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

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

[Out]

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

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Rubi [A]  time = 0.0747416, 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} \[ \frac{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^2}{8 d^3}-\frac{5 (a+b x)^{3/2} \sqrt{c+d x} (b c-a d)}{12 d^2}-\frac{5 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{8 \sqrt{b} d^{7/2}}+\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)/Sqrt[c + d*x],x]

[Out]

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

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{(a+b x)^{5/2}}{\sqrt{c+d x}} \, dx &=\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 d}-\frac{(5 (b c-a d)) \int \frac{(a+b x)^{3/2}}{\sqrt{c+d x}} \, dx}{6 d}\\ &=-\frac{5 (b c-a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 d^2}+\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 d}+\frac{\left (5 (b c-a d)^2\right ) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{8 d^2}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 d^3}-\frac{5 (b c-a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 d^2}+\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 d}-\frac{\left (5 (b c-a d)^3\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{16 d^3}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 d^3}-\frac{5 (b c-a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 d^2}+\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 d}-\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 d^3}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 d^3}-\frac{5 (b c-a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 d^2}+\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 d}-\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 d^3}\\ &=\frac{5 (b c-a d)^2 \sqrt{a+b x} \sqrt{c+d x}}{8 d^3}-\frac{5 (b c-a d) (a+b x)^{3/2} \sqrt{c+d x}}{12 d^2}+\frac{(a+b x)^{5/2} \sqrt{c+d x}}{3 d}-\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 \sqrt{b} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.487239, size = 150, normalized size = 1.01 \[ \frac{\sqrt{d} \sqrt{a+b x} (c+d x) \left (33 a^2 d^2+2 a b d (13 d x-20 c)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )-\frac{15 (b c-a d)^{7/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{b}}{24 d^{7/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(33*a^2*d^2 + 2*a*b*d*(-20*c + 13*d*x) + b^2*(15*c^2 - 10*c*d*x + 8*d^2*x^2))
- (15*(b*c - a*d)^(7/2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/b)/
(24*d^(7/2)*Sqrt[c + d*x])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x+a)^(5/2)*(d*x+c)^(1/2)/d+5/12/d*(b*x+a)^(3/2)*(d*x+c)^(1/2)*a-5/12/d^2*(b*x+a)^(3/2)*(d*x+c)^(1/2)*b*
c+5/8/d*(b*x+a)^(1/2)*(d*x+c)^(1/2)*a^2-5/4/d^2*(b*x+a)^(1/2)*(d*x+c)^(1/2)*a*b*c+5/8/d^3*(b*x+a)^(1/2)*(d*x+c
)^(1/2)*b^2*c^2+5/16*((b*x+a)*(d*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(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^3-15/16/d*((b*x+a)*(d*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(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*b*c+15/16/d^2*((b*x+a
)*(d*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(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*b^2*c^2-5/16/d^3*((b*x+a)*(d*x+c))^(1/2)/(b*x+a)^(1/2)/(d*x+c)^(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)*b^3*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((b*x+a)^(5/2)/(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8927, size = 932, 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} + 15 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} - 2 \,{\left (5 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{96 \, b d^{4}}, \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} + 15 \, b^{3} c^{2} d - 40 \, a b^{2} c d^{2} + 33 \, a^{2} b d^{3} - 2 \,{\left (5 \, b^{3} c d^{2} - 13 \, a b^{2} d^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{48 \, b d^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)/(d*x+c)^(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 + 15*b^3*c^2*d - 40*a*b^2*c*d^2 + 33*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 13*a*b^2*d^3)*x)*sqrt(b*x + a)*
sqrt(d*x + c))/(b*d^4), 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 + 15*b^3*c^2*d - 40*a*b^2*c*d^2 + 33*a^2*b*d^3 - 2*(5*b^3*c*d^2 - 13*a*b^2*d^3)*x)*sqrt(b*x +
a)*sqrt(d*x + c))/(b*d^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28733, size = 267, normalized size = 1.8 \begin{align*} \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 d} - \frac{5 \,{\left (b c d^{3} - a d^{4}\right )}}{b d^{5}}\right )} + \frac{15 \,{\left (b^{2} c^{2} d^{2} - 2 \, a b c d^{3} + a^{2} d^{4}\right )}}{b d^{5}}\right )} + \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 )} \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} d^{3}}\right )} b}{24 \,{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b*x + a)/(b*d) - 5*(b*c*d^3 - a*d^4)/
(b*d^5)) + 15*(b^2*c^2*d^2 - 2*a*b*c*d^3 + a^2*d^4)/(b*d^5)) + 15*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a
^3*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d^3))*b/abs(b)