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

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

[Out]

1/4*(b*x+a)^(5/2)*(d*x+c)^(3/2)/b+3/64*(-a*d+b*c)^4*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(5/
2)/d^(5/2)+1/32*(-a*d+b*c)^2*(b*x+a)^(3/2)*(d*x+c)^(1/2)/b^2/d+1/8*(-a*d+b*c)*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b^2-
3/64*(-a*d+b*c)^3*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^2/d^2

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Rubi [A]  time = 0.10, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, 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 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])

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]

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*}

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Mathematica [A]  time = 0.59, size = 193, normalized size = 1.02 \[ \frac {3 (b c-a d)^{9/2} \sqrt {\frac {b (c+d x)}{b c-a d}} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )-b \sqrt {d} \sqrt {a+b x} (c+d x) \left (3 a^3 d^3-a^2 b d^2 (11 c+2 d x)-a b^2 d \left (11 c^2+44 c d x+24 d^2 x^2\right )+b^3 \left (3 c^3-2 c^2 d x-24 c d^2 x^2-16 d^3 x^3\right )\right )}{64 b^3 d^{5/2} \sqrt {c+d x}} \]

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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fricas [A]  time = 1.17, size = 534, normalized size = 2.83 \[ \left [\frac {3 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (16 \, b^{4} d^{4} x^{3} - 3 \, b^{4} c^{3} d + 11 \, a b^{3} c^{2} d^{2} + 11 \, a^{2} b^{2} c d^{3} - 3 \, a^{3} b d^{4} + 24 \, {\left (b^{4} c d^{3} + a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (b^{4} c^{2} d^{2} + 22 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{256 \, b^{3} d^{3}}, -\frac {3 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (16 \, b^{4} d^{4} x^{3} - 3 \, b^{4} c^{3} d + 11 \, a b^{3} c^{2} d^{2} + 11 \, a^{2} b^{2} c d^{3} - 3 \, a^{3} b d^{4} + 24 \, {\left (b^{4} c d^{3} + a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (b^{4} c^{2} d^{2} + 22 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{128 \, b^{3} d^{3}}\right ] \]

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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giac [B]  time = 2.96, size = 1071, normalized size = 5.67 \[ \text {result too large to display} \]

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]

1/192*(8*(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^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)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d
)*b*d^2))*c*abs(b) - 192*((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))*a^2*c*abs(b)/b^2 + (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^3 + (b^12*c*d^5 - 25*a*b^11*d^6)/(b^14*d^6)) - (5*b^13*
c^2*d^4 + 14*a*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^6)) + 3*(5*b^14*c^3*d^3 + 9*a*b^13*c^2*d^4 + 15*a^2*b^12
*c*d^5 - 93*a^3*b^11*d^6)/(b^14*d^6))*sqrt(b*x + a) + 3*(5*b^4*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 20*a^
3*b*c*d^3 - 35*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^
2*d^3))*d*abs(b) + 16*(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^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)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d
)))/(sqrt(b*d)*b*d^2))*a*d*abs(b)/b + 96*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)
/d^2)*sqrt(b*x + a) + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*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)*d))*a*c*abs(b)/b^2 + 48*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a +
(b*c*d - 5*a*d^2)/d^2)*sqrt(b*x + a) + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*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)*d))*a^2*d*abs(b)/b^3)/b

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maple [B]  time = 0.00, size = 640, normalized size = 3.39 \[ \frac {3 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{4} d^{2} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{128 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, b^{2}}-\frac {3 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} c d \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{32 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, b}+\frac {9 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c^{2} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{64 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}}-\frac {3 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,c^{3} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{32 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, d}+\frac {3 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{4} \ln \left (\frac {b d x +\frac {1}{2} a d +\frac {1}{2} b c}{\sqrt {b d}}+\sqrt {b d \,x^{2}+a c +\left (a d +b c \right ) x}\right )}{128 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, d^{2}}-\frac {3 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{3} d}{64 b^{2}}+\frac {9 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} c}{64 b}-\frac {9 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,c^{2}}{64 d}+\frac {3 \sqrt {d x +c}\, \sqrt {b x +a}\, b \,c^{3}}{64 d^{2}}+\frac {\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}\, a^{2}}{32 b}-\frac {\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}\, a c}{16 d}+\frac {\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}\, b \,c^{2}}{32 d^{2}}+\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {5}{2}} a}{8 d}-\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {5}{2}} b c}{8 d^{2}}+\frac {\left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {5}{2}}}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/128*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)/(b*d)^(1/2)*a^4/b^2*d^2*ln((b*d*x+1/2*a*d+1/2*b*c)/(
b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))-3/32*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)/(b*d)^(1/
2)*a^3/b*c*d*ln((b*d*x+1/2*a*d+1/2*b*c)/(b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))+9/64*((b*x+a)*(d*x+c))^(1
/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)/(b*d)^(1/2)*a^2*c^2*ln((b*d*x+1/2*a*d+1/2*b*c)/(b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b
*c)*x)^(1/2))-3/32*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)/(b*d)^(1/2)*a*b*c^3/d*ln((b*d*x+1/2*a*d
+1/2*b*c)/(b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))+3/128*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/
2)/(b*d)^(1/2)*b^2*c^4/d^2*ln((b*d*x+1/2*a*d+1/2*b*c)/(b*d)^(1/2)+(b*d*x^2+a*c+(a*d+b*c)*x)^(1/2))-3/64*(d*x+c
)^(1/2)*(b*x+a)^(1/2)*a^3/b^2*d+9/64*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^2/b*c-9/64*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a*c^
2/d+3/64*(d*x+c)^(1/2)*(b*x+a)^(1/2)*b*c^3/d^2+1/32*(d*x+c)^(3/2)*(b*x+a)^(1/2)*a^2/b-1/16*(d*x+c)^(3/2)*(b*x+
a)^(1/2)*a*c/d+1/32*(d*x+c)^(3/2)*(b*x+a)^(1/2)*b*c^2/d^2+1/8*(b*x+a)^(1/2)*(d*x+c)^(5/2)*a/d-1/8*(b*x+a)^(1/2
)*(d*x+c)^(5/2)*b*c/d^2+1/4*(b*x+a)^(3/2)*(d*x+c)^(5/2)/d

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {3}{2}}\, dx \]

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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