\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{(c+d x)^{15/2}} \, dx\) [90]

Optimal result

Integrand size = 31, antiderivative size = 229 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{15/2}} \, dx=\frac {(53 B c-5 A d) \sqrt {c^2-d^2 x^2}}{32 d^2 (c+d x)^{5/2}}-\frac {(181 B c-5 A d) \sqrt {c^2-d^2 x^2}}{128 c d^2 (c+d x)^{3/2}}-\frac {(21 B c-5 A d) \left (c^2-d^2 x^2\right )^{3/2}}{24 d^2 (c+d x)^{9/2}}+\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{5/2}}{4 d^2 (c+d x)^{13/2}}+\frac {5 (15 B c+A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{128 \sqrt {2} c^{3/2} d^2} \] Output:

1/32*(-5*A*d+53*B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(5/2)-1/128*(-5*A*d+ 
181*B*c)*(-d^2*x^2+c^2)^(1/2)/c/d^2/(d*x+c)^(3/2)-1/24*(-5*A*d+21*B*c)*(-d 
^2*x^2+c^2)^(3/2)/d^2/(d*x+c)^(9/2)+1/4*(-A*d+B*c)*(-d^2*x^2+c^2)^(5/2)/d^ 
2/(d*x+c)^(13/2)+5/256*(A*d+15*B*c)*arctanh(2^(1/2)*c^(1/2)*(d*x+c)^(1/2)/ 
(-d^2*x^2+c^2)^(1/2))*2^(1/2)/c^(3/2)/d^2
 

Mathematica [A] (verified)

Time = 2.62 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.72 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{15/2}} \, dx=\frac {-\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (A d \left (61 c^3-117 c^2 d x+191 c d^2 x^2-15 d^3 x^3\right )+3 B c \left (49 c^3+183 c^2 d x+187 c d^2 x^2+181 d^3 x^3\right )\right )}{(c+d x)^{9/2}}+15 \sqrt {2} (15 B c+A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{768 c^{3/2} d^2} \] Input:

Integrate[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(15/2),x]
 

Output:

((-2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(A*d*(61*c^3 - 117*c^2*d*x + 191*c*d^2*x^ 
2 - 15*d^3*x^3) + 3*B*c*(49*c^3 + 183*c^2*d*x + 187*c*d^2*x^2 + 181*d^3*x^ 
3)))/(c + d*x)^(9/2) + 15*Sqrt[2]*(15*B*c + A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]* 
Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/(768*c^(3/2)*d^2)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {671, 465, 465, 465, 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.

\(\displaystyle \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{15/2}} \, dx\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {(A d+15 B c) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{13/2}}dx}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{8 c d^2 (c+d x)^{15/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(A d+15 B c) \left (-\frac {5}{6} \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{9/2}}dx-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{3 d (c+d x)^{11/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{8 c d^2 (c+d x)^{15/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(A d+15 B c) \left (-\frac {5}{6} \left (-\frac {3}{4} \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{5/2}}dx-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 d (c+d x)^{7/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{3 d (c+d x)^{11/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{8 c d^2 (c+d x)^{15/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {(A d+15 B c) \left (-\frac {5}{6} \left (-\frac {3}{4} \left (-\frac {1}{2} \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx-\frac {\sqrt {c^2-d^2 x^2}}{d (c+d x)^{3/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 d (c+d x)^{7/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{3 d (c+d x)^{11/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{8 c d^2 (c+d x)^{15/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(A d+15 B c) \left (-\frac {5}{6} \left (-\frac {3}{4} \left (-d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}-\frac {\sqrt {c^2-d^2 x^2}}{d (c+d x)^{3/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 d (c+d x)^{7/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{3 d (c+d x)^{11/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{8 c d^2 (c+d x)^{15/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(A d+15 B c) \left (-\frac {5}{6} \left (-\frac {3}{4} \left (\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {2} \sqrt {c} d}-\frac {\sqrt {c^2-d^2 x^2}}{d (c+d x)^{3/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 d (c+d x)^{7/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{5/2}}{3 d (c+d x)^{11/2}}\right )}{16 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{8 c d^2 (c+d x)^{15/2}}\)

