\(\int (a+b x)^{3/2} (c+d x)^{3/2} \, dx\) [1472]

Optimal result

Integrand size = 19, antiderivative size = 189 \[ \int (a+b x)^{3/2} (c+d x)^{3/2} \, dx=-\frac {3 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}{64 b^2 d^2}+\frac {(b c-a d)^2 (a+b x)^{3/2} \sqrt {c+d x}}{32 b^2 d}+\frac {(b c-a d) (a+b x)^{5/2} \sqrt {c+d x}}{8 b^2}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}+\frac {3 (b c-a d)^4 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{5/2} d^{5/2}} \]

[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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {52, 65, 223, 212} \[ \int (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\frac {3 (b c-a d)^4 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{64 b^{5/2} d^{5/2}}-\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)^3}{64 b^2 d^2}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (b c-a d)^2}{32 b^2 d}+\frac {(a+b x)^{5/2} \sqrt {c+d x} (b c-a d)}{8 b^2}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b} \]

[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 52

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 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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*} \text {integral}& = \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 ) \text {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 ) \text {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] (verified)

Time = 0.30 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.88 \[ \int (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-3 a^3 d^3+a^2 b d^2 (11 c+2 d x)+a b^2 d \left (11 c^2+44 c d x+24 d^2 x^2\right )+b^3 \left (-3 c^3+2 c^2 d x+24 c d^2 x^2+16 d^3 x^3\right )\right )}{64 b^2 d^2}+\frac {3 (b c-a d)^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{64 b^{5/2} d^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {\left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {5}{2}}}{4 d}-\frac {3 \left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {5}{2}}}{3 d}-\frac {\left (-a d +b c \right ) \left (\frac {\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}}{2 b}-\frac {3 \left (a d -b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {d x +c}}{b}-\frac {\left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 b \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}}\right )}{4 b}\right )}{6 d}\right )}{8 d}\)	\(206\)

[In]

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

[Out]

1/4/d*(b*x+a)^(3/2)*(d*x+c)^(5/2)-3/8*(-a*d+b*c)/d*(1/3/d*(b*x+a)^(1/2)*(d*x+c)^(5/2)-1/6*(-a*d+b*c)/d*(1/2*(d
*x+c)^(3/2)*(b*x+a)^(1/2)/b-3/4*(a*d-b*c)/b*((b*x+a)^(1/2)*(d*x+c)^(1/2)/b-1/2*(a*d-b*c)/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)+(b*d*x^2+(a*d+b*c)*x+a*c)^(1/2))/(b*d)
^(1/2))))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 534, normalized size of antiderivative = 2.83 \[ \int (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\left [\frac {3 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (16 \, b^{4} d^{4} x^{3} - 3 \, b^{4} c^{3} d + 11 \, a b^{3} c^{2} d^{2} + 11 \, a^{2} b^{2} c d^{3} - 3 \, a^{3} b d^{4} + 24 \, {\left (b^{4} c d^{3} + a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (b^{4} c^{2} d^{2} + 22 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{256 \, b^{3} d^{3}}, -\frac {3 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (16 \, b^{4} d^{4} x^{3} - 3 \, b^{4} c^{3} d + 11 \, a b^{3} c^{2} d^{2} + 11 \, a^{2} b^{2} c d^{3} - 3 \, a^{3} b d^{4} + 24 \, {\left (b^{4} c d^{3} + a b^{3} d^{4}\right )} x^{2} + 2 \, {\left (b^{4} c^{2} d^{2} + 22 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{128 \, b^{3} d^{3}}\right ] \]

[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)]

Sympy [F]

\[ \int (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\int \left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {3}{2}}\, dx \]

[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)

Maxima [F(-2)]

Exception generated. \[ \int (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\text {Exception raised: ValueError} \]

[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 detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1071 vs. \(2 (151) = 302\).

Time = 0.45 (sec) , antiderivative size = 1071, normalized size of antiderivative = 5.67 \[ \int (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\text {Too large to display} \]

[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

Mupad [F(-1)]

Timed out. \[ \int (a+b x)^{3/2} (c+d x)^{3/2} \, dx=\int {\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{3/2} \,d x \]

[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)