3.5.35 \(\int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx\)

Optimal. Leaf size=128 \[ -\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^2 b^{3/2}}+\frac {c^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}+\frac {d \sqrt {c+d x} (2 a d+b c)}{a b}-\frac {c (c+d x)^{3/2}}{a x} \]

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Rubi [A]  time = 0.17, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {98, 154, 156, 63, 208} \begin {gather*} -\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^2 b^{3/2}}+\frac {c^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}+\frac {d \sqrt {c+d x} (2 a d+b c)}{a b}-\frac {c (c+d x)^{3/2}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(x^2*(a + b*x)),x]

[Out]

(d*(b*c + 2*a*d)*Sqrt[c + d*x])/(a*b) - (c*(c + d*x)^(3/2))/(a*x) + (c^(3/2)*(2*b*c - 5*a*d)*ArcTanh[Sqrt[c +
d*x]/Sqrt[c]])/a^2 - (2*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(a^2*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 98

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

Rule 154

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

Rule 156

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

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]

Rubi steps

\begin {align*} \int \frac {(c+d x)^{5/2}}{x^2 (a+b x)} \, dx &=-\frac {c (c+d x)^{3/2}}{a x}-\frac {\int \frac {\sqrt {c+d x} \left (\frac {1}{2} c (2 b c-5 a d)-\frac {1}{2} d (b c+2 a d) x\right )}{x (a+b x)} \, dx}{a}\\ &=\frac {d (b c+2 a d) \sqrt {c+d x}}{a b}-\frac {c (c+d x)^{3/2}}{a x}-\frac {2 \int \frac {\frac {1}{4} b c^2 (2 b c-5 a d)+\frac {1}{4} d \left (b^2 c^2-6 a b c d+2 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}} \, dx}{a b}\\ &=\frac {d (b c+2 a d) \sqrt {c+d x}}{a b}-\frac {c (c+d x)^{3/2}}{a x}-\frac {\left (c^2 (2 b c-5 a d)\right ) \int \frac {1}{x \sqrt {c+d x}} \, dx}{2 a^2}+\frac {(b c-a d)^3 \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{a^2 b}\\ &=\frac {d (b c+2 a d) \sqrt {c+d x}}{a b}-\frac {c (c+d x)^{3/2}}{a x}-\frac {\left (c^2 (2 b c-5 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{a^2 d}+\frac {\left (2 (b c-a d)^3\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{a^2 b d}\\ &=\frac {d (b c+2 a d) \sqrt {c+d x}}{a b}-\frac {c (c+d x)^{3/2}}{a x}+\frac {c^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}-\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^2 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 114, normalized size = 0.89 \begin {gather*} \frac {-\frac {2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{3/2}}+c^{3/2} (2 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )+\frac {a \sqrt {c+d x} \left (2 a d^2 x-b c^2\right )}{b x}}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/2)/(x^2*(a + b*x)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.40, size = 139, normalized size = 1.09 \begin {gather*} \frac {2 (a d-b c)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x} \sqrt {a d-b c}}{b c-a d}\right )}{a^2 b^{3/2}}+\frac {\left (2 b c^{5/2}-5 a c^{3/2} d\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a^2}+\frac {\sqrt {c+d x} \left (2 a d (c+d x)-2 a c d-b c^2\right )}{a b x} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x)^(5/2)/(x^2*(a + b*x)),x]

[Out]

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

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fricas [A]  time = 1.88, size = 638, normalized size = 4.98 \begin {gather*} \left [\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) - {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {c} x \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) + 2 \, {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{2 \, a^{2} b x}, -\frac {4 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {c} x \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{2 \, a^{2} b x}, -\frac {{\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x + 2 \, b c - a d - 2 \, \sqrt {d x + c} b \sqrt {\frac {b c - a d}{b}}}{b x + a}\right ) - {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{a^{2} b x}, -\frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) + {\left (2 \, b^{2} c^{2} - 5 \, a b c d\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - {\left (2 \, a^{2} d^{2} x - a b c^{2}\right )} \sqrt {d x + c}}{a^{2} b x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^2/(b*x+a),x, algorithm="fricas")

[Out]

