\(\int \frac {A+B x}{(c+d x)^{3/2} (c^2-d^2 x^2)^{5/2}} \, dx\) [113]

Optimal result

Integrand size = 31, antiderivative size = 204 \[ \int \frac {A+B x}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {B c-A d}{6 c d^2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}+\frac {(B c+3 A d) (c+7 d x)}{96 c^3 d^2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}+\frac {35 (B c+3 A d) (c+3 d x)}{768 c^5 d^2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}-\frac {35 (B c+3 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{256 \sqrt {2} c^{11/2} d^2} \] Output:

1/6*(-A*d+B*c)/c/d^2/(d*x+c)^(3/2)/(-d^2*x^2+c^2)^(3/2)+1/96*(3*A*d+B*c)*( 
7*d*x+c)/c^3/d^2/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(3/2)+35/768*(3*A*d+B*c)*(3* 
d*x+c)/c^5/d^2/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2)-35/512*(3*A*d+B*c)*arcta 
nh(2^(1/2)*c^(1/2)*(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2))*2^(1/2)/c^(11/2)/d^ 
2
 

Mathematica [A] (verified)

Time = 2.35 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (A d \left (c^4+612 c^3 d x+378 c^2 d^2 x^2-420 c d^3 x^3-315 d^4 x^4\right )+B c \left (171 c^4+204 c^3 d x+126 c^2 d^2 x^2-140 c d^3 x^3-105 d^4 x^4\right )\right )}{(c-d x)^2 (c+d x)^{7/2}}-105 \sqrt {2} (B c+3 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{1536 c^{11/2} d^2} \] Input:

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

Output:

((2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(A*d*(c^4 + 612*c^3*d*x + 378*c^2*d^2*x^2 
- 420*c*d^3*x^3 - 315*d^4*x^4) + B*c*(171*c^4 + 204*c^3*d*x + 126*c^2*d^2* 
x^2 - 140*c*d^3*x^3 - 105*d^4*x^4)))/((c - d*x)^2*(c + d*x)^(7/2)) - 105*S 
qrt[2]*(B*c + 3*A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^ 
2*x^2]])/(1536*c^(11/2)*d^2)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.35, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {671, 470, 467, 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.

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {(3 A d+B c) \int \frac {1}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}dx}{4 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(3 A d+B c) \left (\frac {7 \int \frac {\sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{5/2}}dx}{8 c}-\frac {1}{4 c d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\right )}{4 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 467

\(\displaystyle \frac {(3 A d+B c) \left (\frac {7 \left (\frac {5 \int \frac {1}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}dx}{6 c}+\frac {\sqrt {c+d x}}{3 c d \left (c^2-d^2 x^2\right )^{3/2}}\right )}{8 c}-\frac {1}{4 c d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\right )}{4 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(3 A d+B c) \left (\frac {7 \left (\frac {5 \left (\frac {3 \int \frac {\sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c}+\frac {\sqrt {c+d x}}{3 c d \left (c^2-d^2 x^2\right )^{3/2}}\right )}{8 c}-\frac {1}{4 c d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\right )}{4 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 467

\(\displaystyle \frac {(3 A d+B c) \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{2 c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c}+\frac {\sqrt {c+d x}}{3 c d \left (c^2-d^2 x^2\right )^{3/2}}\right )}{8 c}-\frac {1}{4 c d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\right )}{4 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(3 A d+B c) \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c}+\frac {\sqrt {c+d x}}{3 c d \left (c^2-d^2 x^2\right )^{3/2}}\right )}{8 c}-\frac {1}{4 c d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\right )}{4 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(3 A d+B c) \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {2} c^{3/2} d}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c}+\frac {\sqrt {c+d x}}{3 c d \left (c^2-d^2 x^2\right )^{3/2}}\right )}{8 c}-\frac {1}{4 c d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}\right )}{4 c d}+\frac {B c-A d}{6 c d^2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}\)

Input:

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

Output:

(B*c - A*d)/(6*c*d^2*(c + d*x)^(3/2)*(c^2 - d^2*x^2)^(3/2)) + ((B*c + 3*A* 
d)*(-1/4*1/(c*d*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(3/2)) + (7*(Sqrt[c + d*x]/( 
3*c*d*(c^2 - d^2*x^2)^(3/2)) + (5*(-1/2*1/(c*d*Sqrt[c + d*x]*Sqrt[c^2 - d^ 
2*x^2]) + (3*(Sqrt[c + d*x]/(c*d*Sqrt[c^2 - d^2*x^2]) - ArcTanh[Sqrt[c^2 - 
 d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])]/(Sqrt[2]*c^(3/2)*d)))/(4*c)))/( 
6*c)))/(8*c)))/(4*c*d)
 

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 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]
 

rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m 
 + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m 
+ p + 1))   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, 
e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p 
 + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p 
 + 1, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(437\) vs. \(2(172)=344\).

