3/2}}{5 b (a+b x)^5}+\frac{\left (3 d^4\right ) \int \frac{1}{(a+b x)^2 \sqrt{c+d x}} \, dx}{128 b^2 (b c-a d)^2}\\ &=-\frac{3 d \sqrt{c+d x}}{40 b^2 (a+b x)^4}-\frac{d^2 \sqrt{c+d x}}{80 b^2 (b c-a d) (a+b x)^3}+\frac{d^3 \sqrt{c+d x}}{64 b^2 (b c-a d)^2 (a+b x)^2}-\frac{3 d^4 \sqrt{c+d x}}{128 b^2 (b c-a d)^3 (a+b x)}-\frac{(c+d x)^{3/2}}{5 b (a+b x)^5}-\frac{\left (3 d^5\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{256 b^2 (b c-a d)^3}\\ &=-\frac{3 d \sqrt{c+d x}}{40 b^2 (a+b x)^4}-\frac{d^2 \sqrt{c+d x}}{80 b^2 (b c-a d) (a+b x)^3}+\frac{d^3 \sqrt{c+d x}}{64 b^2 (b c-a d)^2 (a+b x)^2}-\frac{3 d^4 \sqrt{c+d x}}{128 b^2 (b c-a d)^3 (a+b x)}-\frac{(c+d x)^{3/2}}{5 b (a+b x)^5}-\frac{\left (3 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{128 b^2 (b c-a d)^3}\\ &=-\frac{3 d \sqrt{c+d x}}{40 b^2 (a+b x)^4}-\frac{d^2 \sqrt{c+d x}}{80 b^2 (b c-a d) (a+b x)^3}+\frac{d^3 \sqrt{c+d x}}{64 b^2 (b c-a d)^2 (a+b x)^2}-\frac{3 d^4 \sqrt{c+d x}}{128 b^2 (b c-a d)^3 (a+b x)}-\frac{(c+d x)^{3/2}}{5 b (a+b x)^5}+\frac{3 d^5 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{128 b^{5/2} (b c-a d)^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0178282, size = 52, normalized size = 0.25 \[ \frac{2 d^5 (c+d x)^{5/2} \, _2F_1\left (\frac{5}{2},6;\frac{7}{2};-\frac{b (c+d x)}{a d-b c}\right )}{5 (a d-b c)^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(3/2)/(a + b*x)^6,x]

[Out]

(2*d^5*(c + d*x)^(5/2)*Hypergeometric2F1[5/2, 6, 7/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(5*(-(b*c) + a*d)^6)

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Maple [A]  time = 0.015, size = 300, normalized size = 1.4 \begin{align*}{\frac{3\,{d}^{5}{b}^{2}}{128\, \left ( bdx+ad \right ) ^{5} \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{9}{2}}}}+{\frac{7\,{d}^{5}b}{64\, \left ( bdx+ad \right ) ^{5} \left ({a}^{2}{d}^{2}-2\,abcd+{b}^{2}{c}^{2} \right ) } \left ( dx+c \right ) ^{{\frac{7}{2}}}}+{\frac{{d}^{5}}{5\, \left ( bdx+ad \right ) ^{5} \left ( ad-bc \right ) } \left ( dx+c \right ) ^{{\frac{5}{2}}}}-{\frac{7\,{d}^{5}}{64\, \left ( bdx+ad \right ) ^{5}b} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{d}^{6}a}{128\, \left ( bdx+ad \right ) ^{5}{b}^{2}}\sqrt{dx+c}}+{\frac{3\,{d}^{5}c}{128\, \left ( bdx+ad \right ) ^{5}b}\sqrt{dx+c}}+{\frac{3\,{d}^{5}}{ \left ( 128\,{a}^{3}{d}^{3}-384\,{a}^{2}bc{d}^{2}+384\,a{b}^{2}{c}^{2}d-128\,{b}^{3}{c}^{3} \right ){b}^{2}}\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)^(3/2)/(b*x+a)^6,x)

[Out]

3/128*d^5/(b*d*x+a*d)^5*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)^(9/2)+7/64*d^5/(b*d*x+a*d)^5
*b/(a^2*d^2-2*a*b*c*d+b^2*c^2)*(d*x+c)^(7/2)+1/5*d^5/(b*d*x+a*d)^5/(a*d-b*c)*(d*x+c)^(5/2)-7/64*d^5/(b*d*x+a*d
)^5/b*(d*x+c)^(3/2)-3/128*d^6/(b*d*x+a*d)^5/b^2*(d*x+c)^(1/2)*a+3/128*d^5/(b*d*x+a*d)^5/b*(d*x+c)^(1/2)*c+3/12
8*d^5/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/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((d*x+c)^(3/2)/(b*x+a)^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.06007, size = 3065, normalized size = 14.74 \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)^(3/2)/(b*x+a)^6,x, algorithm="fricas")

