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

Optimal. Leaf size=148 \[ \frac{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^2}{8 b^3}+\frac{5 \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}{12 b^2}+\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{7/2} \sqrt{d}}+\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 b} \]

[Out]

(5*(b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^3) + (5*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(12*b^2)
+ (Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*b) + (5*(b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c +
d*x])])/(8*b^(7/2)*Sqrt[d])

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Rubi [A]  time = 0.069809, antiderivative size = 148, 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 = {50, 63, 217, 206} \[ \frac{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^2}{8 b^3}+\frac{5 \sqrt{a+b x} (c+d x)^{3/2} (b c-a d)}{12 b^2}+\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{7/2} \sqrt{d}}+\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/Sqrt[a + b*x],x]

[Out]

(5*(b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b^3) + (5*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2))/(12*b^2)
+ (Sqrt[a + b*x]*(c + d*x)^(5/2))/(3*b) + (5*(b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c +
d*x])])/(8*b^(7/2)*Sqrt[d])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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}}{\sqrt{a+b x}} \, dx &=\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 b}+\frac{(5 (b c-a d)) \int \frac{(c+d x)^{3/2}}{\sqrt{a+b x}} \, dx}{6 b}\\ &=\frac{5 (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 b^2}+\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 b}+\frac{\left (5 (b c-a d)^2\right ) \int \frac{\sqrt{c+d x}}{\sqrt{a+b x}} \, dx}{8 b^2}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 b^3}+\frac{5 (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 b^2}+\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 b}+\frac{\left (5 (b c-a d)^3\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{16 b^3}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 b^3}+\frac{5 (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 b^2}+\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 b}+\frac{\left (5 (b c-a 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 )}{8 b^4}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 b^3}+\frac{5 (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 b^2}+\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 b}+\frac{\left (5 (b c-a 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 )}{8 b^4}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 b^3}+\frac{5 (b c-a d) \sqrt{a+b x} (c+d x)^{3/2}}{12 b^2}+\frac{\sqrt{a+b x} (c+d x)^{5/2}}{3 b}+\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 b^{7/2} \sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 0.236425, size = 139, normalized size = 0.94 \[ \frac{\sqrt{c+d x} \left (\sqrt{a+b x} \left (15 a^2 d^2-10 a b d (4 c+d x)+b^2 \left (33 c^2+26 c d x+8 d^2 x^2\right )\right )+\frac{15 (b c-a d)^{5/2} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{\sqrt{d} \sqrt{\frac{b (c+d x)}{b c-a d}}}\right )}{24 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(Sqrt[a + b*x]*(15*a^2*d^2 - 10*a*b*d*(4*c + d*x) + b^2*(33*c^2 + 26*c*d*x + 8*d^2*x^2)) + (15*
(b*c - a*d)^(5/2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(Sqrt[d]*Sqrt[(b*(c + d*x))/(b*c - a*d)]))
)/(24*b^3)

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Maple [B]  time = 0.006, size = 465, normalized size = 3.1 \begin{align*}{\frac{1}{3\,b}\sqrt{bx+a} \left ( dx+c \right ) ^{{\frac{5}{2}}}}-{\frac{5\,ad}{12\,{b}^{2}}\sqrt{bx+a} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{5\,c}{12\,b}\sqrt{bx+a} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{a}^{2}{d}^{2}}{8\,{b}^{3}}\sqrt{bx+a}\sqrt{dx+c}}-{\frac{5\,adc}{4\,{b}^{2}}\sqrt{bx+a}\sqrt{dx+c}}+{\frac{5\,{c}^{2}}{8\,b}\sqrt{bx+a}\sqrt{dx+c}}-{\frac{5\,{a}^{3}{d}^{3}}{16\,{b}^{3}}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({ \left ({\frac{ad}{2}}+{\frac{bc}{2}}+bdx \right ){\frac{1}{\sqrt{bd}}}}+\sqrt{d{x}^{2}b+ \left ( ad+bc \right ) x+ac} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{bd}}}}+{\frac{15\,c{a}^{2}{d}^{2}}{16\,{b}^{2}}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({ \left ({\frac{ad}{2}}+{\frac{bc}{2}}+bdx \right ){\frac{1}{\sqrt{bd}}}}+\sqrt{d{x}^{2}b+ \left ( ad+bc \right ) x+ac} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{bd}}}}-{\frac{15\,ad{c}^{2}}{16\,b}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({ \left ({\frac{ad}{2}}+{\frac{bc}{2}}+bdx \right ){\frac{1}{\sqrt{bd}}}}+\sqrt{d{x}^{2}b+ \left ( ad+bc \right ) x+ac} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{bd}}}}+{\frac{5\,{c}^{3}}{16}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\ln \left ({ \left ({\frac{ad}{2}}+{\frac{bc}{2}}+bdx \right ){\frac{1}{\sqrt{bd}}}}+\sqrt{d{x}^{2}b+ \left ( ad+bc \right ) x+ac} \right ){\frac{1}{\sqrt{bx+a}}}{\frac{1}{\sqrt{dx+c}}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)/(b*x+a)^(1/2),x)