[1/2*(2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d - 2*sqrt(d*x + c)*b*sqr
t((b*c - a*d)/b))/(b*x + a)) - (2*b^2*c^2 - 5*a*b*c*d)*sqrt(c)*x*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x)
+ 2*(2*a^2*d^2*x - a*b*c^2)*sqrt(d*x + c))/(a^2*b*x), -1/2*(4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*sqrt(-(b*c - a
*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + (2*b^2*c^2 - 5*a*b*c*d)*sqrt(c)*x*log((d*x
- 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(2*a^2*d^2*x - a*b*c^2)*sqrt(d*x + c))/(a^2*b*x), -((2*b^2*c^2 - 5*a*b
*c*d)*sqrt(-c)*x*arctan(sqrt(d*x + c)*sqrt(-c)/c) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*sqrt((b*c - a*d)/b)*log(
(b*d*x + 2*b*c - a*d - 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) - (2*a^2*d^2*x - a*b*c^2)*sqrt(d*x +
c))/(a^2*b*x), -(2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c -
a*d)/b)/(b*c - a*d)) + (2*b^2*c^2 - 5*a*b*c*d)*sqrt(-c)*x*arctan(sqrt(d*x + c)*sqrt(-c)/c) - (2*a^2*d^2*x - a
*b*c^2)*sqrt(d*x + c))/(a^2*b*x)]

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giac [A]  time = 1.28, size = 152, normalized size = 1.19 \begin {gather*} \frac {2 \, \sqrt {d x + c} d^{2}}{b} - \frac {\sqrt {d x + c} c^{2}}{a x} - \frac {{\left (2 \, b c^{3} - 5 \, a c^{2} d\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{a^{2} \sqrt {-c}} + \frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^2/(b*x+a),x, algorithm="giac")

[Out]

2*sqrt(d*x + c)*d^2/b - sqrt(d*x + c)*c^2/(a*x) - (2*b*c^3 - 5*a*c^2*d)*arctan(sqrt(d*x + c)/sqrt(-c))/(a^2*sq
rt(-c)) + 2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(
sqrt(-b^2*c + a*b*d)*a^2*b)

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maple [B]  time = 0.02, size = 249, normalized size = 1.95 \begin {gather*} -\frac {2 a \,d^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, b}-\frac {6 b \,c^{2} d \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a}+\frac {2 b^{2} c^{3} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}\, a^{2}}+\frac {6 c \,d^{2} \arctan \left (\frac {\sqrt {d x +c}\, b}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\frac {5 c^{\frac {3}{2}} d \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a}+\frac {2 b \,c^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a^{2}}+\frac {2 \sqrt {d x +c}\, d^{2}}{b}-\frac {\sqrt {d x +c}\, c^{2}}{a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(5/2)/x^2/(b*x+a),x)

[Out]

2*d^2/b*(d*x+c)^(1/2)-2*d^3/b*a/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)+6*d^2/((a*d-b*
c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*c-6*d*b/a/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a
*d-b*c)*b)^(1/2)*b)*c^2+2*b^2/a^2/((a*d-b*c)*b)^(1/2)*arctan((d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2)*b)*c^3-c^2/a*(d
*x+c)^(1/2)/x-5*d*c^(3/2)/a*arctanh((d*x+c)^(1/2)/c^(1/2))+2*c^(5/2)/a^2*arctanh((d*x+c)^(1/2)/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)^(5/2)/x^2/(b*x+a),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 positive or negative?