Input:

Int[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(15/2),x]
 

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(7/2))/(8*c*d^2*(c + d*x)^(15/2)) + ((15*B*c 
+ A*d)*(-1/3*(c^2 - d^2*x^2)^(5/2)/(d*(c + d*x)^(11/2)) - (5*(-1/2*(c^2 - 
d^2*x^2)^(3/2)/(d*(c + d*x)^(7/2)) - (3*(-(Sqrt[c^2 - d^2*x^2]/(d*(c + d*x 
)^(3/2))) + ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])]/( 
Sqrt[2]*Sqrt[c]*d)))/4))/6))/(16*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 465

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + p + 1))), x] - Simp[b*(p/(d^2*(n + 
 p + 1)))   Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, 
b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LtQ[n, -2] || EqQ[n 
+ 2*p + 1, 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. \(477\) vs. \(2(191)=382\).

Time = 0.37 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.09

method	result	size
default	\(\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (15 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{5} x^{4}+225 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{4} x^{4}+60 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{4} x^{3}+900 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{3} x^{3}+90 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{3} x^{2}+30 A \,d^{4} x^{3} \sqrt {-d x +c}\, \sqrt {c}+1350 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{2} x^{2}-1086 B \,c^{\frac {3}{2}} d^{3} x^{3} \sqrt {-d x +c}+60 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{2} x -382 A \,c^{\frac {3}{2}} d^{3} x^{2} \sqrt {-d x +c}+900 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d x -1122 B \,c^{\frac {5}{2}} d^{2} x^{2} \sqrt {-d x +c}+15 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d +234 A \,c^{\frac {5}{2}} d^{2} x \sqrt {-d x +c}+225 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5}-1098 B \,c^{\frac {7}{2}} d x \sqrt {-d x +c}-122 A \,c^{\frac {7}{2}} \sqrt {-d x +c}\, d -294 B \,c^{\frac {9}{2}} \sqrt {-d x +c}\right )}{768 c^{\frac {3}{2}} \left (d x +c \right )^{\frac {9}{2}} \sqrt {-d x +c}\, d^{2}}\)	\(478\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(15/2),x,method=_RETURNVERBOSE)
 

Output:

1/768*(-d^2*x^2+c^2)^(1/2)/c^(3/2)*(15*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2 
)*2^(1/2)/c^(1/2))*d^5*x^4+225*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2 
)/c^(1/2))*c*d^4*x^4+60*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/ 
2))*c*d^4*x^3+900*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^ 
2*d^3*x^3+90*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^3 
*x^2+30*A*d^4*x^3*(-d*x+c)^(1/2)*c^(1/2)+1350*B*2^(1/2)*arctanh(1/2*(-d*x+ 
c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d^2*x^2-1086*B*c^(3/2)*d^3*x^3*(-d*x+c)^(1/2 
)+60*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d^2*x-382*A 
*c^(3/2)*d^3*x^2*(-d*x+c)^(1/2)+900*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2 
^(1/2)/c^(1/2))*c^4*d*x-1122*B*c^(5/2)*d^2*x^2*(-d*x+c)^(1/2)+15*A*2^(1/2) 
*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4*d+234*A*c^(5/2)*d^2*x*(-d 
*x+c)^(1/2)+225*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^5- 
1098*B*c^(7/2)*d*x*(-d*x+c)^(1/2)-122*A*c^(7/2)*(-d*x+c)^(1/2)*d-294*B*c^( 
9/2)*(-d*x+c)^(1/2))/(d*x+c)^(9/2)/(-d*x+c)^(1/2)/d^2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 695, normalized size of antiderivative = 3.03 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{15/2}} \, dx=\left [\frac {15 \, \sqrt {2} {\left (15 \, B c^{6} + A c^{5} d + {\left (15 \, B c d^{5} + A d^{6}\right )} x^{5} + 5 \, {\left (15 \, B c^{2} d^{4} + A c d^{5}\right )} x^{4} + 10 \, {\left (15 \, B c^{3} d^{3} + A c^{2} d^{4}\right )} x^{3} + 10 \, {\left (15 \, B c^{4} d^{2} + A c^{3} d^{3}\right )} x^{2} + 5 \, {\left (15 \, B c^{5} d + A c^{4} d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {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}}\right ) - 4 \, {\left (147 \, B c^{5} + 61 \, A c^{4} d + 3 \, {\left (181 \, B c^{2} d^{3} - 5 \, A c d^{4}\right )} x^{3} + {\left (561 \, B c^{3} d^{2} + 191 \, A c^{2} d^{3}\right )} x^{2} + 9 \, {\left (61 \, B c^{4} d - 13 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{1536 \, {\left (c^{2} d^{7} x^{5} + 5 \, c^{3} d^{6} x^{4} + 10 \, c^{4} d^{5} x^{3} + 10 \, c^{5} d^{4} x^{2} + 5 \, c^{6} d^{3} x + c^{7} d^{2}\right )}}, -\frac {15 \, \sqrt {2} {\left (15 \, B c^{6} + A c^{5} d + {\left (15 \, B c d^{5} + A d^{6}\right )} x^{5} + 5 \, {\left (15 \, B c^{2} d^{4} + A c d^{5}\right )} x^{4} + 10 \, {\left (15 \, B c^{3} d^{3} + A c^{2} d^{4}\right )} x^{3} + 10 \, {\left (15 \, B c^{4} d^{2} + A c^{3} d^{3}\right )} x^{2} + 5 \, {\left (15 \, B c^{5} d + A c^{4} d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + 2 \, {\left (147 \, B c^{5} + 61 \, A c^{4} d + 3 \, {\left (181 \, B c^{2} d^{3} - 5 \, A c d^{4}\right )} x^{3} + {\left (561 \, B c^{3} d^{2} + 191 \, A c^{2} d^{3}\right )} x^{2} + 9 \, {\left (61 \, B c^{4} d - 13 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{768 \, {\left (c^{2} d^{7} x^{5} + 5 \, c^{3} d^{6} x^{4} + 10 \, c^{4} d^{5} x^{3} + 10 \, c^{5} d^{4} x^{2} + 5 \, c^{6} d^{3} x + c^{7} d^{2}\right )}}\right ] \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(15/2),x, algorithm="fricas 
")
 

Output:

[1/1536*(15*sqrt(2)*(15*B*c^6 + A*c^5*d + (15*B*c*d^5 + A*d^6)*x^5 + 5*(15 
*B*c^2*d^4 + A*c*d^5)*x^4 + 10*(15*B*c^3*d^3 + A*c^2*d^4)*x^3 + 10*(15*B*c 
^4*d^2 + A*c^3*d^3)*x^2 + 5*(15*B*c^5*d + A*c^4*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*(147*B*c^5 + 61*A*c^4*d + 3*(181*B*c^2* 
d^3 - 5*A*c*d^4)*x^3 + (561*B*c^3*d^2 + 191*A*c^2*d^3)*x^2 + 9*(61*B*c^4*d 
 - 13*A*c^3*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c^2*d^7*x^5 + 5*c 
^3*d^6*x^4 + 10*c^4*d^5*x^3 + 10*c^5*d^4*x^2 + 5*c^6*d^3*x + c^7*d^2), -1/ 
768*(15*sqrt(2)*(15*B*c^6 + A*c^5*d + (15*B*c*d^5 + A*d^6)*x^5 + 5*(15*B*c 
^2*d^4 + A*c*d^5)*x^4 + 10*(15*B*c^3*d^3 + A*c^2*d^4)*x^3 + 10*(15*B*c^4*d 
^2 + A*c^3*d^3)*x^2 + 5*(15*B*c^5*d + A*c^4*d^2)*x)*sqrt(-c)*arctan(1/2*sq 
rt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + c^2)) + 2*(147* 
B*c^5 + 61*A*c^4*d + 3*(181*B*c^2*d^3 - 5*A*c*d^4)*x^3 + (561*B*c^3*d^2 + 
191*A*c^2*d^3)*x^2 + 9*(61*B*c^4*d - 13*A*c^3*d^2)*x)*sqrt(-d^2*x^2 + c^2) 
*sqrt(d*x + c))/(c^2*d^7*x^5 + 5*c^3*d^6*x^4 + 10*c^4*d^5*x^3 + 10*c^5*d^4 
*x^2 + 5*c^6*d^3*x + c^7*d^2)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{15/2}} \, dx=\text {Timed out} \] Input:

integrate((B*x+A)*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**(15/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{15/2}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {15}{2}}} \,d x } \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(15/2),x, algorithm="maxima 
")
 

Output:

integrate((-d^2*x^2 + c^2)^(5/2)*(B*x + A)/(d*x + c)^(15/2), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{15/2}} \, dx=-\frac {\frac {15 \, \sqrt {2} {\left (15 \, B c + A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c} + \frac {2 \, {\left (543 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} B c + 2190 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B c^{2} - 3300 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c^{3} + 1800 \, \sqrt {-d x + c} B c^{4} - 15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} A d + 146 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} A c d - 220 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c^{2} d + 120 \, \sqrt {-d x + c} A c^{3} d\right )}}{{\left (d x + c\right )}^{4} c}}{768 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(15/2),x, algorithm="giac")
 

Output:

-1/768*(15*sqrt(2)*(15*B*c + A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(- 
c))/(sqrt(-c)*c) + 2*(543*(d*x - c)^3*sqrt(-d*x + c)*B*c + 2190*(d*x - c)^ 
2*sqrt(-d*x + c)*B*c^2 - 3300*(-d*x + c)^(3/2)*B*c^3 + 1800*sqrt(-d*x + c) 
*B*c^4 - 15*(d*x - c)^3*sqrt(-d*x + c)*A*d + 146*(d*x - c)^2*sqrt(-d*x + c 
)*A*c*d - 220*(-d*x + c)^(3/2)*A*c^2*d + 120*sqrt(-d*x + c)*A*c^3*d)/((d*x 
 + c)^4*c))/d^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{15/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^{15/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(15/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(15/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.26 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{15/2}} \, dx =\text {Too large to display} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(15/2),x)
 

Output:

( - 122*sqrt(c - d*x)*a*c**4*d + 234*sqrt(c - d*x)*a*c**3*d**2*x - 382*sqr 
t(c - d*x)*a*c**2*d**3*x**2 + 30*sqrt(c - d*x)*a*c*d**4*x**3 - 294*sqrt(c 
- d*x)*b*c**5 - 1098*sqrt(c - d*x)*b*c**4*d*x - 1122*sqrt(c - d*x)*b*c**3* 
d**2*x**2 - 1086*sqrt(c - d*x)*b*c**2*d**3*x**3 - 15*sqrt(c)*sqrt(2)*log(t 
an(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**4*d - 60*sqrt(c)*sqrt(2) 
*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**3*d**2*x - 90*sqrt 
(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**2*d**3* 
x**2 - 60*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2) 
)*a*c*d**4*x**3 - 15*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*s 
qrt(2)))/2))*a*d**5*x**4 - 225*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/ 
(sqrt(c)*sqrt(2)))/2))*b*c**5 - 900*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + 
d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**4*d*x - 1350*sqrt(c)*sqrt(2)*log(tan(asin 
(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**3*d**2*x**2 - 900*sqrt(c)*sqrt( 
2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**2*d**3*x**3 - 22 
5*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c*d* 
*4*x**4)/(768*c**2*d**2*(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x 
**3 + d**4*x**4))