3.15.73 \(\int \sqrt {a+b x} (c+d x)^{3/2} \, dx\) [1473]

Optimal. Leaf size=151 \[ \frac {(b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 b^2 d}+\frac {(b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{4 b^2}+\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 b}-\frac {(b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{5/2} d^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {52, 65, 223, 212} \begin {gather*} -\frac {(b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{5/2} d^{3/2}}+\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}{8 b^2 d}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{4 b^2}+\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]*(c + d*x)^(3/2),x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]
time = 0.24, size = 127, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} \left (-3 a^2 d^2+2 a b d (4 c+d x)+b^2 \left (3 c^2+14 c d x+8 d^2 x^2\right )\right )}{24 b^2 d}-\frac {(b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{5/2} d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*Sqrt[c + d*x]*(-3*a^2*d^2 + 2*a*b*d*(4*c + d*x) + b^2*(3*c^2 + 14*c*d*x + 8*d^2*x^2)))/(24*b^2*
d) - ((b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(8*b^(5/2)*d^(3/2))

________________________________________________________________________________________

Mathics [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[(a + b*x)^(1/2)*(c + d*x)^(3/2),x]')

[Out]

Timed out

________________________________________________________________________________________

Maple [A]
time = 0.16, size = 173, normalized size = 1.15

method	result	size
default	\(\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {5}{2}}}{3 d}-\frac {\left (-a d +b c \right ) \left (\frac {\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}}{2 b}-\frac {3 \left (a d -b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {d x +c}}{b}-\frac {\left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 b \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}}\right )}{4 b}\right )}{6 d}\)	\(173\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(1/2)*(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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)))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(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

________________________________________________________________________________________

Fricas [A]
time = 0.33, size = 410, normalized size = 2.72 \begin {gather*} \left [-\frac {3 \, {\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} + 3 \, b^{3} c^{2} d + 8 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} + 2 \, {\left (7 \, b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{3} d^{2}}, \frac {3 \, {\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} + 3 \, b^{3} c^{2} d + 8 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} + 2 \, {\left (7 \, b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{3} d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(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 + 3*b^3*c^2*d + 8*a*b^2*c*d^2 - 3*a^2*b*d^3 + 2*(7*b^3*c*d^2 + a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*
x + c))/(b^3*d^2), 1/48*(3*(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 + 3*b^3*c^2*d + 8*a*b^2*c*d^2 - 3*a^2*b*d^3 + 2*(7*b^3*c*d^2 + a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x
+ c))/(b^3*d^2)]

________________________________________________________________________________________

Sympy [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(1/2)*(d*x+c)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 576 vs. \(2 (119) = 238\).
time = 0.06, size = 768, normalized size = 5.09 \begin {gather*} \frac {\frac {2 b d \left |b\right | \left (2 \left (\left (\frac {\frac {1}{2304}\cdot 192 b^{5} d^{4} \sqrt {a+b x} \sqrt {a+b x}}{b^{7} d^{4}}-\frac {\frac {1}{2304} \left (-48 b^{6} d^{3} c+624 b^{5} d^{4} a\right )}{b^{7} d^{4}}\right ) \sqrt {a+b x} \sqrt {a+b x}-\frac {\frac {1}{2304} \left (72 b^{7} d^{2} c^{2}+144 b^{6} d^{3} a c-792 b^{5} d^{4} a^{2}\right )}{b^{7} d^{4}}\right ) \sqrt {a+b x} \sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}+\frac {2 \left (5 a^{3} d^{3}-3 a^{2} b c d^{2}-a b^{2} c^{2} d-b^{3} c^{3}\right ) \ln \left |\sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}-\sqrt {b d} \sqrt {a+b x}\right |}{32 b d^{2} \sqrt {b d}}\right )}{b^{2}}+\frac {2 a d \left |b\right | \left (2 \left (\frac {\frac {1}{64}\cdot 8 d^{2} \sqrt {a+b x} \sqrt {a+b x}}{d^{2}}-\frac {\frac {1}{64} \left (-4 b d c+20 d^{2} a\right )}{d^{2}}\right ) \sqrt {a+b x} \sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}+\frac {2 \left (-3 a^{2} b d^{2}+2 a b^{2} c d+b^{3} c^{2}\right ) \ln \left |\sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}-\sqrt {b d} \sqrt {a+b x}\right |}{16 d \sqrt {b d}}\right )}{b^{2} b}+\frac {2 b c \left |b\right | \left (2 \left (\frac {\frac {1}{64}\cdot 8 d^{2} \sqrt {a+b x} \sqrt {a+b x}}{d^{2}}-\frac {\frac {1}{64} \left (-4 b d c+20 d^{2} a\right )}{d^{2}}\right ) \sqrt {a+b x} \sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}+\frac {2 \left (-3 a^{2} b d^{2}+2 a b^{2} c d+b^{3} c^{2}\right ) \ln \left |\sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}-\sqrt {b d} \sqrt {a+b x}\right |}{16 d \sqrt {b d}}\right )}{b^{2} b}+\frac {2 a c \left |b\right | \left (\frac {1}{2} \sqrt {a+b x} \sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}+\frac {2 \left (a b d-b^{2} c\right ) \ln \left |\sqrt {-a b d+b^{2} c+b d \left (a+b x\right )}-\sqrt {b d} \sqrt {a+b x}\right |}{4 \sqrt {b d}}\right )}{b^{2}}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/2)*(d*x+c)^(3/2),x)

[Out]

-1/24*(24*((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*c*abs(b)/b^2 - (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqr
t(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)*sq
rt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b*d^2))*d*abs(b)/b - 6*(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(a
bs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d))*c*abs(b)/b^2 - 6*(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*d*abs(b)/b^
3)/b

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(1/2)*(c + d*x)^(3/2),x)

[Out]

int((a + b*x)^(1/2)*(c + d*x)^(3/2), x)

________________________________________________________________________________________