3.1409 \(\int \frac{(c+d x)^{5/2}}{(a+b x)^5} \, dx\)

Optimal. Leaf size=162 \[ -\frac{5 d^3 \sqrt{c+d x}}{64 b^3 (a+b x) (b c-a d)}-\frac{5 d^2 \sqrt{c+d x}}{32 b^3 (a+b x)^2}+\frac{5 d^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{64 b^{7/2} (b c-a d)^{3/2}}-\frac{5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(c+d x)^{5/2}}{4 b (a+b x)^4} \]

[Out]

(-5*d^2*Sqrt[c + d*x])/(32*b^3*(a + b*x)^2) - (5*d^3*Sqrt[c + d*x])/(64*b^3*(b*c - a*d)*(a + b*x)) - (5*d*(c +
d*x)^(3/2))/(24*b^2*(a + b*x)^3) - (c + d*x)^(5/2)/(4*b*(a + b*x)^4) + (5*d^4*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])
/Sqrt[b*c - a*d]])/(64*b^(7/2)*(b*c - a*d)^(3/2))

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Rubi [A]  time = 0.0703069, antiderivative size = 162, 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} \[ -\frac{5 d^3 \sqrt{c+d x}}{64 b^3 (a+b x) (b c-a d)}-\frac{5 d^2 \sqrt{c+d x}}{32 b^3 (a+b x)^2}+\frac{5 d^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{64 b^{7/2} (b c-a d)^{3/2}}-\frac{5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(c+d x)^{5/2}}{4 b (a+b x)^4} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(a + b*x)^5,x]

[Out]

(-5*d^2*Sqrt[c + d*x])/(32*b^3*(a + b*x)^2) - (5*d^3*Sqrt[c + d*x])/(64*b^3*(b*c - a*d)*(a + b*x)) - (5*d*(c +
d*x)^(3/2))/(24*b^2*(a + b*x)^3) - (c + d*x)^(5/2)/(4*b*(a + b*x)^4) + (5*d^4*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])
/Sqrt[b*c - a*d]])/(64*b^(7/2)*(b*c - a*d)^(3/2))

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)^{5/2}}{(a+b x)^5} \, dx &=-\frac{(c+d x)^{5/2}}{4 b (a+b x)^4}+\frac{(5 d) \int \frac{(c+d x)^{3/2}}{(a+b x)^4} \, dx}{8 b}\\ &=-\frac{5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(c+d x)^{5/2}}{4 b (a+b x)^4}+\frac{\left (5 d^2\right ) \int \frac{\sqrt{c+d x}}{(a+b x)^3} \, dx}{16 b^2}\\ &=-\frac{5 d^2 \sqrt{c+d x}}{32 b^3 (a+b x)^2}-\frac{5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(c+d x)^{5/2}}{4 b (a+b x)^4}+\frac{\left (5 d^3\right ) \int \frac{1}{(a+b x)^2 \sqrt{c+d x}} \, dx}{64 b^3}\\ &=-\frac{5 d^2 \sqrt{c+d x}}{32 b^3 (a+b x)^2}-\frac{5 d^3 \sqrt{c+d x}}{64 b^3 (b c-a d) (a+b x)}-\frac{5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(c+d x)^{5/2}}{4 b (a+b x)^4}-\frac{\left (5 d^4\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{128 b^3 (b c-a d)}\\ &=-\frac{5 d^2 \sqrt{c+d x}}{32 b^3 (a+b x)^2}-\frac{5 d^3 \sqrt{c+d x}}{64 b^3 (b c-a d) (a+b x)}-\frac{5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(c+d x)^{5/2}}{4 b (a+b x)^4}-\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^3 (b c-a d)}\\ &=-\frac{5 d^2 \sqrt{c+d x}}{32 b^3 (a+b x)^2}-\frac{5 d^3 \sqrt{c+d x}}{64 b^3 (b c-a d) (a+b x)}-\frac{5 d (c+d x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(c+d x)^{5/2}}{4 b (a+b x)^4}+\frac{5 d^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{64 b^{7/2} (b c-a d)^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0177148, size = 52, normalized size = 0.32 \[ \frac{2 d^4 (c+d x)^{7/2} \, _2F_1\left (\frac{7}{2},5;\frac{9}{2};-\frac{b (c+d x)}{a d-b c}\right )}{7 (a d-b c)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/2)/(a + b*x)^5,x]

[Out]

(2*d^4*(c + d*x)^(7/2)*Hypergeometric2F1[7/2, 5, 9/2, -((b*(c + d*x))/(-(b*c) + a*d))])/(7*(-(b*c) + a*d)^5)

