3.15.6 \(\int \frac {(c+d x)^{5/2}}{(a+b x)^2} \, dx\) [1406]

Optimal. Leaf size=110 \[ \frac {5 d (b c-a d) \sqrt {c+d x}}{b^3}+\frac {5 d (c+d x)^{3/2}}{3 b^2}-\frac {(c+d x)^{5/2}}{b (a+b x)}-\frac {5 d (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{7/2}} \]

[Out]

5/3*d*(d*x+c)^(3/2)/b^2-(d*x+c)^(5/2)/b/(b*x+a)-5*d*(-a*d+b*c)^(3/2)*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^
(1/2))/b^(7/2)+5*d*(-a*d+b*c)*(d*x+c)^(1/2)/b^3

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Rubi [A]
time = 0.04, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {43, 52, 65, 214} \begin {gather*} -\frac {5 d (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{7/2}}+\frac {5 d \sqrt {c+d x} (b c-a d)}{b^3}-\frac {(c+d x)^{5/2}}{b (a+b x)}+\frac {5 d (c+d x)^{3/2}}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(a + b*x)^2,x]

[Out]

(5*d*(b*c - a*d)*Sqrt[c + d*x])/b^3 + (5*d*(c + d*x)^(3/2))/(3*b^2) - (c + d*x)^(5/2)/(b*(a + b*x)) - (5*d*(b*
c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/b^(7/2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 214

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

Rubi steps

\begin {align*} \int \frac {(c+d x)^{5/2}}{(a+b x)^2} \, dx &=-\frac {(c+d x)^{5/2}}{b (a+b x)}+\frac {(5 d) \int \frac {(c+d x)^{3/2}}{a+b x} \, dx}{2 b}\\ &=\frac {5 d (c+d x)^{3/2}}{3 b^2}-\frac {(c+d x)^{5/2}}{b (a+b x)}+\frac {(5 d (b c-a d)) \int \frac {\sqrt {c+d x}}{a+b x} \, dx}{2 b^2}\\ &=\frac {5 d (b c-a d) \sqrt {c+d x}}{b^3}+\frac {5 d (c+d x)^{3/2}}{3 b^2}-\frac {(c+d x)^{5/2}}{b (a+b x)}+\frac {\left (5 d (b c-a d)^2\right ) \int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx}{2 b^3}\\ &=\frac {5 d (b c-a d) \sqrt {c+d x}}{b^3}+\frac {5 d (c+d x)^{3/2}}{3 b^2}-\frac {(c+d x)^{5/2}}{b (a+b x)}+\frac {\left (5 (b c-a d)^2\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{b^3}\\ &=\frac {5 d (b c-a d) \sqrt {c+d x}}{b^3}+\frac {5 d (c+d x)^{3/2}}{3 b^2}-\frac {(c+d x)^{5/2}}{b (a+b x)}-\frac {5 d (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 116, normalized size = 1.05 \begin {gather*} -\frac {\sqrt {c+d x} \left (15 a^2 d^2+10 a b d (-2 c+d x)+b^2 \left (3 c^2-14 c d x-2 d^2 x^2\right )\right )}{3 b^3 (a+b x)}+\frac {5 d (-b c+a d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Sqrt[c + d*x]*(15*a^2*d^2 + 10*a*b*d*(-2*c + d*x) + b^2*(3*c^2 - 14*c*d*x - 2*d^2*x^2)))/(b^3*(a + b*x))
+ (5*d*(-(b*c) + a*d)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/b^(7/2)

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Maple [A]
time = 0.19, size = 152, normalized size = 1.38