[Out]

[-1/1280*(15*(b^5*d^5*x^5 + 5*a*b^4*d^5*x^4 + 10*a^2*b^3*d^5*x^3 + 10*a^3*b^2*d^5*x^2 + 5*a^4*b*d^5*x + a^5*d^
5)*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*(128*b^6
*c^5 - 464*a*b^5*c^4*d + 584*a^2*b^4*c^3*d^2 - 258*a^3*b^3*c^2*d^3 - 5*a^4*b^2*c*d^4 + 15*a^5*b*d^5 + 15*(b^6*
c*d^4 - a*b^5*d^5)*x^4 - 10*(b^6*c^2*d^3 - 8*a*b^5*c*d^4 + 7*a^2*b^4*d^5)*x^3 + 2*(4*b^6*c^3*d^2 - 27*a*b^5*c^
2*d^3 + 87*a^2*b^4*c*d^4 - 64*a^3*b^3*d^5)*x^2 + 2*(88*b^6*c^4*d - 344*a*b^5*c^3*d^2 + 489*a^2*b^4*c^2*d^3 - 2
68*a^3*b^3*c*d^4 + 35*a^4*b^2*d^5)*x)*sqrt(d*x + c))/(a^5*b^7*c^4 - 4*a^6*b^6*c^3*d + 6*a^7*b^5*c^2*d^2 - 4*a^
8*b^4*c*d^3 + a^9*b^3*d^4 + (b^12*c^4 - 4*a*b^11*c^3*d + 6*a^2*b^10*c^2*d^2 - 4*a^3*b^9*c*d^3 + a^4*b^8*d^4)*x
^5 + 5*(a*b^11*c^4 - 4*a^2*b^10*c^3*d + 6*a^3*b^9*c^2*d^2 - 4*a^4*b^8*c*d^3 + a^5*b^7*d^4)*x^4 + 10*(a^2*b^10*
c^4 - 4*a^3*b^9*c^3*d + 6*a^4*b^8*c^2*d^2 - 4*a^5*b^7*c*d^3 + a^6*b^6*d^4)*x^3 + 10*(a^3*b^9*c^4 - 4*a^4*b^8*c
^3*d + 6*a^5*b^7*c^2*d^2 - 4*a^6*b^6*c*d^3 + a^7*b^5*d^4)*x^2 + 5*(a^4*b^8*c^4 - 4*a^5*b^7*c^3*d + 6*a^6*b^6*c
^2*d^2 - 4*a^7*b^5*c*d^3 + a^8*b^4*d^4)*x), -1/640*(15*(b^5*d^5*x^5 + 5*a*b^4*d^5*x^4 + 10*a^2*b^3*d^5*x^3 + 1
0*a^3*b^2*d^5*x^2 + 5*a^4*b*d^5*x + a^5*d^5)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b
*d*x + b*c)) + (128*b^6*c^5 - 464*a*b^5*c^4*d + 584*a^2*b^4*c^3*d^2 - 258*a^3*b^3*c^2*d^3 - 5*a^4*b^2*c*d^4 +
15*a^5*b*d^5 + 15*(b^6*c*d^4 - a*b^5*d^5)*x^4 - 10*(b^6*c^2*d^3 - 8*a*b^5*c*d^4 + 7*a^2*b^4*d^5)*x^3 + 2*(4*b^
6*c^3*d^2 - 27*a*b^5*c^2*d^3 + 87*a^2*b^4*c*d^4 - 64*a^3*b^3*d^5)*x^2 + 2*(88*b^6*c^4*d - 344*a*b^5*c^3*d^2 +
489*a^2*b^4*c^2*d^3 - 268*a^3*b^3*c*d^4 + 35*a^4*b^2*d^5)*x)*sqrt(d*x + c))/(a^5*b^7*c^4 - 4*a^6*b^6*c^3*d + 6
*a^7*b^5*c^2*d^2 - 4*a^8*b^4*c*d^3 + a^9*b^3*d^4 + (b^12*c^4 - 4*a*b^11*c^3*d + 6*a^2*b^10*c^2*d^2 - 4*a^3*b^9
*c*d^3 + a^4*b^8*d^4)*x^5 + 5*(a*b^11*c^4 - 4*a^2*b^10*c^3*d + 6*a^3*b^9*c^2*d^2 - 4*a^4*b^8*c*d^3 + a^5*b^7*d
^4)*x^4 + 10*(a^2*b^10*c^4 - 4*a^3*b^9*c^3*d + 6*a^4*b^8*c^2*d^2 - 4*a^5*b^7*c*d^3 + a^6*b^6*d^4)*x^3 + 10*(a^
3*b^9*c^4 - 4*a^4*b^8*c^3*d + 6*a^5*b^7*c^2*d^2 - 4*a^6*b^6*c*d^3 + a^7*b^5*d^4)*x^2 + 5*(a^4*b^8*c^4 - 4*a^5*
b^7*c^3*d + 6*a^6*b^6*c^2*d^2 - 4*a^7*b^5*c*d^3 + a^8*b^4*d^4)*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((d*x+c)**(3/2)/(b*x+a)**6,x)