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Maple [A]  time = 0.014, size = 246, normalized size = 1.5 \begin{align*}{\frac{5\,{d}^{4}}{64\, \left ( bdx+ad \right ) ^{4} \left ( ad-bc \right ) } \left ( dx+c \right ) ^{{\frac{7}{2}}}}-{\frac{73\,{d}^{4}}{192\, \left ( bdx+ad \right ) ^{4}b} \left ( dx+c \right ) ^{{\frac{5}{2}}}}-{\frac{55\,{d}^{5}a}{192\, \left ( bdx+ad \right ) ^{4}{b}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{55\,{d}^{4}c}{192\, \left ( bdx+ad \right ) ^{4}b} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{d}^{6}{a}^{2}}{64\, \left ( bdx+ad \right ) ^{4}{b}^{3}}\sqrt{dx+c}}+{\frac{5\,{d}^{5}ac}{32\, \left ( bdx+ad \right ) ^{4}{b}^{2}}\sqrt{dx+c}}-{\frac{5\,{d}^{4}{c}^{2}}{64\, \left ( bdx+ad \right ) ^{4}b}\sqrt{dx+c}}+{\frac{5\,{d}^{4}}{ \left ( 64\,ad-64\,bc \right ){b}^{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((d*x+c)^(5/2)/(b*x+a)^5,x)

[Out]

5/64*d^4/(b*d*x+a*d)^4/(a*d-b*c)*(d*x+c)^(7/2)-73/192*d^4/(b*d*x+a*d)^4/b*(d*x+c)^(5/2)-55/192*d^5/(b*d*x+a*d)
^4/b^2*(d*x+c)^(3/2)*a+55/192*d^4/(b*d*x+a*d)^4/b*(d*x+c)^(3/2)*c-5/64*d^6/(b*d*x+a*d)^4/b^3*(d*x+c)^(1/2)*a^2
+5/32*d^5/(b*d*x+a*d)^4/b^2*(d*x+c)^(1/2)*a*c-5/64*d^4/(b*d*x+a*d)^4/b*(d*x+c)^(1/2)*c^2+5/64*d^4/(a*d-b*c)/b^
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((d*x+c)^(5/2)/(b*x+a)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.06212, size = 1859, normalized size = 11.48 \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)^(5/2)/(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 - 56*a*b^4*c^3*d -
2*a^2*b^3*c^2*d^2 - 5*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 + (118*b^5*c^2*d^2 - 191*a
*b^4*c*d^3 + 73*a^2*b^3*d^4)*x^2 + (136*b^5*c^3*d - 172*a*b^4*c^2*d^2 - 19*a^2*b^3*c*d^3 + 55*a^3*b^2*d^4)*x)*
sqrt(d*x + c))/(a^4*b^6*c^2 - 2*a^5*b^5*c*d + a^6*b^4*d^2 + (b^10*c^2 - 2*a*b^9*c*d + a^2*b^8*d^2)*x^4 + 4*(a*
b^9*c^2 - 2*a^2*b^8*c*d + a^3*b^7*d^2)*x^3 + 6*(a^2*b^8*c^2 - 2*a^3*b^7*c*d + a^4*b^6*d^2)*x^2 + 4*(a^3*b^7*c^
2 - 2*a^4*b^6*c*d + a^5*b^5*d^2)*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)*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 -
56*a*b^4*c^3*d - 2*a^2*b^3*c^2*d^2 - 5*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 + (118*b^
5*c^2*d^2 - 191*a*b^4*c*d^3 + 73*a^2*b^3*d^4)*x^2 + (136*b^5*c^3*d - 172*a*b^4*c^2*d^2 - 19*a^2*b^3*c*d^3 + 55
*a^3*b^2*d^4)*x)*sqrt(d*x + c))/(a^4*b^6*c^2 - 2*a^5*b^5*c*d + a^6*b^4*d^2 + (b^10*c^2 - 2*a*b^9*c*d + a^2*b^8
*d^2)*x^4 + 4*(a*b^9*c^2 - 2*a^2*b^8*c*d + a^3*b^7*d^2)*x^3 + 6*(a^2*b^8*c^2 - 2*a^3*b^7*c*d + a^4*b^6*d^2)*x^
2 + 4*(a^3*b^7*c^2 - 2*a^4*b^6*c*d + a^5*b^5*d^2)*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)**(5/2)/(b*x+a)**5,x)

[Out]

Timed out

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Giac [A]  time = 1.11564, size = 350, normalized size = 2.16 \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 - a b^{3} d\right )} \sqrt{-b^{2} c + a b d}} - \frac{15 \,{\left (d x + c\right )}^{\frac{7}{2}} b^{3} d^{4} + 73 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{3} c d^{4} - 55 \,{\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} - 73 \,{\left (d x + c\right )}^{\frac{5}{2}} a b^{2} d^{5} + 110 \,{\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} - 55 \,{\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 - a b^{3} d\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)^(5/2)/(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 - a*b^3*d)*sqrt(-b^2*c + a*b*d)) - 1/192*(15*(d
*x + c)^(7/2)*b^3*d^4 + 73*(d*x + c)^(5/2)*b^3*c*d^4 - 55*(d*x + c)^(3/2)*b^3*c^2*d^4 + 15*sqrt(d*x + c)*b^3*c
^3*d^4 - 73*(d*x + c)^(5/2)*a*b^2*d^5 + 110*(d*x + c)^(3/2)*a*b^2*c*d^5 - 45*sqrt(d*x + c)*a*b^2*c^2*d^5 - 55*
(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 - a*b^3*d)*((d*x
+ c)*b - b*c + a*d)^4)