[Out]

1/3*(d*x+c)^(5/2)*(b*x+a)^(1/2)/b-5/12/b^2*(d*x+c)^(3/2)*(b*x+a)^(1/2)*a*d+5/12/b*(d*x+c)^(3/2)*(b*x+a)^(1/2)*
c+5/8/b^3*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^2*d^2-5/4/b^2*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a*d*c+5/8/b*(d*x+c)^(1/2)*(b
*x+a)^(1/2)*c^2-5/16/b^3*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(b*d)^
(1/2)+(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*a^3*d^3+15/16/b^2*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*
x+a)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(b*d)^(1/2)+(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*a^2*d^2*c-15/16
/b*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(b*d)^(1/2)+(d*x^2*b+(a*d+b*
c)*x+a*c)^(1/2))/(b*d)^(1/2)*a*d*c^2+5/16*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*ln((1/2*a*d+1/2*
b*c+b*d*x)/(b*d)^(1/2)+(d*x^2*b+(a*d+b*c)*x+a*c)^(1/2))/(b*d)^(1/2)*c^3

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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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.38559, size = 933, normalized size = 6.3 \begin{align*} \left [-\frac{15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{b d} \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 d x + b c + a d\right )} \sqrt{b d} \sqrt{b x + a} \sqrt{d x + c} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \,{\left (8 \, b^{3} d^{3} x^{2} + 33 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \,{\left (13 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{96 \, b^{4} d}, -\frac{15 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-b d} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{-b d} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (b^{2} d^{2} x^{2} + a b c d +{\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \,{\left (8 \, b^{3} d^{3} x^{2} + 33 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \,{\left (13 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{48 \, b^{4} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(b*d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c
*d + a^2*d^2 - 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(8
*b^3*d^3*x^2 + 33*b^3*c^2*d - 40*a*b^2*c*d^2 + 15*a^2*b*d^3 + 2*(13*b^3*c*d^2 - 5*a*b^2*d^3)*x)*sqrt(b*x + a)*
sqrt(d*x + c))/(b^4*d), -1/48*(15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-b*d)*arctan(1/2*(2
*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) -
2*(8*b^3*d^3*x^2 + 33*b^3*c^2*d - 40*a*b^2*c*d^2 + 15*a^2*b*d^3 + 2*(13*b^3*c*d^2 - 5*a*b^2*d^3)*x)*sqrt(b*x +
a)*sqrt(d*x + c))/(b^4*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d x\right )^{\frac{5}{2}}}{\sqrt{a + b x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/2)/(b*x+a)**(1/2),x)

[Out]

Integral((c + d*x)**(5/2)/sqrt(a + b*x), x)

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Giac [B]  time = 1.40422, size = 614, normalized size = 4.15 \begin{align*} -\frac{\frac{24 \,{\left (\frac{{\left (b^{2} c - a b d\right )} \log \left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d}} - \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}\right )} c^{2}{\left | b \right |}}{b^{2}} - \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (2 \,{\left (b x + a\right )}{\left (\frac{4 \,{\left (b x + a\right )}}{b^{2}} + \frac{b^{6} c d^{3} - 13 \, a b^{5} d^{4}}{b^{7} d^{4}}\right )} - \frac{3 \,{\left (b^{7} c^{2} d^{2} + 2 \, a b^{6} c d^{3} - 11 \, a^{2} b^{5} d^{4}\right )}}{b^{7} d^{4}}\right )} - \frac{3 \,{\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d} b d^{2}}\right )} d^{2}{\left | b \right |}}{b^{2}} - \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )}}{b^{4} d^{2}} + \frac{b c d - 5 \, a d^{2}}{b^{4} d^{4}}\right )} + \frac{{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d} b^{3} d^{3}}\right )} c d{\left | b \right |}}{b^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(24*((b^2*c - a*b*d)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/sqrt(b*d)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a))*c^2*abs(b)/b^2 - (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqr
t(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/b^2 + (b^6*c*d^3 - 13*a*b^5*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 + 2*a*b^6*c*
d^3 - 11*a^2*b^5*d^4)/(b^7*d^4)) - 3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*log(abs(-sqrt(b*d)*sq
rt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b*d^2))*d^2*abs(b)/b^2 - (sqrt(b^2*c + (b*x + a
)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)/(b^4*d^2) + (b*c*d - 5*a*d^2)/(b^4*d^4)) + (b^2*c^2 + 2*a*b*c*d - 3*
a^2*d^2)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^3*d^3))*c*d*abs
(b)/b^3)/b