3.1472 \(\int (a+b x)^{3/2} (c+d x)^{3/2} \, dx\)

Optimal. Leaf size=189 \[ -\frac{3 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^3}{64 b^2 d^2}+\frac{3 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{5/2} d^{5/2}}+\frac{(a+b x)^{3/2} \sqrt{c+d x} (b c-a d)^2}{32 b^2 d}+\frac{(a+b x)^{5/2} \sqrt{c+d x} (b c-a d)}{8 b^2}+\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{4 b} \]

[Out]

(-3*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^2*d^2) + ((b*c - a*d)^2*(a + b*x)^(3/2)*Sqrt[c + d*x])/(3
2*b^2*d) + ((b*c - a*d)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(8*b^2) + ((a + b*x)^(5/2)*(c + d*x)^(3/2))/(4*b) + (3*
(b*c - a*d)^4*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(5/2)*d^(5/2))

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Rubi [A]  time = 0.0898839, antiderivative size = 189, 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} \[ -\frac{3 \sqrt{a+b x} \sqrt{c+d x} (b c-a d)^3}{64 b^2 d^2}+\frac{3 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{5/2} d^{5/2}}+\frac{(a+b x)^{3/2} \sqrt{c+d x} (b c-a d)^2}{32 b^2 d}+\frac{(a+b x)^{5/2} \sqrt{c+d x} (b c-a d)}{8 b^2}+\frac{(a+b x)^{5/2} (c+d x)^{3/2}}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*c - a*d)^3*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^2*d^2) + ((b*c - a*d)^2*(a + b*x)^(3/2)*Sqrt[c + d*x])/(3
2*b^2*d) + ((b*c - a*d)*(a + b*x)^(5/2)*Sqrt[c + d*x])/(8*b^2) + ((a + b*x)^(5/2)*(c + d*x)^(3/2))/(4*b) + (3*
(b*c - a*d)^4*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(5/2)*d^(5/2))

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

Mathematica [A]  time = 0.536016, size = 193, normalized size = 1.02 \[ \frac{3 (b c-a d)^{9/2} \sqrt{\frac{b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b c-a d}}\right )-b \sqrt{d} \sqrt{a+b x} (c+d x) \left (-a^2 b d^2 (11 c+2 d x)+3 a^3 d^3-a b^2 d \left (11 c^2+44 c d x+24 d^2 x^2\right )+b^3 \left (-2 c^2 d x+3 c^3-24 c d^2 x^2-16 d^3 x^3\right )\right )}{64 b^3 d^{5/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(3*a^3*d^3 - a^2*b*d^2*(11*c + 2*d*x) - a*b^2*d*(11*c^2 + 44*c*d*x + 24*d
^2*x^2) + b^3*(3*c^3 - 2*c^2*d*x - 24*c*d^2*x^2 - 16*d^3*x^3))) + 3*(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]])/(64*b^3*d^(5/2)*Sqrt[c + d*x])

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Maple [B]  time = 0.004, size = 640, 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)^(3/2)*(d*x+c)^(3/2),x)

[Out]

1/4/d*(b*x+a)^(3/2)*(d*x+c)^(5/2)+1/8/d*(b*x+a)^(1/2)*(d*x+c)^(5/2)*a-1/8/d^2*(b*x+a)^(1/2)*(d*x+c)^(5/2)*b*c+
1/32/b*(d*x+c)^(3/2)*(b*x+a)^(1/2)*a^2-1/16/d*(d*x+c)^(3/2)*(b*x+a)^(1/2)*a*c+1/32/d^2*(d*x+c)^(3/2)*(b*x+a)^(
1/2)*c^2*b-3/64*d/b^2*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^3+9/64/b*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^2*c-9/64/d*(d*x+c)^
(1/2)*(b*x+a)^(1/2)*a*c^2+3/64/d^2*(d*x+c)^(1/2)*(b*x+a)^(1/2)*c^3*b+3/128*d^2/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
^4-3/32*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^3*c+9/64*((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^2-3/32/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*c^3*b+3/128/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)*c^4*b^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((b*x+a)^(3/2)*(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.17454, size = 1177, normalized size = 6.23 \begin{align*} \left [\frac{3 \,{\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 (16 \, b^{4} d^{4} x^{3} - 3 \, b^{4} c^{3} d + 11 \, a b^{3} c^{2} d^{2} + 11 \, a^{2} b^{2} c d^{3} - 3 \, a^{3} b d^{4} + 24 \,{\left (b^{4} c d^{3} + a b^{3} d^{4}\right )} x^{2} + 2 \,{\left (b^{4} c^{2} d^{2} + 22 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{256 \, b^{3} d^{3}}, -\frac{3 \,{\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 (16 \, b^{4} d^{4} x^{3} - 3 \, b^{4} c^{3} d + 11 \, a b^{3} c^{2} d^{2} + 11 \, a^{2} b^{2} c d^{3} - 3 \, a^{3} b d^{4} + 24 \,{\left (b^{4} c d^{3} + a b^{3} d^{4}\right )} x^{2} + 2 \,{\left (b^{4} c^{2} d^{2} + 22 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{128 \, b^{3} d^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[1/256*(3*(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*(16*b^4*d^4*x^3 - 3*b^4*c^3*d + 11*a*b^3*c^2*d^2 + 11*a^2*b^2*c*d^3 - 3*a^3*b*d^4 + 24*(b^4*c
*d^3 + a*b^3*d^4)*x^2 + 2*(b^4*c^2*d^2 + 22*a*b^3*c*d^3 + a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^
3), -1/128*(3*(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)*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*(16*b^4*d^4*x^3 - 3*b^4*c^3*d + 11*a*b^3*c^2*d^2 + 11*a^2*b^2*c*d^3 - 3*a^3*b*d^4 + 24*(b^4*c*d^3 + a*b^3*d^
4)*x^2 + 2*(b^4*c^2*d^2 + 22*a*b^3*c*d^3 + a^2*b^2*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b x\right )^{\frac{3}{2}} \left (c + d x\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(d*x+c)**(3/2),x)

[Out]

Integral((a + b*x)**(3/2)*(c + d*x)**(3/2), x)

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Giac [B]  time = 1.28398, size = 1107, normalized size = 5.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(d*x+c)^(3/2),x, algorithm="giac")

[Out]

1/1920*(20*(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*c*abs(b)/b^2 + 10*(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)))/(sqrt(b*d)*b*d^3))*d*abs(b)/b + (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))*c*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*d*abs(b)/b^3)/b