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

Optimal. Leaf size=192 \[ -\frac{5 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{3/2} d^{7/2}}+\frac{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^3}{64 b d^3}-\frac{5 (a+b x)^{3/2} \sqrt{c+d x} (b c-a d)^2}{96 b d^2}+\frac{(a+b x)^{5/2} \sqrt{c+d x} (b c-a d)}{24 b d}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b} \]

[Out]

(5*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b*d^3) - (5*(b*c - a*d)^2*(a + b*x)^(3/2)*Sqrt[c + d*x])/(96
*b*d^2) + ((b*c - a*d)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(24*b*d) + ((a + b*x)^(7/2)*Sqrt[c + d*x])/(4*b) - (5*(b
*c - a*d)^4*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(3/2)*d^(7/2))

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Rubi [A]  time = 0.100166, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {50, 63, 217, 206} \[ -\frac{5 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{3/2} d^{7/2}}+\frac{5 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^3}{64 b d^3}-\frac{5 (a+b x)^{3/2} \sqrt{c+d x} (b c-a d)^2}{96 b d^2}+\frac{(a+b x)^{5/2} \sqrt{c+d x} (b c-a d)}{24 b d}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(5/2)*Sqrt[c + d*x],x]

[Out]

(5*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b*d^3) - (5*(b*c - a*d)^2*(a + b*x)^(3/2)*Sqrt[c + d*x])/(96
*b*d^2) + ((b*c - a*d)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(24*b*d) + ((a + b*x)^(7/2)*Sqrt[c + d*x])/(4*b) - (5*(b
*c - a*d)^4*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(3/2)*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 (a+b x)^{5/2} \sqrt{c+d x} \, dx &=\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b}+\frac{(b c-a d) \int \frac{(a+b x)^{5/2}}{\sqrt{c+d x}} \, dx}{8 b}\\ &=\frac{(b c-a d) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b}-\frac{\left (5 (b c-a d)^2\right ) \int \frac{(a+b x)^{3/2}}{\sqrt{c+d x}} \, dx}{48 b d}\\ &=-\frac{5 (b c-a d)^2 (a+b x)^{3/2} \sqrt{c+d x}}{96 b d^2}+\frac{(b c-a d) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b}+\frac{\left (5 (b c-a d)^3\right ) \int \frac{\sqrt{a+b x}}{\sqrt{c+d x}} \, dx}{64 b d^2}\\ &=\frac{5 (b c-a d)^3 \sqrt{a+b x} \sqrt{c+d x}}{64 b d^3}-\frac{5 (b c-a d)^2 (a+b x)^{3/2} \sqrt{c+d x}}{96 b d^2}+\frac{(b c-a d) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b}-\frac{\left (5 (b c-a d)^4\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{c+d x}} \, dx}{128 b d^3}\\ &=\frac{5 (b c-a d)^3 \sqrt{a+b x} \sqrt{c+d x}}{64 b d^3}-\frac{5 (b c-a d)^2 (a+b x)^{3/2} \sqrt{c+d x}}{96 b d^2}+\frac{(b c-a d) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b}-\frac{\left (5 (b c-a d)^4\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 )}{64 b^2 d^3}\\ &=\frac{5 (b c-a d)^3 \sqrt{a+b x} \sqrt{c+d x}}{64 b d^3}-\frac{5 (b c-a d)^2 (a+b x)^{3/2} \sqrt{c+d x}}{96 b d^2}+\frac{(b c-a d) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b}-\frac{\left (5 (b c-a d)^4\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 )}{64 b^2 d^3}\\ &=\frac{5 (b c-a d)^3 \sqrt{a+b x} \sqrt{c+d x}}{64 b d^3}-\frac{5 (b c-a d)^2 (a+b x)^{3/2} \sqrt{c+d x}}{96 b d^2}+\frac{(b c-a d) (a+b x)^{5/2} \sqrt{c+d x}}{24 b d}+\frac{(a+b x)^{7/2} \sqrt{c+d x}}{4 b}-\frac{5 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{3/2} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.570298, size = 190, normalized size = 0.99 \[ \frac{b \sqrt{d} \sqrt{a+b x} (c+d x) \left (a^2 b d^2 (73 c+118 d x)+15 a^3 d^3+a b^2 d \left (-55 c^2+36 c d x+136 d^2 x^2\right )+b^3 \left (-10 c^2 d x+15 c^3+8 c d^2 x^2+48 d^3 x^3\right )\right )-15 (b c-a d)^{9/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )}{192 b^2 d^{7/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(15*a^3*d^3 + a^2*b*d^2*(73*c + 118*d*x) + a*b^2*d*(-55*c^2 + 36*c*d*x + 13
6*d^2*x^2) + b^3*(15*c^3 - 10*c^2*d*x + 8*c*d^2*x^2 + 48*d^3*x^3)) - 15*(b*c - a*d)^(9/2)*Sqrt[(b*(c + d*x))/(
b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(192*b^2*d^(7/2)*Sqrt[c + d*x])

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Maple [B]  time = 0., size = 645, normalized size = 3.4 \begin{align*} \text{result too large to display} \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]

