3.1487 \(\int \frac{(c+d x)^{5/2}}{(a+b x)^{7/2}} \, dx\)

Optimal. Leaf size=120 \[ -\frac{2 d^2 \sqrt{c+d x}}{b^3 \sqrt{a+b x}}+\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{7/2}}-\frac{2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac{2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0519115, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {47, 63, 217, 206} \[ -\frac{2 d^2 \sqrt{c+d x}}{b^3 \sqrt{a+b x}}+\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{7/2}}-\frac{2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac{2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^2*Sqrt[c + d*x])/(b^3*Sqrt[a + b*x]) - (2*d*(c + d*x)^(3/2))/(3*b^2*(a + b*x)^(3/2)) - (2*(c + d*x)^(5/2
))/(5*b*(a + b*x)^(5/2)) + (2*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/b^(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 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(c+d x)^{5/2}}{(a+b x)^{7/2}} \, dx &=-\frac{2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac{d \int \frac{(c+d x)^{3/2}}{(a+b x)^{5/2}} \, dx}{b}\\ &=-\frac{2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac{2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac{d^2 \int \frac{\sqrt{c+d x}}{(a+b x)^{3/2}} \, dx}{b^2}\\ &=-\frac{2 d^2 \sqrt{c+d x}}{b^3 \sqrt{a+b x}}-\frac{2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac{2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac{d^3 \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{b^3}\\ &=-\frac{2 d^2 \sqrt{c+d x}}{b^3 \sqrt{a+b x}}-\frac{2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac{2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac{\left (2 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c-\frac{a d}{b}+\frac{d x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{b^4}\\ &=-\frac{2 d^2 \sqrt{c+d x}}{b^3 \sqrt{a+b x}}-\frac{2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac{2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac{\left (2 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{d x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{b^4}\\ &=-\frac{2 d^2 \sqrt{c+d x}}{b^3 \sqrt{a+b x}}-\frac{2 d (c+d x)^{3/2}}{3 b^2 (a+b x)^{3/2}}-\frac{2 (c+d x)^{5/2}}{5 b (a+b x)^{5/2}}+\frac{2 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0686998, size = 73, normalized size = 0.61 \[ -\frac{2 (c+d x)^{5/2} \, _2F_1\left (-\frac{5}{2},-\frac{5}{2};-\frac{3}{2};\frac{d (a+b x)}{a d-b c}\right )}{5 b (a+b x)^{5/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.035, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dx+c \right ) ^{{\frac{5}{2}}} \left ( bx+a \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)/(b*x+a)^(7/2),x)

[Out]

int((d*x+c)^(5/2)/(b*x+a)^(7/2),x)

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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)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 7.97312, size = 1015, normalized size = 8.46 \begin{align*} \left [\frac{15 \,{\left (b^{3} d^{2} x^{3} + 3 \, a b^{2} d^{2} x^{2} + 3 \, a^{2} b d^{2} x + a^{3} d^{2}\right )} \sqrt{\frac{d}{b}} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \,{\left (2 \, b^{2} d x + b^{2} c + a b d\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{\frac{d}{b}} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \,{\left (23 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 5 \, a b c d + 15 \, a^{2} d^{2} +{\left (11 \, b^{2} c d + 35 \, a b d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{30 \,{\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}}, -\frac{15 \,{\left (b^{3} d^{2} x^{3} + 3 \, a b^{2} d^{2} x^{2} + 3 \, a^{2} b d^{2} x + a^{3} d^{2}\right )} \sqrt{-\frac{d}{b}} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{-\frac{d}{b}}}{2 \,{\left (b d^{2} x^{2} + a c d +{\left (b c d + a d^{2}\right )} x\right )}}\right ) + 2 \,{\left (23 \, b^{2} d^{2} x^{2} + 3 \, b^{2} c^{2} + 5 \, a b c d + 15 \, a^{2} d^{2} +{\left (11 \, b^{2} c d + 35 \, a b d^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{15 \,{\left (b^{6} x^{3} + 3 \, a b^{5} x^{2} + 3 \, a^{2} b^{4} x + a^{3} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/(b*x+a)^(7/2),x, algorithm="fricas")

[Out]

[1/30*(15*(b^3*d^2*x^3 + 3*a*b^2*d^2*x^2 + 3*a^2*b*d^2*x + a^3*d^2)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*
a*b*c*d + a^2*d^2 + 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2
)*x) - 4*(23*b^2*d^2*x^2 + 3*b^2*c^2 + 5*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d + 35*a*b*d^2)*x)*sqrt(b*x + a)*sqr
t(d*x + c))/(b^6*x^3 + 3*a*b^5*x^2 + 3*a^2*b^4*x + a^3*b^3), -1/15*(15*(b^3*d^2*x^3 + 3*a*b^2*d^2*x^2 + 3*a^2*
b*d^2*x + a^3*d^2)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-d/b)/(b*d^2*x
^2 + a*c*d + (b*c*d + a*d^2)*x)) + 2*(23*b^2*d^2*x^2 + 3*b^2*c^2 + 5*a*b*c*d + 15*a^2*d^2 + (11*b^2*c*d + 35*a
*b*d^2)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^6*x^3 + 3*a*b^5*x^2 + 3*a^2*b^4*x + a^3*b^3)]

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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)**(7/2),x)

[Out]

Timed out

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Giac [B]  time = 1.71627, size = 1384, normalized size = 11.53 \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)^(7/2),x, algorithm="giac")

[Out]

-sqrt(b*d)*d^2*abs(b)*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/b^5 - 4/15*(23*sq
rt(b*d)*b^9*c^5*d^2*abs(b) - 115*sqrt(b*d)*a*b^8*c^4*d^3*abs(b) + 230*sqrt(b*d)*a^2*b^7*c^3*d^4*abs(b) - 230*s
qrt(b*d)*a^3*b^6*c^2*d^5*abs(b) + 115*sqrt(b*d)*a^4*b^5*c*d^6*abs(b) - 23*sqrt(b*d)*a^5*b^4*d^7*abs(b) - 70*sq
rt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^7*c^4*d^2*abs(b) + 280*sqrt(b*d)*(
sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^6*c^3*d^3*abs(b) - 420*sqrt(b*d)*(sqrt(b*
d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^5*c^2*d^4*abs(b) + 280*sqrt(b*d)*(sqrt(b*d)*sq
rt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^4*c*d^5*abs(b) - 70*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x +
a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^3*d^6*abs(b) + 140*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqr
t(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^5*c^3*d^2*abs(b) - 420*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c +
(b*x + a)*b*d - a*b*d))^4*a*b^4*c^2*d^3*abs(b) + 420*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x +
a)*b*d - a*b*d))^4*a^2*b^3*c*d^4*abs(b) - 140*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d
- a*b*d))^4*a^3*b^2*d^5*abs(b) - 90*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))
^6*b^3*c^2*d^2*abs(b) + 180*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^2*
c*d^3*abs(b) - 90*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b*d^4*abs(b)
+ 45*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*b*c*d^2*abs(b) - 45*sqrt(b*d
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^8*a*d^3*abs(b))/((b^2*c - a*b*d - (sqrt(b*d)
*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)^5*b^4)