Time = 0.36 (sec) , antiderivative size = 438, normalized size of antiderivative = 2.15

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (-315 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{5} x^{4} \sqrt {-d x +c}-105 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{4} x^{4} \sqrt {-d x +c}-630 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{4} x^{3} \sqrt {-d x +c}-210 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{3} x^{3} \sqrt {-d x +c}+630 A \sqrt {c}\, d^{5} x^{4}+210 B \,c^{\frac {3}{2}} d^{4} x^{4}+630 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{2} x \sqrt {-d x +c}+840 A \,c^{\frac {3}{2}} d^{4} x^{3}+210 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d x \sqrt {-d x +c}+280 B \,c^{\frac {5}{2}} d^{3} x^{3}+315 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d \sqrt {-d x +c}-756 A \,c^{\frac {5}{2}} d^{3} x^{2}+105 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5} \sqrt {-d x +c}-252 B \,c^{\frac {7}{2}} d^{2} x^{2}-1224 A \,c^{\frac {7}{2}} d^{2} x -408 B \,c^{\frac {9}{2}} d x -2 A \,c^{\frac {9}{2}} d -342 B \,c^{\frac {11}{2}}\right )}{1536 \left (d x +c \right )^{\frac {7}{2}} \left (-d x +c \right )^{2} d^{2} c^{\frac {11}{2}}}\)	\(438\)

Input:

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

Output:

-1/1536*(-d^2*x^2+c^2)^(1/2)*(-315*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^ 
(1/2)/c^(1/2))*d^5*x^4*(-d*x+c)^(1/2)-105*B*2^(1/2)*arctanh(1/2*(-d*x+c)^( 
1/2)*2^(1/2)/c^(1/2))*c*d^4*x^4*(-d*x+c)^(1/2)-630*A*2^(1/2)*arctanh(1/2*( 
-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^4*x^3*(-d*x+c)^(1/2)-210*B*2^(1/2)*arct 
anh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^3*x^3*(-d*x+c)^(1/2)+630*A*c 
^(1/2)*d^5*x^4+210*B*c^(3/2)*d^4*x^4+630*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1 
/2)*2^(1/2)/c^(1/2))*c^3*d^2*x*(-d*x+c)^(1/2)+840*A*c^(3/2)*d^4*x^3+210*B* 
2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4*d*x*(-d*x+c)^(1/2) 
+280*B*c^(5/2)*d^3*x^3+315*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^ 
(1/2))*c^4*d*(-d*x+c)^(1/2)-756*A*c^(5/2)*d^3*x^2+105*B*2^(1/2)*arctanh(1/ 
2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^5*(-d*x+c)^(1/2)-252*B*c^(7/2)*d^2*x^2 
-1224*A*c^(7/2)*d^2*x-408*B*c^(9/2)*d*x-2*A*c^(9/2)*d-342*B*c^(11/2))/(d*x 
+c)^(7/2)/(-d*x+c)^2/d^2/c^(11/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (172) = 344\).