method	result	size
derivativedivides	\(2 d \left (-\frac {-\frac {b \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a d \sqrt {d x +c}-2 b c \sqrt {d x +c}}{b^{3}}+\frac {\frac {\left (-\frac {1}{2} a^{2} d^{2}+a b c d -\frac {1}{2} b^{2} c^{2}\right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {5 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}}{b^{3}}\right )\)	\(152\)
default	\(2 d \left (-\frac {-\frac {b \left (d x +c \right )^{\frac {3}{2}}}{3}+2 a d \sqrt {d x +c}-2 b c \sqrt {d x +c}}{b^{3}}+\frac {\frac {\left (-\frac {1}{2} a^{2} d^{2}+a b c d -\frac {1}{2} b^{2} c^{2}\right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {5 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}}{b^{3}}\right )\)	\(152\)
risch	\(-\frac {2 d \left (-b d x +6 a d -7 b c \right ) \sqrt {d x +c}}{3 b^{3}}-\frac {d^{3} \sqrt {d x +c}\, a^{2}}{b^{3} \left (b d x +a d \right )}+\frac {2 d^{2} \sqrt {d x +c}\, a c}{b^{2} \left (b d x +a d \right )}-\frac {d \sqrt {d x +c}\, c^{2}}{b \left (b d x +a d \right )}+\frac {5 d^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right ) a^{2}}{b^{3} \sqrt {\left (a d -b c \right ) b}}-\frac {10 d^{2} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right ) a c}{b^{2} \sqrt {\left (a d -b c \right ) b}}+\frac {5 d \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right ) c^{2}}{b \sqrt {\left (a d -b c \right ) b}}\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*d*(-1/b^3*(-1/3*b*(d*x+c)^(3/2)+2*a*d*(d*x+c)^(1/2)-2*b*c*(d*x+c)^(1/2))+1/b^3*((-1/2*a^2*d^2+a*b*c*d-1/2*b^
2*c^2)*(d*x+c)^(1/2)/((d*x+c)*b+a*d-b*c)+5/2*(a^2*d^2-2*a*b*c*d+b^2*c^2)/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^
(1/2)/((a*d-b*c)*b)^(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)^(5/2)/(b*x+a)^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 = 0.42, size = 330, normalized size = 3.00 \begin {gather*} \left [-\frac {15 \, {\left (a b c d - a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x\right )} \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 ) - 2 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c^{2} + 20 \, a b c d - 15 \, a^{2} d^{2} + 2 \, {\left (7 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, -\frac {15 \, {\left (a b c d - a^{2} d^{2} + {\left (b^{2} c d - a b d^{2}\right )} x\right )} \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} d^{2} x^{2} - 3 \, b^{2} c^{2} + 20 \, a b c d - 15 \, a^{2} d^{2} + 2 \, {\left (7 \, b^{2} c d - 5 \, a b d^{2}\right )} x\right )} \sqrt {d x + c}}{3 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(15*(a*b*c*d - a^2*d^2 + (b^2*c*d - a*b*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*(2*b^2*d^2*x^2 - 3*b^2*c^2 + 20*a*b*c*d - 15*a^2*d^2 + 2*(7*b^2*c
*d - 5*a*b*d^2)*x)*sqrt(d*x + c))/(b^4*x + a*b^3), -1/3*(15*(a*b*c*d - a^2*d^2 + (b^2*c*d - a*b*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*d^2*x^2 - 3*b^2*c^2 + 20*a*b
*c*d - 15*a^2*d^2 + 2*(7*b^2*c*d - 5*a*b*d^2)*x)*sqrt(d*x + c))/(b^4*x + a*b^3)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1312 vs. \(2 (97) = 194\).
time = 124.93, size = 1312, normalized size = 11.93 \begin {gather*} - \frac {2 a^{3} d^{4} \sqrt {c + d x}}{2 a^{2} b^{3} d^{2} - 2 a b^{4} c d + 2 a b^{4} d^{2} x - 2 b^{5} c d x} + \frac {a^{3} d^{4} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (- a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2 b^{3}} - \frac {a^{3} d^{4} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2 b^{3}} + \frac {6 a^{2} c d^{3} \sqrt {c + d x}}{2 a^{2} b^{2} d^{2} - 2 a b^{3} c d + 2 a b^{3} d^{2} x - 2 b^{4} c d x} - \frac {3 a^{2} c d^{3} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (- a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2 b^{2}} + \frac {3 a^{2} c d^{3} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2 b^{2}} + \frac {6 a^{2} d^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d}{b} - c}} \right )}}{b^{4} \sqrt {\frac {a d}{b} - c}} - \frac {6 a c^{2} d^{2} \sqrt {c + d x}}{2 a^{2} b d^{2} - 2 a b^{2} c d + 2 a b^{2} d^{2} x - 2 b^{3} c d x} + \frac {3 a c^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (- a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2 b} - \frac {3 a c^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2 b} - \frac {12 a c d^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d}{b} - c}} \right )}}{b^{3} \sqrt {\frac {a d}{b} - c}} - \frac {4 a d^{2} \sqrt {c + d x}}{b^{3}} - \frac {c^{3} d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (- a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2} + \frac {c^{3} d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} \log {\left (a^{2} d^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} - 2 a b c d \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + b^{2} c^{2} \sqrt {- \frac {1}{b \left (a d - b c\right )^{3}}} + \sqrt {c + d x} \right )}}{2} + \frac {2 c^{3} d \sqrt {c + d x}}{2 a^{2} d^{2} - 2 a b c d + 2 a b d^{2} x - 2 b^{2} c d x} + \frac {6 c^{2} d \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {\frac {a d}{b} - c}} \right )}}{b^{2} \sqrt {\frac {a d}{b} - c}} + \frac {4 c d \sqrt {c + d x}}{b^{2}} + \frac {2 d \left (c + d x\right )^{\frac {3}{2}}}{3 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**3*d**4*sqrt(c + d*x)/(2*a**2*b**3*d**2 - 2*a*b**4*c*d + 2*a*b**4*d**2*x - 2*b**5*c*d*x) + a**3*d**4*sqrt
(-1/(b*(a*d - b*c)**3))*log(-a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) + 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) - b
**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/(2*b**3) - a**3*d**4*sqrt(-1/(b*(a*d - b*c)**3))*log(a**
2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) - 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) + b**2*c**2*sqrt(-1/(b*(a*d - b*c)*
*3)) + sqrt(c + d*x))/(2*b**3) + 6*a**2*c*d**3*sqrt(c + d*x)/(2*a**2*b**2*d**2 - 2*a*b**3*c*d + 2*a*b**3*d**2*
x - 2*b**4*c*d*x) - 3*a**2*c*d**3*sqrt(-1/(b*(a*d - b*c)**3))*log(-a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) + 2*a
*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) - b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/(2*b**2) + 3*a**2*
c*d**3*sqrt(-1/(b*(a*d - b*c)**3))*log(a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) - 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c
)**3)) + b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/(2*b**2) + 6*a**2*d**3*atan(sqrt(c + d*x)/sqrt
(a*d/b - c))/(b**4*sqrt(a*d/b - c)) - 6*a*c**2*d**2*sqrt(c + d*x)/(2*a**2*b*d**2 - 2*a*b**2*c*d + 2*a*b**2*d**
2*x - 2*b**3*c*d*x) + 3*a*c**2*d**2*sqrt(-1/(b*(a*d - b*c)**3))*log(-a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) + 2
*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) - b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/(2*b) - 3*a*c**2
*d**2*sqrt(-1/(b*(a*d - b*c)**3))*log(a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) - 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)
**3)) + b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/(2*b) - 12*a*c*d**2*atan(sqrt(c + d*x)/sqrt(a*d
/b - c))/(b**3*sqrt(a*d/b - c)) - 4*a*d**2*sqrt(c + d*x)/b**3 - c**3*d*sqrt(-1/(b*(a*d - b*c)**3))*log(-a**2*d
**2*sqrt(-1/(b*(a*d - b*c)**3)) + 2*a*b*c*d*sqrt(-1/(b*(a*d - b*c)**3)) - b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)
) + sqrt(c + d*x))/2 + c**3*d*sqrt(-1/(b*(a*d - b*c)**3))*log(a**2*d**2*sqrt(-1/(b*(a*d - b*c)**3)) - 2*a*b*c*
d*sqrt(-1/(b*(a*d - b*c)**3)) + b**2*c**2*sqrt(-1/(b*(a*d - b*c)**3)) + sqrt(c + d*x))/2 + 2*c**3*d*sqrt(c + d
*x)/(2*a**2*d**2 - 2*a*b*c*d + 2*a*b*d**2*x - 2*b**2*c*d*x) + 6*c**2*d*atan(sqrt(c + d*x)/sqrt(a*d/b - c))/(b*
*2*sqrt(a*d/b - c)) + 4*c*d*sqrt(c + d*x)/b**2 + 2*d*(c + d*x)**(3/2)/(3*b**2)

