3.15.74 \(\int \frac {(c+d x)^{3/2}}{\sqrt {a+b x}} \, dx\) [1474]

Optimal. Leaf size=113 \[ \frac {3 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^2}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b}+\frac {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} \sqrt {d}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} \sqrt {d}}+\frac {3 \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}{4 b^2}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)/Sqrt[a + b*x],x]

[Out]

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

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

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Mathematica [A]
time = 0.17, size = 94, normalized size = 0.83 \begin {gather*} \frac {\sqrt {a+b x} \sqrt {c+d x} (5 b c-3 a d+2 b d x)}{4 b^2}+\frac {3 (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} \sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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[(c + d*x)^(3/2)/(a + b*x)^(1/2),x]')

[Out]

Timed out

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Maple [A]
time = 0.16, size = 140, normalized size = 1.24

method	result	size
default	\(\frac {\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}}{2 b}-\frac {3 \left (a d -b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {d x +c}}{b}-\frac {\left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {b d}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 b \sqrt {d x +c}\, \sqrt {b x +a}\, \sqrt {b d}}\right )}{4 b}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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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((d*x+c)^(3/2)/(b*x+a)^(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 detail

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Fricas [A]
time = 0.31, size = 306, normalized size = 2.71 \begin {gather*} \left [\frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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 (2 \, b^{2} d^{2} x + 5 \, b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{16 \, b^{3} d}, -\frac {3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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 (2 \, b^{2} d^{2} x + 5 \, b^{2} c d - 3 \, a b d^{2}\right )} \sqrt {b x + a} \sqrt {d x + c}}{8 \, b^{3} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*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*(2*b^2*d^2*x + 5*b^2*c*d
- 3*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d), -1/8*(3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*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*(2*b^2*d^2*x + 5*b^2*c*d - 3*a*b*d^2)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d)]

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

[Out]

Timed out

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Giac [A]
time = 0.02, size = 186, normalized size = 1.65 \begin {gather*} \frac {d^{2} \left (2 \left (\frac {\frac {1}{8}\cdot 2 b^{2} \sqrt {c+d x} \sqrt {c+d x}}{b^{3} d}-\frac {\frac {1}{8} \left (-3 b^{2} c+3 b d a\right )}{b^{3} d}\right ) \sqrt {c+d x} \sqrt {a d^{2}-b c d+b d \left (c+d x\right )}+\frac {2 \left (-3 a^{2} d^{2}+6 a b c d-3 b^{2} c^{2}\right ) \ln \left |\sqrt {a d^{2}-b c d+b d \left (c+d x\right )}-\sqrt {b d} \sqrt {c+d x}\right |}{8 b^{2} \sqrt {b d}}\right )}{\left |d\right | d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt((d*x + c)*b*d - b*c*d + a*d^2)*sqrt(d*x + c)*(2*(d*x + c)/(b*d) + 3*(b^2*c - a*b*d)/(b^3*d)) - 3*(b^
2*c^2 - 2*a*b*c*d + a^2*d^2)*log(abs(-sqrt(b*d)*sqrt(d*x + c) + sqrt((d*x + c)*b*d - b*c*d + a*d^2)))/(sqrt(b*
d)*b^2))*d/abs(d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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