\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{(c+d x)^{11/2}} \, dx\) [88]

Optimal result

Integrand size = 31, antiderivative size = 219 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{11/2}} \, dx=-\frac {5 (11 B c-3 A d) \sqrt {c^2-d^2 x^2}}{4 d^2 \sqrt {c+d x}}-\frac {(13 B c-5 A d) \left (c^2-d^2 x^2\right )^{3/2}}{4 d^2 (c+d x)^{5/2}}-\frac {2 B \left (c^2-d^2 x^2\right )^{3/2}}{3 d^2 (c+d x)^{3/2}}+\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{5/2}}{2 d^2 (c+d x)^{9/2}}+\frac {5 \sqrt {c} (11 B c-3 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{2 \sqrt {2} d^2} \] Output:

-5/4*(-3*A*d+11*B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^(1/2)-1/4*(-5*A*d+13 
*B*c)*(-d^2*x^2+c^2)^(3/2)/d^2/(d*x+c)^(5/2)-2/3*B*(-d^2*x^2+c^2)^(3/2)/d^ 
2/(d*x+c)^(3/2)+1/2*(-A*d+B*c)*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^(9/2)+5/4* 
c^(1/2)*(-3*A*d+11*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)/d^2
 

Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.68 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{11/2}} \, dx=\frac {-\frac {2 \sqrt {c^2-d^2 x^2} \left (-3 A d \left (9 c^2+17 c d x+4 d^2 x^2\right )+B \left (103 c^3+175 c^2 d x+56 c d^2 x^2-4 d^3 x^3\right )\right )}{(c+d x)^{5/2}}+15 \sqrt {2} \sqrt {c} (11 B c-3 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{12 d^2} \] Input:

Integrate[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(11/2),x]
 

Output:

((-2*Sqrt[c^2 - d^2*x^2]*(-3*A*d*(9*c^2 + 17*c*d*x + 4*d^2*x^2) + B*(103*c 
^3 + 175*c^2*d*x + 56*c*d^2*x^2 - 4*d^3*x^3)))/(c + d*x)^(5/2) + 15*Sqrt[2 
]*Sqrt[c]*(11*B*c - 3*A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^ 
2 - d^2*x^2]])/(12*d^2)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {671, 465, 466, 466, 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.

\(\displaystyle \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{11/2}} \, dx\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {(11 B c-3 A d) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{9/2}}dx}{8 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{4 c d^2 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 465

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 471

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

\(\Big \downarrow \) 221

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

Input:

Int[((A + B*x)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^(11/2),x]
 

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(7/2))/(4*c*d^2*(c + d*x)^(11/2)) + ((11*B*c 
- 3*A*d)*(-((c^2 - d^2*x^2)^(5/2)/(d*(c + d*x)^(7/2))) - (5*((2*(c^2 - d^2 
*x^2)^(3/2))/(3*d*(c + d*x)^(3/2)) + 2*c*((2*Sqrt[c^2 - d^2*x^2])/(d*Sqrt[ 
c + d*x]) - (2*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c 
]*Sqrt[c + d*x])])/d)))/2))/(8*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 466

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 + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 
2*(n + 2*p + 1)))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr 
eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 
] || EqQ[n + p + 1, 0]) && NeQ[n + 2*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 [A] (verified)

Time = 0.39 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {2 \left (B d x +3 A d -16 B c \right ) \sqrt {-d x +c}\, \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{3 d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {4 c \left (\frac {\left (-\frac {9 A d}{8}+\frac {17 B c}{8}\right ) \left (-d x +c \right )^{\frac {3}{2}}+\left (\frac {7}{4} A c d -\frac {15}{4} B \,c^{2}\right ) \sqrt {-d x +c}}{\left (-d x -c \right )^{2}}-\frac {5 \left (3 A d -11 B c \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}\right ) \sqrt {\frac {-d^{2} x^{2}+c^{2}}{d x +c}}\, \sqrt {d x +c}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(206\)
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (45 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{3} x^{2}-165 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x^{2}-8 B \,d^{3} x^{3} \sqrt {-d x +c}\, \sqrt {c}+90 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{2} x -24 A \,d^{3} x^{2} \sqrt {-d x +c}\, \sqrt {c}-330 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d x +112 B \,c^{\frac {3}{2}} d^{2} x^{2} \sqrt {-d x +c}+45 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d -102 A \,c^{\frac {3}{2}} d^{2} x \sqrt {-d x +c}-165 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}+350 B \,c^{\frac {5}{2}} d x \sqrt {-d x +c}-54 A \sqrt {-d x +c}\, c^{\frac {5}{2}} d +206 B \sqrt {-d x +c}\, c^{\frac {7}{2}}\right )}{12 \left (d x +c \right )^{\frac {5}{2}} \sqrt {-d x +c}\, d^{2} \sqrt {c}}\)	\(335\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(11/2),x,method=_RETURNVERBOSE)
 