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Giac [A]
time = 1.95, size = 181, normalized size = 1.65 \begin {gather*} \frac {5 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} 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} b^{3}} - \frac {\sqrt {d x + c} b^{2} c^{2} d - 2 \, \sqrt {d x + c} a b c d^{2} + \sqrt {d x + c} a^{2} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{3}} + \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {3}{2}} b^{4} d + 6 \, \sqrt {d x + c} b^{4} c d - 6 \, \sqrt {d x + c} a b^{3} d^{2}\right )}}{3 \, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^3)
- (sqrt(d*x + c)*b^2*c^2*d - 2*sqrt(d*x + c)*a*b*c*d^2 + sqrt(d*x + c)*a^2*d^3)/(((d*x + c)*b - b*c + a*d)*b^3
) + 2/3*((d*x + c)^(3/2)*b^4*d + 6*sqrt(d*x + c)*b^4*c*d - 6*sqrt(d*x + c)*a*b^3*d^2)/b^6

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Mupad [B]
time = 0.12, size = 161, normalized size = 1.46 \begin {gather*} \frac {2\,d\,{\left (c+d\,x\right )}^{3/2}}{3\,b^2}-\frac {\sqrt {c+d\,x}\,\left (a^2\,d^3-2\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{b^4\,\left (c+d\,x\right )-b^4\,c+a\,b^3\,d}+\frac {5\,d\,\mathrm {atan}\left (\frac {\sqrt {b}\,d\,{\left (a\,d-b\,c\right )}^{3/2}\,\sqrt {c+d\,x}}{a^2\,d^3-2\,a\,b\,c\,d^2+b^2\,c^2\,d}\right )\,{\left (a\,d-b\,c\right )}^{3/2}}{b^{7/2}}+\frac {2\,d\,\left (2\,b^2\,c-2\,a\,b\,d\right )\,\sqrt {c+d\,x}}{b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(5/2)/(a + b*x)^2,x)

[Out]

(2*d*(c + d*x)^(3/2))/(3*b^2) - ((c + d*x)^(1/2)*(a^2*d^3 + b^2*c^2*d - 2*a*b*c*d^2))/(b^4*(c + d*x) - b^4*c +
a*b^3*d) + (5*d*atan((b^(1/2)*d*(a*d - b*c)^(3/2)*(c + d*x)^(1/2))/(a^2*d^3 + b^2*c^2*d - 2*a*b*c*d^2))*(a*d
- b*c)^(3/2))/b^(7/2) + (2*d*(2*b^2*c - 2*a*b*d)*(c + d*x)^(1/2))/b^4

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