1/4/d*(b*x+a)^(5/2)*(d*x+c)^(3/2)+5/24/d*(b*x+a)^(3/2)*(d*x+c)^(3/2)*a-5/24/d^2*(b*x+a)^(3/2)*(d*x+c)^(3/2)*b*
c+5/32/d*(b*x+a)^(1/2)*(d*x+c)^(3/2)*a^2-5/16/d^2*(b*x+a)^(1/2)*(d*x+c)^(3/2)*a*b*c+5/32/d^3*(b*x+a)^(1/2)*(d*
x+c)^(3/2)*b^2*c^2+5/64/b*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^3-15/64/d*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^2*c+15/64/d^2*
(d*x+c)^(1/2)*(b*x+a)^(1/2)*a*c^2*b-5/64/d^3*(d*x+c)^(1/2)*(b*x+a)^(1/2)*c^3*b^2-5/128*d/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^4+5/32*((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*c-15/64/d*((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*c^2*b+5/32/d^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*c^3*b^2-5/128/d^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)*c^4*b^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 = 1.94235, size = 1203, normalized size = 6.27 \begin{align*} \left [\frac{15 \,{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{3} + 15 \, b^{4} c^{3} d - 55 \, a b^{3} c^{2} d^{2} + 73 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} + 8 \,{\left (b^{4} c d^{3} + 17 \, a b^{3} d^{4}\right )} x^{2} - 2 \,{\left (5 \, b^{4} c^{2} d^{2} - 18 \, a b^{3} c d^{3} - 59 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{768 \, b^{2} d^{4}}, \frac{15 \,{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\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 (48 \, b^{4} d^{4} x^{3} + 15 \, b^{4} c^{3} d - 55 \, a b^{3} c^{2} d^{2} + 73 \, a^{2} b^{2} c d^{3} + 15 \, a^{3} b d^{4} + 8 \,{\left (b^{4} c d^{3} + 17 \, a b^{3} d^{4}\right )} x^{2} - 2 \,{\left (5 \, b^{4} c^{2} d^{2} - 18 \, a b^{3} c d^{3} - 59 \, a^{2} b^{2} d^{4}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{384 \, b^{2} 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]

[1/768*(15*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*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*(48*b^4*d^4*x^3 + 15*b^4*c^3*d - 55*a*b^3*c^2*d^2 + 73*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(b^4
*c*d^3 + 17*a*b^3*d^4)*x^2 - 2*(5*b^4*c^2*d^2 - 18*a*b^3*c*d^3 - 59*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c
))/(b^2*d^4), 1/384*(15*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*sqrt(-b*d)*arc
tan(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*(48*b^4*d^4*x^3 + 15*b^4*c^3*d - 55*a*b^3*c^2*d^2 + 73*a^2*b^2*c*d^3 + 15*a^3*b*d^4 + 8*(b^4*c*d^3
+ 17*a*b^3*d^4)*x^2 - 2*(5*b^4*c^2*d^2 - 18*a*b^3*c*d^3 - 59*a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^
2*d^4)]

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

[Out]

Timed out

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Giac [B]  time = 1.45777, size = 842, normalized size = 4.39 \begin{align*} \frac{5 \,{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}{\left (2 \,{\left (b x + a\right )}{\left (4 \,{\left (b x + a\right )}{\left (\frac{6 \,{\left (b x + a\right )}}{b^{2}} + \frac{b^{7} c d^{5} - 17 \, a b^{6} d^{6}}{b^{8} d^{6}}\right )} - \frac{5 \, b^{8} c^{2} d^{4} + 6 \, a b^{7} c d^{5} - 59 \, a^{2} b^{6} d^{6}}{b^{8} d^{6}}\right )} + \frac{3 \,{\left (5 \, b^{9} c^{3} d^{3} + a b^{8} c^{2} d^{4} - a^{2} b^{7} c d^{5} - 5 \, a^{3} b^{6} d^{6}\right )}}{b^{8} d^{6}}\right )} \sqrt{b x + a} + \frac{3 \,{\left (5 \, b^{4} c^{4} - 4 \, a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + 5 \, a^{4} d^{4}\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^{3}}\right )}{\left | b \right |} + \frac{10 \,{\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 - a d^{2}}{b^{4} d^{4}}\right )} + \frac{{\left (b^{2} c^{2} - 2 \, a b c d + 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 )} a^{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^{6} d^{2}} + \frac{b c d^{3} - 7 \, a d^{4}}{b^{6} d^{6}}\right )} - \frac{3 \,{\left (b^{2} c^{2} d^{2} - a^{2} d^{4}\right )}}{b^{6} d^{6}}\right )} - \frac{3 \,{\left (b^{3} c^{3} - a b^{2} c^{2} d - 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} b^{5} d^{4}}\right )} a{\left | b \right |}}{b^{2}}}{960 \, b} \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]

1/960*(5*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^2 + (b^7*c*d^5 - 17*a*b
^6*d^6)/(b^8*d^6)) - (5*b^8*c^2*d^4 + 6*a*b^7*c*d^5 - 59*a^2*b^6*d^6)/(b^8*d^6)) + 3*(5*b^9*c^3*d^3 + a*b^8*c^
2*d^4 - a^2*b^7*c*d^5 - 5*a^3*b^6*d^6)/(b^8*d^6))*sqrt(b*x + a) + 3*(5*b^4*c^4 - 4*a*b^3*c^3*d - 2*a^2*b^2*c^2
*d^2 - 4*a^3*b*c*d^3 + 5*a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sq
rt(b*d)*b*d^3))*abs(b) + 10*(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
- a*d^2)/(b^4*d^4)) + (b^2*c^2 - 2*a*b*c*d + 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))*a^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)*(4*(b*x + a)/(b^6*d^2) + (b*c*d^3 - 7*a*d^4)/(b^6*d^6)) - 3*(b^2*c^2*d^2 - a^2*d^4)/(b^6*d^6)) - 3*(
b^3*c^3 - a*b^2*c^2*d - 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)*b^5*d^4))*a*abs(b)/b^2)/b