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mupad [B]  time = 0.77, size = 1402, normalized size = 10.95 \begin {gather*} \frac {2\,d^2\,\sqrt {c+d\,x}}{b}+\frac {b\,c^2\,d\,\sqrt {c+d\,x}}{a\,\left (b\,c-b\,\left (c+d\,x\right )\right )}-\frac {\mathrm {atan}\left (\frac {c^3\,d^5\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b^3\,d^5+5\,a^4\,b^4\,c\,d^4-10\,a^3\,b^5\,c^2\,d^3+10\,a^2\,b^6\,c^3\,d^2-5\,a\,b^7\,c^4\,d+b^8\,c^5}\,160{}\mathrm {i}}{740\,a\,b^3\,c^5\,d^6-340\,b^4\,c^6\,d^5-128\,a^4\,c^2\,d^9+448\,a^3\,b\,c^3\,d^8+\frac {16\,a^5\,c\,d^{10}}{b}-796\,a^2\,b^2\,c^4\,d^7+\frac {60\,b^5\,c^7\,d^4}{a}}-\frac {c^2\,d^6\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b^3\,d^5+5\,a^4\,b^4\,c\,d^4-10\,a^3\,b^5\,c^2\,d^3+10\,a^2\,b^6\,c^3\,d^2-5\,a\,b^7\,c^4\,d+b^8\,c^5}\,80{}\mathrm {i}}{16\,a^4\,c\,d^{10}+740\,b^4\,c^5\,d^6-796\,a\,b^3\,c^4\,d^7-128\,a^3\,b\,c^2\,d^9+448\,a^2\,b^2\,c^3\,d^8-\frac {340\,b^5\,c^6\,d^5}{a}+\frac {60\,b^6\,c^7\,d^4}{a^2}}-\frac {c^4\,d^4\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b^3\,d^5+5\,a^4\,b^4\,c\,d^4-10\,a^3\,b^5\,c^2\,d^3+10\,a^2\,b^6\,c^3\,d^2-5\,a\,b^7\,c^4\,d+b^8\,c^5}\,60{}\mathrm {i}}{448\,a^4\,c^3\,d^8+60\,b^4\,c^7\,d^4-340\,a\,b^3\,c^6\,d^5-796\,a^3\,b\,c^4\,d^7+\frac {16\,a^6\,c\,d^{10}}{b^2}+740\,a^2\,b^2\,c^5\,d^6-\frac {128\,a^5\,c^2\,d^9}{b}}+\frac {a\,c\,d^7\,\sqrt {c+d\,x}\,\sqrt {-a^5\,b^3\,d^5+5\,a^4\,b^4\,c\,d^4-10\,a^3\,b^5\,c^2\,d^3+10\,a^2\,b^6\,c^3\,d^2-5\,a\,b^7\,c^4\,d+b^8\,c^5}\,16{}\mathrm {i}}{740\,b^5\,c^5\,d^6-796\,a\,b^4\,c^4\,d^7+448\,a^2\,b^3\,c^3\,d^8-128\,a^3\,b^2\,c^2\,d^9-\frac {340\,b^6\,c^6\,d^5}{a}+\frac {60\,b^7\,c^7\,d^4}{a^2}+16\,a^4\,b\,c\,d^{10}}\right )\,\sqrt {-b^3\,{\left (a\,d-b\,c\right )}^5}\,2{}\mathrm {i}}{a^2\,b^3}+\frac {\mathrm {atan}\left (\frac {a^3\,d^9\,\sqrt {c^3}\,\sqrt {c+d\,x}\,40{}\mathrm {i}}{40\,a^3\,c^2\,d^9-790\,b^3\,c^5\,d^6+696\,a\,b^2\,c^4\,d^7-256\,a^2\,b\,c^3\,d^8+\frac {370\,b^4\,c^6\,d^5}{a}-\frac {60\,b^5\,c^7\,d^4}{a^2}}+\frac {b^2\,c^3\,d^6\,\sqrt {c^3}\,\sqrt {c+d\,x}\,790{}\mathrm {i}}{256\,a^2\,c^3\,d^8+790\,b^2\,c^5\,d^6-\frac {370\,b^3\,c^6\,d^5}{a}-\frac {40\,a^3\,c^2\,d^9}{b}+\frac {60\,b^4\,c^7\,d^4}{a^2}-696\,a\,b\,c^4\,d^7}-\frac {b^3\,c^4\,d^5\,\sqrt {c^3}\,\sqrt {c+d\,x}\,370{}\mathrm {i}}{256\,a^3\,c^3\,d^8-370\,b^3\,c^6\,d^5+790\,a\,b^2\,c^5\,d^6-696\,a^2\,b\,c^4\,d^7+\frac {60\,b^4\,c^7\,d^4}{a}-\frac {40\,a^4\,c^2\,d^9}{b}}+\frac {b^4\,c^5\,d^4\,\sqrt {c^3}\,\sqrt {c+d\,x}\,60{}\mathrm {i}}{256\,a^4\,c^3\,d^8+60\,b^4\,c^7\,d^4-370\,a\,b^3\,c^6\,d^5-696\,a^3\,b\,c^4\,d^7+790\,a^2\,b^2\,c^5\,d^6-\frac {40\,a^5\,c^2\,d^9}{b}}+\frac {a^2\,c\,d^8\,\sqrt {c^3}\,\sqrt {c+d\,x}\,256{}\mathrm {i}}{256\,a^2\,c^3\,d^8+790\,b^2\,c^5\,d^6-\frac {370\,b^3\,c^6\,d^5}{a}-\frac {40\,a^3\,c^2\,d^9}{b}+\frac {60\,b^4\,c^7\,d^4}{a^2}-696\,a\,b\,c^4\,d^7}-\frac {a\,b\,c^2\,d^7\,\sqrt {c^3}\,\sqrt {c+d\,x}\,696{}\mathrm {i}}{256\,a^2\,c^3\,d^8+790\,b^2\,c^5\,d^6-\frac {370\,b^3\,c^6\,d^5}{a}-\frac {40\,a^3\,c^2\,d^9}{b}+\frac {60\,b^4\,c^7\,d^4}{a^2}-696\,a\,b\,c^4\,d^7}\right )\,\left (5\,a\,d-2\,b\,c\right )\,\sqrt {c^3}\,1{}\mathrm {i}}{a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(x^2*(a + b*x)),x)