Time = 0.12 (sec) , antiderivative size = 803, normalized size of antiderivative = 3.94 \[ \int \frac {A+B x}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/3072*(105*sqrt(2)*(B*c^7 + 3*A*c^6*d + (B*c*d^6 + 3*A*d^7)*x^6 + 2*(B*c 
^2*d^5 + 3*A*c*d^6)*x^5 - (B*c^3*d^4 + 3*A*c^2*d^5)*x^4 - 4*(B*c^4*d^3 + 3 
*A*c^3*d^4)*x^3 - (B*c^5*d^2 + 3*A*c^4*d^3)*x^2 + 2*(B*c^6*d + 3*A*c^5*d^2 
)*x)*sqrt(c)*log(-(d^2*x^2 - 2*c*d*x + 2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt 
(d*x + c)*sqrt(c) - 3*c^2)/(d^2*x^2 + 2*c*d*x + c^2)) + 4*(171*B*c^6 + A*c 
^5*d - 105*(B*c^2*d^4 + 3*A*c*d^5)*x^4 - 140*(B*c^3*d^3 + 3*A*c^2*d^4)*x^3 
 + 126*(B*c^4*d^2 + 3*A*c^3*d^3)*x^2 + 204*(B*c^5*d + 3*A*c^4*d^2)*x)*sqrt 
(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^6*d^8*x^6 + 2*c^7*d^7*x^5 - c^8*d^6*x^4 
 - 4*c^9*d^5*x^3 - c^10*d^4*x^2 + 2*c^11*d^3*x + c^12*d^2), 1/1536*(105*sq 
rt(2)*(B*c^7 + 3*A*c^6*d + (B*c*d^6 + 3*A*d^7)*x^6 + 2*(B*c^2*d^5 + 3*A*c* 
d^6)*x^5 - (B*c^3*d^4 + 3*A*c^2*d^5)*x^4 - 4*(B*c^4*d^3 + 3*A*c^3*d^4)*x^3 
 - (B*c^5*d^2 + 3*A*c^4*d^3)*x^2 + 2*(B*c^6*d + 3*A*c^5*d^2)*x)*sqrt(-c)*a 
rctan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + c^2 
)) + 2*(171*B*c^6 + A*c^5*d - 105*(B*c^2*d^4 + 3*A*c*d^5)*x^4 - 140*(B*c^3 
*d^3 + 3*A*c^2*d^4)*x^3 + 126*(B*c^4*d^2 + 3*A*c^3*d^3)*x^2 + 204*(B*c^5*d 
 + 3*A*c^4*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^6*d^8*x^6 + 2*c^ 
7*d^7*x^5 - c^8*d^6*x^4 - 4*c^9*d^5*x^3 - c^10*d^4*x^2 + 2*c^11*d^3*x + c^ 
12*d^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\frac {105 \, \sqrt {2} {\left (B c + 3 \, A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{5} d} + \frac {2 \, {\left (105 \, {\left (d x - c\right )}^{4} B c + 560 \, {\left (d x - c\right )}^{3} B c^{2} + 924 \, {\left (d x - c\right )}^{2} B c^{3} + 384 \, {\left (d x - c\right )} B c^{4} - 256 \, B c^{5} + 315 \, {\left (d x - c\right )}^{4} A d + 1680 \, {\left (d x - c\right )}^{3} A c d + 2772 \, {\left (d x - c\right )}^{2} A c^{2} d + 1152 \, {\left (d x - c\right )} A c^{3} d - 256 \, A c^{4} d\right )}}{{\left ({\left (-d x + c\right )}^{\frac {3}{2}} - 2 \, \sqrt {-d x + c} c\right )}^{3} c^{5} d}}{1536 \, d} \] Input:

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

Output:

1/1536*(105*sqrt(2)*(B*c + 3*A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(- 
c))/(sqrt(-c)*c^5*d) + 2*(105*(d*x - c)^4*B*c + 560*(d*x - c)^3*B*c^2 + 92 
4*(d*x - c)^2*B*c^3 + 384*(d*x - c)*B*c^4 - 256*B*c^5 + 315*(d*x - c)^4*A* 
d + 1680*(d*x - c)^3*A*c*d + 2772*(d*x - c)^2*A*c^2*d + 1152*(d*x - c)*A*c 
^3*d - 256*A*c^4*d)/(((-d*x + c)^(3/2) - 2*sqrt(-d*x + c)*c)^3*c^5*d))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 701, normalized size of antiderivative = 3.44 \[ \int \frac {A+B x}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(315*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*a* 
c**4*d + 630*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqr 
t(2))*a*c**3*d**2*x - 630*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) 
- sqrt(c)*sqrt(2))*a*c*d**4*x**3 - 315*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(s 
qrt(c - d*x) - sqrt(c)*sqrt(2))*a*d**5*x**4 + 105*sqrt(c)*sqrt(c - d*x)*sq 
rt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c**5 + 210*sqrt(c)*sqrt(c - d 
*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c**4*d*x - 210*sqrt(c)* 
sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c**2*d**3*x** 
3 - 105*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2)) 
*b*c*d**4*x**4 - 315*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqr 
t(c)*sqrt(2))*a*c**4*d - 630*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d* 
x) + sqrt(c)*sqrt(2))*a*c**3*d**2*x + 630*sqrt(c)*sqrt(c - d*x)*sqrt(2)*lo 
g(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*c*d**4*x**3 + 315*sqrt(c)*sqrt(c - d* 
x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*d**5*x**4 - 105*sqrt(c)* 
sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**5 - 210*sq 
rt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**4*d* 
x + 210*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2)) 
*b*c**2*d**3*x**3 + 105*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + 
sqrt(c)*sqrt(2))*b*c*d**4*x**4 + 4*a*c**5*d + 2448*a*c**4*d**2*x + 1512*a* 
c**3*d**3*x**2 - 1680*a*c**2*d**4*x**3 - 1260*a*c*d**5*x**4 + 684*b*c**...