[Out]

Timed out

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Giac [B]  time = 1.12158, size = 554, normalized size = 2.66 \begin{align*} -\frac{3 \, d^{5} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{128 \,{\left (b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} \sqrt{-b^{2} c + a b d}} - \frac{15 \,{\left (d x + c\right )}^{\frac{9}{2}} b^{4} d^{5} - 70 \,{\left (d x + c\right )}^{\frac{7}{2}} b^{4} c d^{5} + 128 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{4} c^{2} d^{5} + 70 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{4} c^{3} d^{5} - 15 \, \sqrt{d x + c} b^{4} c^{4} d^{5} + 70 \,{\left (d x + c\right )}^{\frac{7}{2}} a b^{3} d^{6} - 256 \,{\left (d x + c\right )}^{\frac{5}{2}} a b^{3} c d^{6} - 210 \,{\left (d x + c\right )}^{\frac{3}{2}} a b^{3} c^{2} d^{6} + 60 \, \sqrt{d x + c} a b^{3} c^{3} d^{6} + 128 \,{\left (d x + c\right )}^{\frac{5}{2}} a^{2} b^{2} d^{7} + 210 \,{\left (d x + c\right )}^{\frac{3}{2}} a^{2} b^{2} c d^{7} - 90 \, \sqrt{d x + c} a^{2} b^{2} c^{2} d^{7} - 70 \,{\left (d x + c\right )}^{\frac{3}{2}} a^{3} b d^{8} + 60 \, \sqrt{d x + c} a^{3} b c d^{8} - 15 \, \sqrt{d x + c} a^{4} d^{9}}{640 \,{\left (b^{5} c^{3} - 3 \, a b^{4} c^{2} d + 3 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )}{\left ({\left (d x + c\right )} b - b c + a d\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(3/2)/(b*x+a)^6,x, algorithm="giac")

[Out]

-3/128*d^5*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^5*c^3 - 3*a*b^4*c^2*d + 3*a^2*b^3*c*d^2 - a^3*b^2*
d^3)*sqrt(-b^2*c + a*b*d)) - 1/640*(15*(d*x + c)^(9/2)*b^4*d^5 - 70*(d*x + c)^(7/2)*b^4*c*d^5 + 128*(d*x + c)^
(5/2)*b^4*c^2*d^5 + 70*(d*x + c)^(3/2)*b^4*c^3*d^5 - 15*sqrt(d*x + c)*b^4*c^4*d^5 + 70*(d*x + c)^(7/2)*a*b^3*d
^6 - 256*(d*x + c)^(5/2)*a*b^3*c*d^6 - 210*(d*x + c)^(3/2)*a*b^3*c^2*d^6 + 60*sqrt(d*x + c)*a*b^3*c^3*d^6 + 12
8*(d*x + c)^(5/2)*a^2*b^2*d^7 + 210*(d*x + c)^(3/2)*a^2*b^2*c*d^7 - 90*sqrt(d*x + c)*a^2*b^2*c^2*d^7 - 70*(d*x
+ c)^(3/2)*a^3*b*d^8 + 60*sqrt(d*x + c)*a^3*b*c*d^8 - 15*sqrt(d*x + c)*a^4*d^9)/((b^5*c^3 - 3*a*b^4*c^2*d + 3
*a^2*b^3*c*d^2 - a^3*b^2*d^3)*((d*x + c)*b - b*c + a*d)^5)