[Out]

(2*d^2*(c + d*x)^(1/2))/b - (atan((c^3*d^5*(c + d*x)^(1/2)*(b^8*c^5 - a^5*b^3*d^5 + 5*a^4*b^4*c*d^4 + 10*a^2*b
^6*c^3*d^2 - 10*a^3*b^5*c^2*d^3 - 5*a*b^7*c^4*d)^(1/2)*160i)/(740*a*b^3*c^5*d^6 - 340*b^4*c^6*d^5 - 128*a^4*c^
2*d^9 + 448*a^3*b*c^3*d^8 + (16*a^5*c*d^10)/b - 796*a^2*b^2*c^4*d^7 + (60*b^5*c^7*d^4)/a) - (c^2*d^6*(c + d*x)
^(1/2)*(b^8*c^5 - a^5*b^3*d^5 + 5*a^4*b^4*c*d^4 + 10*a^2*b^6*c^3*d^2 - 10*a^3*b^5*c^2*d^3 - 5*a*b^7*c^4*d)^(1/
2)*80i)/(16*a^4*c*d^10 + 740*b^4*c^5*d^6 - 796*a*b^3*c^4*d^7 - 128*a^3*b*c^2*d^9 + 448*a^2*b^2*c^3*d^8 - (340*
b^5*c^6*d^5)/a + (60*b^6*c^7*d^4)/a^2) - (c^4*d^4*(c + d*x)^(1/2)*(b^8*c^5 - a^5*b^3*d^5 + 5*a^4*b^4*c*d^4 + 1
0*a^2*b^6*c^3*d^2 - 10*a^3*b^5*c^2*d^3 - 5*a*b^7*c^4*d)^(1/2)*60i)/(448*a^4*c^3*d^8 + 60*b^4*c^7*d^4 - 340*a*b
^3*c^6*d^5 - 796*a^3*b*c^4*d^7 + (16*a^6*c*d^10)/b^2 + 740*a^2*b^2*c^5*d^6 - (128*a^5*c^2*d^9)/b) + (a*c*d^7*(
c + d*x)^(1/2)*(b^8*c^5 - a^5*b^3*d^5 + 5*a^4*b^4*c*d^4 + 10*a^2*b^6*c^3*d^2 - 10*a^3*b^5*c^2*d^3 - 5*a*b^7*c^
4*d)^(1/2)*16i)/(740*b^5*c^5*d^6 - 796*a*b^4*c^4*d^7 + 448*a^2*b^3*c^3*d^8 - 128*a^3*b^2*c^2*d^9 - (340*b^6*c^
6*d^5)/a + (60*b^7*c^7*d^4)/a^2 + 16*a^4*b*c*d^10))*(-b^3*(a*d - b*c)^5)^(1/2)*2i)/(a^2*b^3) + (atan((a^3*d^9*
(c^3)^(1/2)*(c + d*x)^(1/2)*40i)/(40*a^3*c^2*d^9 - 790*b^3*c^5*d^6 + 696*a*b^2*c^4*d^7 - 256*a^2*b*c^3*d^8 + (
370*b^4*c^6*d^5)/a - (60*b^5*c^7*d^4)/a^2) + (b^2*c^3*d^6*(c^3)^(1/2)*(c + d*x)^(1/2)*790i)/(256*a^2*c^3*d^8 +
790*b^2*c^5*d^6 - (370*b^3*c^6*d^5)/a - (40*a^3*c^2*d^9)/b + (60*b^4*c^7*d^4)/a^2 - 696*a*b*c^4*d^7) - (b^3*c
^4*d^5*(c^3)^(1/2)*(c + d*x)^(1/2)*370i)/(256*a^3*c^3*d^8 - 370*b^3*c^6*d^5 + 790*a*b^2*c^5*d^6 - 696*a^2*b*c^
4*d^7 + (60*b^4*c^7*d^4)/a - (40*a^4*c^2*d^9)/b) + (b^4*c^5*d^4*(c^3)^(1/2)*(c + d*x)^(1/2)*60i)/(256*a^4*c^3*
d^8 + 60*b^4*c^7*d^4 - 370*a*b^3*c^6*d^5 - 696*a^3*b*c^4*d^7 + 790*a^2*b^2*c^5*d^6 - (40*a^5*c^2*d^9)/b) + (a^
2*c*d^8*(c^3)^(1/2)*(c + d*x)^(1/2)*256i)/(256*a^2*c^3*d^8 + 790*b^2*c^5*d^6 - (370*b^3*c^6*d^5)/a - (40*a^3*c
^2*d^9)/b + (60*b^4*c^7*d^4)/a^2 - 696*a*b*c^4*d^7) - (a*b*c^2*d^7*(c^3)^(1/2)*(c + d*x)^(1/2)*696i)/(256*a^2*
c^3*d^8 + 790*b^2*c^5*d^6 - (370*b^3*c^6*d^5)/a - (40*a^3*c^2*d^9)/b + (60*b^4*c^7*d^4)/a^2 - 696*a*b*c^4*d^7)
)*(5*a*d - 2*b*c)*(c^3)^(1/2)*1i)/a^2 + (b*c^2*d*(c + d*x)^(1/2))/(a*(b*c - b*(c + d*x)))

