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

Optimal. Leaf size=154 \[ \frac {(b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{3/2} d^{5/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}{8 b d^2}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{12 b d}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b} \]

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Rubi [A]  time = 0.07, antiderivative size = 154, 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 = {50, 63, 217, 206} \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^{3/2} d^{5/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} (b c-a d)^2}{8 b d^2}+\frac {(a+b x)^{3/2} \sqrt {c+d x} (b c-a d)}{12 b d}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)*Sqrt[c + d*x],x]

[Out]

-((b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(8*b*d^2) + ((b*c - a*d)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(12*b*d)
+ ((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*b) + ((b*c - a*d)^3*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x
])])/(8*b^(3/2)*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} \sqrt {c+d x} \, dx &=\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}+\frac {(b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}} \, dx}{6 b}\\ &=\frac {(b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}-\frac {(b c-a d)^2 \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx}{8 b d}\\ &=-\frac {(b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 b d^2}+\frac {(b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}+\frac {(b c-a d)^3 \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx}{16 b d^2}\\ &=-\frac {(b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 b d^2}+\frac {(b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}+\frac {(b c-a d)^3 \operatorname {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^2 d^2}\\ &=-\frac {(b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 b d^2}+\frac {(b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}+\frac {(b c-a d)^3 \operatorname {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^2 d^2}\\ &=-\frac {(b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}{8 b d^2}+\frac {(b c-a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b d}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{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^{3/2} d^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.51, size = 151, normalized size = 0.98 \begin {gather*} \frac {b \sqrt {d} \sqrt {a+b x} (c+d x) \left (3 a^2 d^2+2 a b d (4 c+7 d x)+b^2 \left (-3 c^2+2 c d x+8 d^2 x^2\right )\right )+3 (b c-a d)^{7/2} \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 )}{24 b^2 d^{5/2} \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[d]*Sqrt[a + b*x]*(c + d*x)*(3*a^2*d^2 + 2*a*b*d*(4*c + 7*d*x) + b^2*(-3*c^2 + 2*c*d*x + 8*d^2*x^2)) +
3*(b*c - a*d)^(7/2)*Sqrt[(b*(c + d*x))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/(24*b^2*
d^(5/2)*Sqrt[c + d*x])

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IntegrateAlgebraic [A]  time = 0.00, size = 154, normalized size = 1.00 \begin {gather*} \frac {(b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{3/2} d^{5/2}}-\frac {\sqrt {c+d x} (b c-a d)^3 \left (\frac {3 b^2 (c+d x)^2}{(a+b x)^2}-\frac {8 b d (c+d x)}{a+b x}-3 d^2\right )}{24 b d^2 \sqrt {a+b x} \left (\frac {b (c+d x)}{a+b x}-d\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*((b*c - a*d)^3*Sqrt[c + d*x]*(-3*d^2 - (8*b*d*(c + d*x))/(a + b*x) + (3*b^2*(c + d*x)^2)/(a + b*x)^2))/(
b*d^2*Sqrt[a + b*x]*(-d + (b*(c + d*x))/(a + b*x))^3) + ((b*c - a*d)^3*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d
]*Sqrt[a + b*x])])/(8*b^(3/2)*d^(5/2))

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fricas [A]  time = 1.55, size = 410, normalized size = 2.66 \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 (b^{3} c d^{2} + 7 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{2} d^{3}}, -\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 (b^{3} c d^{2} + 7 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{2} d^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(d*x+c)^(1/2),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*(b^3*c*d^2 + 7*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*
x + c))/(b^2*d^3), -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*(b^3*c*d^2 + 7*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x
+ c))/(b^2*d^3)]

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giac [B]  time = 1.20, size = 438, normalized size = 2.84 \begin {gather*} \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^{2}} + \frac {b^{6} c d^{3} - 13 \, a b^{5} d^{4}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{7} c^{2} d^{2} + 2 \, a b^{6} c d^{3} - 11 \, a^{2} b^{5} d^{4}\right )}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, 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 d^{2}}\right )} {\left | b \right |} - \frac {24 \, {\left (\frac {{\left (b^{2} c - a b d\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}} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a}\right )} a^{2} {\left | b \right |}}{b^{2}} + \frac {12 \, {\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b 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} d}\right )} a {\left | b \right |}}{b^{2}}}{24 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((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))*abs(b) - 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)))/s
qrt(b*d) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a))*a^2*abs(b)/b^2 + 12*(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*abs(b)/b^2)/b

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maple [B]  time = 0.01, size = 460, normalized size = 2.99 \begin {gather*} -\frac {\sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{3} 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 )}{16 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, b}+\frac {3 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} c \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 )}{16 \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^{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 )}{16 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, d}+\frac {\sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} 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 )}{16 \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}\, d^{2}}+\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, a^{2}}{8 b}-\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, a c}{4 d}+\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, b \,c^{2}}{8 d^{2}}+\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{2}} a}{4 d}-\frac {\sqrt {b x +a}\, \left (d x +c \right )^{\frac {3}{2}} b c}{4 d^{2}}+\frac {\left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {3}{2}}}{3 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/d*(b*x+a)^(3/2)*(d*x+c)^(3/2)+1/4/d*(b*x+a)^(1/2)*(d*x+c)^(3/2)*a-1/4/d^2*(b*x+a)^(1/2)*(d*x+c)^(3/2)*b*c+
1/8/b*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a^2-1/4/d*(d*x+c)^(1/2)*(b*x+a)^(1/2)*a*c+1/8/d^2*(d*x+c)^(1/2)*(b*x+a)^(1/2
)*c^2*b-1/16*d/b*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*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))/(b*d)^(1/2)*a^3+3/16*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*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))/(b*d)^(1/2)*a^2*c-3/16/d*((b*x+a)*(d*x+c))^(
1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*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))/(b*d)
^(1/2)*a*c^2*b+1/16/d^2*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+a)^(1/2)*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))/(b*d)^(1/2)*c^3*b^2

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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)^(3/2)*(d*x+c)^(1/2),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 \begin {gather*} \int {\left (a+b\,x\right )}^{3/2}\,\sqrt {c+d\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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