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

Optimal. Leaf size=116 \[ \frac {\sqrt {d} (3 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}}-\frac {2 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {d \sqrt {a+b x} \sqrt {c+d x}}{b} \]

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Rubi [A]  time = 0.08, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {102, 157, 63, 217, 206, 93, 208} \begin {gather*} \frac {\sqrt {d} (3 b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}}-\frac {2 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {d \sqrt {a+b x} \sqrt {c+d x}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 102

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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 208

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

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

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Mathematica [A]  time = 0.97, size = 171, normalized size = 1.47 \begin {gather*} \frac {b \left (\sqrt {a} d \sqrt {a+b x} (c+d x)-2 b c^{3/2} \sqrt {c+d x} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )+\sqrt {a} \sqrt {d} \sqrt {b c-a d} (3 b c-a d) \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 )}{\sqrt {a} b^2 \sqrt {c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.29, size = 147, normalized size = 1.27 \begin {gather*} \frac {\left (3 b c \sqrt {d}-a d^{3/2}\right ) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2}}-\frac {2 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {a}}+\frac {d \sqrt {a+b x} (b c-a d)}{b \sqrt {c+d x} \left (b-\frac {d (a+b x)}{c+d x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 3.05, size = 844, normalized size = 7.28 \begin {gather*} \left [\frac {2 \, b c \sqrt {\frac {c}{a}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - {\left (3 \, b c - a d\right )} \sqrt {\frac {d}{b}} \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^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, \sqrt {b x + a} \sqrt {d x + c} d}{4 \, b}, \frac {b c \sqrt {\frac {c}{a}} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{2} c + {\left (a b c + a^{2} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {c}{a}} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - {\left (3 \, b c - a d\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) + 2 \, \sqrt {b x + a} \sqrt {d x + c} d}{2 \, b}, \frac {4 \, b c \sqrt {-\frac {c}{a}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) - {\left (3 \, b c - a d\right )} \sqrt {\frac {d}{b}} \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^{2} d x + b^{2} c + a b d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {\frac {d}{b}} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, \sqrt {b x + a} \sqrt {d x + c} d}{4 \, b}, \frac {2 \, b c \sqrt {-\frac {c}{a}} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {c}{a}}}{2 \, {\left (b c d x^{2} + a c^{2} + {\left (b c^{2} + a c d\right )} x\right )}}\right ) - {\left (3 \, b c - a d\right )} \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {b x + a} \sqrt {d x + c} \sqrt {-\frac {d}{b}}}{2 \, {\left (b d^{2} x^{2} + a c d + {\left (b c d + a d^{2}\right )} x\right )}}\right ) + 2 \, \sqrt {b x + a} \sqrt {d x + c} d}{2 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*b*c*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*s
qrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - (3*b*c - a*d)*sqrt(d/b)*log(8*b^2*d^2*x
^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b^2*d*x + b^2*c + a*b*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(
b^2*c*d + a*b*d^2)*x) + 4*sqrt(b*x + a)*sqrt(d*x + c)*d)/b, 1/2*(b*c*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a
*b*c*d + a^2*d^2)*x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a
^2*c*d)*x)/x^2) - (3*b*c - a*d)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-
d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) + 2*sqrt(b*x + a)*sqrt(d*x + c)*d)/b, 1/4*(4*b*c*sqrt(-c/a)*arct
an(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (b*c^2 + a*c*d)*x))
- (3*b*c - a*d)*sqrt(d/b)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b^2*d*x + b^2*c + a*b*d)*s
qrt(b*x + a)*sqrt(d*x + c)*sqrt(d/b) + 8*(b^2*c*d + a*b*d^2)*x) + 4*sqrt(b*x + a)*sqrt(d*x + c)*d)/b, 1/2*(2*b
*c*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*x^2 + a*c^2 + (
b*c^2 + a*c*d)*x)) - (3*b*c - a*d)*sqrt(-d/b)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqr
t(-d/b)/(b*d^2*x^2 + a*c*d + (b*c*d + a*d^2)*x)) + 2*sqrt(b*x + a)*sqrt(d*x + c)*d)/b]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:

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maple [B]  time = 0.02, size = 219, normalized size = 1.89 \begin {gather*} -\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (\sqrt {a c}\, a \,d^{2} \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+2 \sqrt {b d}\, b \,c^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-3 \sqrt {a c}\, b c d \ln \left (\frac {2 b d x +a d +b c +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}}{2 \sqrt {b d}}\right )-2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, d \right )}{2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}\, \sqrt {a c}\, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*
a*d^2*(a*c)^(1/2)-3*ln(1/2*(2*b*d*x+a*d+b*c+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2))/(b*d)^(1/2))*b*c*d*(a*c)^(1
/2)+2*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*b*c^2*(b*d)^(1/2)-2*d*((b*x+a)*(d*x+c))^
(1/2)*(b*d)^(1/2)*(a*c)^(1/2))/((b*x+a)*(d*x+c))^(1/2)/(b*d)^(1/2)/(a*c)^(1/2)/b

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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)/x/(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 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 \frac {{\left (c+d\,x\right )}^{3/2}}{x\,\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)/(x*(a + b*x)^(1/2)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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