\(\int \frac {(d+e x)^{3/2}}{(c d^2-c e^2 x^2)^{5/2}} \, dx\) [200]

Optimal result

Integrand size = 29, antiderivative size = 153 \[ \int \frac {(d+e x)^{3/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {d+e x}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}+\frac {d+3 e x}{6 c^2 d^2 e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-c e^2 x^2}}\right )}{2 \sqrt {2} c^{5/2} d^{5/2} e} \] Output:

2/3*(e*x+d)^(1/2)/c/e/(-c*e^2*x^2+c*d^2)^(3/2)+1/6*(3*e*x+d)/c^2/d^2/e/(e* 
x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(1/2)-1/4*arctanh(2^(1/2)*c^(1/2)*d^(1/2)*(e 
*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(1/2))*2^(1/2)/c^(5/2)/d^(5/2)/e
 

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.99 \[ \int \frac {(d+e x)^{3/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {d} \left (5 d^2+2 d e x-3 e^2 x^2\right )-3 \sqrt {2} (d-e x) \sqrt {d+e x} \sqrt {d^2-e^2 x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{12 c^2 d^{5/2} e (d-e x) \sqrt {d+e x} \sqrt {c \left (d^2-e^2 x^2\right )}} \] Input:

Integrate[(d + e*x)^(3/2)/(c*d^2 - c*e^2*x^2)^(5/2),x]
 

Output:

(2*Sqrt[d]*(5*d^2 + 2*d*e*x - 3*e^2*x^2) - 3*Sqrt[2]*(d - e*x)*Sqrt[d + e* 
x]*Sqrt[d^2 - e^2*x^2]*ArcTanh[(Sqrt[2]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[d^2 - 
e^2*x^2]])/(12*c^2*d^(5/2)*e*(d - e*x)*Sqrt[d + e*x]*Sqrt[c*(d^2 - e^2*x^2 
)])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.30, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {468, 470, 467, 471, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(d+e x)^{3/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 468

\(\displaystyle \frac {2 \int \frac {1}{\sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}dx}{3 c}+\frac {2 \sqrt {d+e x}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {2 \left (\frac {3 \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}}dx}{4 d}-\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}\right )}{3 c}+\frac {2 \sqrt {d+e x}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 467

\(\displaystyle \frac {2 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}dx}{2 c d}+\frac {\sqrt {d+e x}}{c d e \sqrt {c d^2-c e^2 x^2}}\right )}{4 d}-\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}\right )}{3 c}+\frac {2 \sqrt {d+e x}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {2 \left (\frac {3 \left (\frac {e \int \frac {1}{\frac {e^2 \left (c d^2-c e^2 x^2\right )}{d+e x}-2 c d e^2}d\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}}}{c d}+\frac {\sqrt {d+e x}}{c d e \sqrt {c d^2-c e^2 x^2}}\right )}{4 d}-\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}\right )}{3 c}+\frac {2 \sqrt {d+e x}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \left (\frac {3 \left (\frac {\sqrt {d+e x}}{c d e \sqrt {c d^2-c e^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {2} c^{3/2} d^{3/2} e}\right )}{4 d}-\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}\right )}{3 c}+\frac {2 \sqrt {d+e x}}{3 c e \left (c d^2-c e^2 x^2\right )^{3/2}}\)

Input:

Int[(d + e*x)^(3/2)/(c*d^2 - c*e^2*x^2)^(5/2),x]
 

Output:

(2*Sqrt[d + e*x])/(3*c*e*(c*d^2 - c*e^2*x^2)^(3/2)) + (2*(-1/2*1/(c*d*e*Sq 
rt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2]) + (3*(Sqrt[d + e*x]/(c*d*e*Sqrt[c*d^2 
 - c*e^2*x^2]) - ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]* 
Sqrt[d + e*x])]/(Sqrt[2]*c^(3/2)*d^(3/2)*e)))/(4*d)))/(3*c)
 

Defintions of rubi rules used

rule 221

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]
 

rule 467

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-c)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*a*d*(p + 1))), x] + Simp[c*((n + 2 
*p + 2)/(2*a*(p + 1)))   Int[(c + d*x)^(n - 1)*(a + b*x^2)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[p, -1] && LtQ[0, 
n, 1] && IntegerQ[2*p]
 

rule 468

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] - Simp[d^2*((n + 
p)/(b*(p + 1)))   Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1), x], x] /; Free 
Q[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[p, -1] && GtQ[n, 1] && I 
ntegerQ[2*p]
 

rule 470

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 
2*p + 2)/(2*c*(n + p + 1))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; 
 FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + 
 p + 1, 0] && IntegerQ[2*p]
 

rule 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {\sqrt {c \left (-e^{2} x^{2}+d^{2}\right )}\, \left (-3 \sqrt {2}\, \sqrt {c \left (-e x +d \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) e x +3 \sqrt {2}\, \sqrt {c \left (-e x +d \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) d +6 \sqrt {c d}\, e x -10 \sqrt {c d}\, d \right )}{12 c^{3} \sqrt {e x +d}\, \left (-e x +d \right )^{2} e \,d^{2} \sqrt {c d}}\)	\(141\)