Output:

2/3*(B*d*x+3*A*d-16*B*c)*(-d*x+c)^(1/2)/d^2*((-d^2*x^2+c^2)/(d*x+c))^(1/2) 
/(-d^2*x^2+c^2)^(1/2)*(d*x+c)^(1/2)+4*c/d^2*(((-9/8*A*d+17/8*B*c)*(-d*x+c) 
^(3/2)+(7/4*A*c*d-15/4*B*c^2)*(-d*x+c)^(1/2))/(-d*x-c)^2-5/16*(3*A*d-11*B* 
c)*2^(1/2)/c^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2)))*((-d^2*x^2 
+c^2)/(d*x+c))^(1/2)/(-d^2*x^2+c^2)^(1/2)*(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.37 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{11/2}} \, dx=\left [-\frac {15 \, \sqrt {\frac {1}{2}} {\left (11 \, B c^{4} - 3 \, A c^{3} d + {\left (11 \, B c d^{3} - 3 \, A d^{4}\right )} x^{3} + 3 \, {\left (11 \, B c^{2} d^{2} - 3 \, A c d^{3}\right )} x^{2} + 3 \, {\left (11 \, B c^{3} d - 3 \, A c^{2} d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 4 \, \sqrt {\frac {1}{2}} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 2 \, {\left (4 \, B d^{3} x^{3} - 103 \, B c^{3} + 27 \, A c^{2} d - 4 \, {\left (14 \, B c d^{2} - 3 \, A d^{3}\right )} x^{2} - {\left (175 \, B c^{2} d - 51 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{12 \, {\left (d^{5} x^{3} + 3 \, c d^{4} x^{2} + 3 \, c^{2} d^{3} x + c^{3} d^{2}\right )}}, -\frac {15 \, \sqrt {\frac {1}{2}} {\left (11 \, B c^{4} - 3 \, A c^{3} d + {\left (11 \, B c d^{3} - 3 \, A d^{4}\right )} x^{3} + 3 \, {\left (11 \, B c^{2} d^{2} - 3 \, A c d^{3}\right )} x^{2} + 3 \, {\left (11 \, B c^{3} d - 3 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {\frac {1}{2}} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{c d x + c^{2}}\right ) - {\left (4 \, B d^{3} x^{3} - 103 \, B c^{3} + 27 \, A c^{2} d - 4 \, {\left (14 \, B c d^{2} - 3 \, A d^{3}\right )} x^{2} - {\left (175 \, B c^{2} d - 51 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{6 \, {\left (d^{5} x^{3} + 3 \, c d^{4} x^{2} + 3 \, c^{2} d^{3} x + c^{3} d^{2}\right )}}\right ] \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(11/2),x, algorithm="fricas 
")
 

Output:

[-1/12*(15*sqrt(1/2)*(11*B*c^4 - 3*A*c^3*d + (11*B*c*d^3 - 3*A*d^4)*x^3 + 
3*(11*B*c^2*d^2 - 3*A*c*d^3)*x^2 + 3*(11*B*c^3*d - 3*A*c^2*d^2)*x)*sqrt(c) 
*log(-(d^2*x^2 - 2*c*d*x + 4*sqrt(1/2)*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)) - 2*(4*B*d^3*x^3 - 103*B*c^3 + 
 27*A*c^2*d - 4*(14*B*c*d^2 - 3*A*d^3)*x^2 - (175*B*c^2*d - 51*A*c*d^2)*x) 
*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x 
+ c^3*d^2), -1/6*(15*sqrt(1/2)*(11*B*c^4 - 3*A*c^3*d + (11*B*c*d^3 - 3*A*d 
^4)*x^3 + 3*(11*B*c^2*d^2 - 3*A*c*d^3)*x^2 + 3*(11*B*c^3*d - 3*A*c^2*d^2)* 
x)*sqrt(-c)*arctan(sqrt(1/2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/( 
c*d*x + c^2)) - (4*B*d^3*x^3 - 103*B*c^3 + 27*A*c^2*d - 4*(14*B*c*d^2 - 3* 
A*d^3)*x^2 - (175*B*c^2*d - 51*A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + 
 c))/(d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2)]
 

Sympy [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{11/2}} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (A + B x\right )}{\left (c + d x\right )^{\frac {11}{2}}}\, dx \] Input:

integrate((B*x+A)*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**(11/2),x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(5/2)*(A + B*x)/(c + d*x)**(11/2), x)
 