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sympy [B]  time = 108.41, size = 333, normalized size = 2.60 \begin {gather*} - \frac {2 a d^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d}{b} - c}} \right )}}{b^{2} \sqrt {\frac {a d}{b} - c}} + \frac {6 c d^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d}{b} - c}} \right )}}{b \sqrt {\frac {a d}{b} - c}} + \frac {2 d^{2} \sqrt {c + d x}}{b} - \frac {c^{3} d \sqrt {\frac {1}{c^{3}}} \log {\left (- c^{2} \sqrt {\frac {1}{c^{3}}} + \sqrt {c + d x} \right )}}{2 a} + \frac {c^{3} d \sqrt {\frac {1}{c^{3}}} \log {\left (c^{2} \sqrt {\frac {1}{c^{3}}} + \sqrt {c + d x} \right )}}{2 a} - \frac {6 c^{2} d \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d}{b} - c}} \right )}}{a \sqrt {\frac {a d}{b} - c}} + \frac {6 c^{2} d \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{a \sqrt {- c}} - \frac {c^{2} \sqrt {c + d x}}{a x} + \frac {2 b c^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d}{b} - c}} \right )}}{a^{2} \sqrt {\frac {a d}{b} - c}} - \frac {2 b c^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{a^{2} \sqrt {- c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(5/2)/x**2/(b*x+a),x)

[Out]

-2*a*d**3*atan(sqrt(c + d*x)/sqrt(a*d/b - c))/(b**2*sqrt(a*d/b - c)) + 6*c*d**2*atan(sqrt(c + d*x)/sqrt(a*d/b
- c))/(b*sqrt(a*d/b - c)) + 2*d**2*sqrt(c + d*x)/b - c**3*d*sqrt(c**(-3))*log(-c**2*sqrt(c**(-3)) + sqrt(c + d
*x))/(2*a) + c**3*d*sqrt(c**(-3))*log(c**2*sqrt(c**(-3)) + sqrt(c + d*x))/(2*a) - 6*c**2*d*atan(sqrt(c + d*x)/
sqrt(a*d/b - c))/(a*sqrt(a*d/b - c)) + 6*c**2*d*atan(sqrt(c + d*x)/sqrt(-c))/(a*sqrt(-c)) - c**2*sqrt(c + d*x)
/(a*x) + 2*b*c**3*atan(sqrt(c + d*x)/sqrt(a*d/b - c))/(a**2*sqrt(a*d/b - c)) - 2*b*c**3*atan(sqrt(c + d*x)/sqr
t(-c))/(a**2*sqrt(-c))

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