Input:

int((e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/12*(c*(-e^2*x^2+d^2))^(1/2)/c^3*(-3*2^(1/2)*(c*(-e*x+d))^(1/2)*arctanh( 
1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*e*x+3*2^(1/2)*(c*(-e*x+d))^(1/ 
2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*d+6*(c*d)^(1/2)*e*x 
-10*(c*d)^(1/2)*d)/(e*x+d)^(1/2)/(-e*x+d)^2/e/d^2/(c*d)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 381, normalized size of antiderivative = 2.49 \[ \int \frac {(d+e x)^{3/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (e^{3} x^{3} - d e^{2} x^{2} - d^{2} e x + d^{3}\right )} \sqrt {c d} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 4 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (3 \, d e x - 5 \, d^{2}\right )} \sqrt {e x + d}}{24 \, {\left (c^{3} d^{3} e^{4} x^{3} - c^{3} d^{4} e^{3} x^{2} - c^{3} d^{5} e^{2} x + c^{3} d^{6} e\right )}}, \frac {3 \, \sqrt {2} {\left (e^{3} x^{3} - d e^{2} x^{2} - d^{2} e x + d^{3}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{2 \, {\left (c d e x + c d^{2}\right )}}\right ) - 2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (3 \, d e x - 5 \, d^{2}\right )} \sqrt {e x + d}}{12 \, {\left (c^{3} d^{3} e^{4} x^{3} - c^{3} d^{4} e^{3} x^{2} - c^{3} d^{5} e^{2} x + c^{3} d^{6} e\right )}}\right ] \] Input:

integrate((e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(5/2),x, algorithm="fricas")
 

Output:

[1/24*(3*sqrt(2)*(e^3*x^3 - d*e^2*x^2 - d^2*e*x + d^3)*sqrt(c*d)*log(-(c*e 
^2*x^2 - 2*c*d*e*x - 3*c*d^2 + 2*sqrt(2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(c*d 
)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) - 4*sqrt(-c*e^2*x^2 + c*d^2)*( 
3*d*e*x - 5*d^2)*sqrt(e*x + d))/(c^3*d^3*e^4*x^3 - c^3*d^4*e^3*x^2 - c^3*d 
^5*e^2*x + c^3*d^6*e), 1/12*(3*sqrt(2)*(e^3*x^3 - d*e^2*x^2 - d^2*e*x + d^ 
3)*sqrt(-c*d)*arctan(1/2*sqrt(2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(-c*d)*sqrt( 
e*x + d)/(c*d*e*x + c*d^2)) - 2*sqrt(-c*e^2*x^2 + c*d^2)*(3*d*e*x - 5*d^2) 
*sqrt(e*x + d))/(c^3*d^3*e^4*x^3 - c^3*d^4*e^3*x^2 - c^3*d^5*e^2*x + c^3*d 
^6*e)]
 

Sympy [F]

\[ \int \frac {(d+e x)^{3/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {3}{2}}}{\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((e*x+d)**(3/2)/(-c*e**2*x**2+c*d**2)**(5/2),x)
 

Output:

Integral((d + e*x)**(3/2)/(-c*(-d + e*x)*(d + e*x))**(5/2), x)
 

Maxima [F]

\[ \int \frac {(d+e x)^{3/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {3}{2}}}{{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate((e*x + d)^(3/2)/(-c*e^2*x^2 + c*d^2)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.65 \[ \int \frac {(d+e x)^{3/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\frac {\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d} c^{2} d^{2}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )} c - 8 \, c d\right )}}{{\left ({\left (e x + d\right )} c - 2 \, c d\right )} \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{2} d^{2}}}{12 \, e} \] Input:

integrate((e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(5/2),x, algorithm="giac")
 

Output:

1/12*(3*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c*d))/ 
(sqrt(-c*d)*c^2*d^2) + 2*(3*(e*x + d)*c - 8*c*d)/(((e*x + d)*c - 2*c*d)*sq 
rt(-(e*x + d)*c + 2*c*d)*c^2*d^2))/e
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{3/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^{3/2}}{{\left (c\,d^2-c\,e^2\,x^2\right )}^{5/2}} \,d x \] Input:

int((d + e*x)^(3/2)/(c*d^2 - c*e^2*x^2)^(5/2),x)
 

Output:

int((d + e*x)^(3/2)/(c*d^2 - c*e^2*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.03 \[ \int \frac {(d+e x)^{3/2}}{\left (c d^2-c e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (3 \sqrt {d}\, \sqrt {-e x +d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}-\sqrt {d}\, \sqrt {2}\right ) d -3 \sqrt {d}\, \sqrt {-e x +d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}-\sqrt {d}\, \sqrt {2}\right ) e x -3 \sqrt {d}\, \sqrt {-e x +d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}+\sqrt {d}\, \sqrt {2}\right ) d +3 \sqrt {d}\, \sqrt {-e x +d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}+\sqrt {d}\, \sqrt {2}\right ) e x +20 d^{2}-12 d e x \right )}{24 \sqrt {-e x +d}\, c^{3} d^{3} e \left (-e x +d \right )} \] Input:

int((e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(5/2),x)
 

Output:

(sqrt(c)*(3*sqrt(d)*sqrt(d - e*x)*sqrt(2)*log(sqrt(d - e*x) - sqrt(d)*sqrt 
(2))*d - 3*sqrt(d)*sqrt(d - e*x)*sqrt(2)*log(sqrt(d - e*x) - sqrt(d)*sqrt( 
2))*e*x - 3*sqrt(d)*sqrt(d - e*x)*sqrt(2)*log(sqrt(d - e*x) + sqrt(d)*sqrt 
(2))*d + 3*sqrt(d)*sqrt(d - e*x)*sqrt(2)*log(sqrt(d - e*x) + sqrt(d)*sqrt( 
2))*e*x + 20*d**2 - 12*d*e*x))/(24*sqrt(d - e*x)*c**3*d**3*e*(d - e*x))