Maxima [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{11/2}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} {\left (B x + A\right )}}{{\left (d x + c\right )}^{\frac {11}{2}}} \,d x } \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(11/2),x, algorithm="maxima 
")
 

Output:

integrate((-d^2*x^2 + c^2)^(5/2)*(B*x + A)/(d*x + c)^(11/2), x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.68 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{11/2}} \, dx=-\frac {8 \, {\left (-d x + c\right )}^{\frac {3}{2}} B + 120 \, \sqrt {-d x + c} B c - 24 \, \sqrt {-d x + c} A d + \frac {15 \, \sqrt {2} {\left (11 \, B c^{2} - 3 \, A c d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} - \frac {6 \, {\left (17 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c^{2} - 30 \, \sqrt {-d x + c} B c^{3} - 9 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c d + 14 \, \sqrt {-d x + c} A c^{2} d\right )}}{{\left (d x + c\right )}^{2}}}{12 \, d^{2}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(11/2),x, algorithm="giac")
 

Output:

-1/12*(8*(-d*x + c)^(3/2)*B + 120*sqrt(-d*x + c)*B*c - 24*sqrt(-d*x + c)*A 
*d + 15*sqrt(2)*(11*B*c^2 - 3*A*c*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqr 
t(-c))/sqrt(-c) - 6*(17*(-d*x + c)^(3/2)*B*c^2 - 30*sqrt(-d*x + c)*B*c^3 - 
 9*(-d*x + c)^(3/2)*A*c*d + 14*sqrt(-d*x + c)*A*c^2*d)/(d*x + c)^2)/d^2
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{11/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^{11/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(11/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x))/(c + d*x)^(11/2), x)
 

Reduce [B] (verification not implemented)

Time = 1.10 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.80 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^{11/2}} \, dx=\frac {432 \sqrt {-d x +c}\, a \,c^{2} d +816 \sqrt {-d x +c}\, a c \,d^{2} x +192 \sqrt {-d x +c}\, a \,d^{3} x^{2}-1648 \sqrt {-d x +c}\, b \,c^{3}-2800 \sqrt {-d x +c}\, b \,c^{2} d x -896 \sqrt {-d x +c}\, b c \,d^{2} x^{2}+64 \sqrt {-d x +c}\, b \,d^{3} x^{3}+360 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a \,c^{2} d +720 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a c \,d^{2} x +360 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a \,d^{3} x^{2}-1320 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{3}-2640 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{2} d x -1320 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b c \,d^{2} x^{2}-327 \sqrt {c}\, \sqrt {2}\, a \,c^{2} d -654 \sqrt {c}\, \sqrt {2}\, a c \,d^{2} x -327 \sqrt {c}\, \sqrt {2}\, a \,d^{3} x^{2}+1367 \sqrt {c}\, \sqrt {2}\, b \,c^{3}+2734 \sqrt {c}\, \sqrt {2}\, b \,c^{2} d x +1367 \sqrt {c}\, \sqrt {2}\, b c \,d^{2} x^{2}}{96 d^{2} \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^(11/2),x)
 

Output:

(432*sqrt(c - d*x)*a*c**2*d + 816*sqrt(c - d*x)*a*c*d**2*x + 192*sqrt(c - 
d*x)*a*d**3*x**2 - 1648*sqrt(c - d*x)*b*c**3 - 2800*sqrt(c - d*x)*b*c**2*d 
*x - 896*sqrt(c - d*x)*b*c*d**2*x**2 + 64*sqrt(c - d*x)*b*d**3*x**3 + 360* 
sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**2*d 
 + 720*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a 
*c*d**2*x + 360*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2 
)))/2))*a*d**3*x**2 - 1320*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqr 
t(c)*sqrt(2)))/2))*b*c**3 - 2640*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x 
)/(sqrt(c)*sqrt(2)))/2))*b*c**2*d*x - 1320*sqrt(c)*sqrt(2)*log(tan(asin(sq 
rt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c*d**2*x**2 - 327*sqrt(c)*sqrt(2)*a*c 
**2*d - 654*sqrt(c)*sqrt(2)*a*c*d**2*x - 327*sqrt(c)*sqrt(2)*a*d**3*x**2 + 
 1367*sqrt(c)*sqrt(2)*b*c**3 + 2734*sqrt(c)*sqrt(2)*b*c**2*d*x + 1367*sqrt 
(c)*sqrt(2)*b*c*d**2*x**2)/(96*d**2*(c**2 + 2*c*d*x + d**2*x**2))