3.4.34 \(\int (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [334]

3.4.34.1 Optimal result

Integrand size = 25, antiderivative size = 188 \[ \int (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {(a-i b)^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 \left (2 a A b+a^2 B-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (A b+a B) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 d} \]

output

-(a-I*b)^(5/2)*(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+(a+ 
I*b)^(5/2)*(I*A-B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d+2*(2*A* 
a*b+B*a^2-B*b^2)*(a+b*tan(d*x+c))^(1/2)/d+2/3*(A*b+B*a)*(a+b*tan(d*x+c))^( 
3/2)/d+2/5*B*(a+b*tan(d*x+c))^(5/2)/d
 

3.4.34.2 Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.24 \[ \int (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {i \left ((A-i B) \left (\frac {2}{5} (a+b \tan (c+d x))^{5/2}+\frac {2}{3} (a-i b) \left (-3 (a-i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+\sqrt {a+b \tan (c+d x)} (4 a-3 i b+b \tan (c+d x))\right )\right )-(A+i B) \left (\frac {2}{5} (a+b \tan (c+d x))^{5/2}+\frac {2}{3} (a+i b) \left (-3 (a+i b)^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+\sqrt {a+b \tan (c+d x)} (4 a+3 i b+b \tan (c+d x))\right )\right )\right )}{2 d} \]

input

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

output

((I/2)*((A - I*B)*((2*(a + b*Tan[c + d*x])^(5/2))/5 + (2*(a - I*b)*(-3*(a 
- I*b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + Sqrt[a + b* 
Tan[c + d*x]]*(4*a - (3*I)*b + b*Tan[c + d*x])))/3) - (A + I*B)*((2*(a + b 
*Tan[c + d*x])^(5/2))/5 + (2*(a + I*b)*(-3*(a + I*b)^(3/2)*ArcTanh[Sqrt[a 
+ b*Tan[c + d*x]]/Sqrt[a + I*b]] + Sqrt[a + b*Tan[c + d*x]]*(4*a + (3*I)*b 
 + b*Tan[c + d*x])))/3)))/d
 

3.4.34.3 Rubi [A] (warning: unable to verify)

Time = 0.99 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.90, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4011, 3042, 4011, 3042, 4011, 3042, 4022, 3042, 4020, 25, 73, 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.

\(\displaystyle \int (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4011

\(\displaystyle \int (a+b \tan (c+d x))^{3/2} (a A-b B+(A b+a B) \tan (c+d x))dx+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (c+d x))^{3/2} (a A-b B+(A b+a B) \tan (c+d x))dx+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

\(\displaystyle \frac {i (a-i b)^3 (A-i B) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a+i b)^3 (A+i B) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a B+A b) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {i (a-i b)^3 (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (a+i b)^3 (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a B+A b) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a-i b)^3 (A-i B) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b)^3 (A+i B) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a B+A b) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \left (a^2 B+2 a A b-b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {(a-i b)^{5/2} (A-i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} (A+i B) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}+\frac {2 (a B+A b) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+b \tan (c+d x))^{5/2}}{5 d}\)

input

Int[(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]
 

output

((a - I*b)^(5/2)*(A - I*B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d + ((a + I 
*b)^(5/2)*(A + I*B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d + (2*(2*a*A*b + 
a^2*B - b^2*B)*Sqrt[a + b*Tan[c + d*x]])/d + (2*(A*b + a*B)*(a + b*Tan[c + 
 d*x])^(3/2))/(3*d) + (2*B*(a + b*Tan[c + d*x])^(5/2))/(5*d)
 

3.4.34.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4011

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int 
[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 
 0] && GtQ[m, 0]
 

rule 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

3.4.34.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2386\) vs. \(2(160)=320\).

Time = 0.11 (sec) , antiderivative size = 2387, normalized size of antiderivative = 12.70

method	result	size
parts	\(\text {Expression too large to display}\)	\(2387\)
derivativedivides	\(\text {Expression too large to display}\)	\(2405\)
default	\(\text {Expression too large to display}\)	\(2405\)

input

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

output

A*(2/3*b*(a+b*tan(d*x+c))^(3/2)/d+4*b/d*(a+b*tan(d*x+c))^(1/2)*a-1/4/b/d*l 
n(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2 
+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a^2+1/4*b/d*ln( 
b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b 
^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)+1/4/b/d*ln(b*tan( 
d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1 
/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3-3/4*b/d*ln(b*tan(d*x+c)+a+(a+b*tan( 
d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^ 
(1/2)+2*a)^(1/2)*a-2*b/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan( 
d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2) 
)*(a^2+b^2)^(1/2)*a+3*b/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan 
(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2 
))*a^2-b^3/d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2 
)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+1/4/b/d*ln 
((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+ 
b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a^2-1/4*b/d*ln(( 
a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^ 
2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)-1/4/b/d*ln((a+b*ta 
n(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/ 
2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3+3/4*b/d*ln((a+b*tan(d*x+c))^(1/2)...
 

3.4.34.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4855 vs. \(2 (154) = 308\).

Time = 0.82 (sec) , antiderivative size = 4855, normalized size of antiderivative = 25.82 \[ \int (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

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

output

-1/30*(15*d*sqrt((10*A*B*a^4*b - 20*A*B*a^2*b^3 + 2*A*B*b^5 - (A^2 - B^2)* 
a^5 + 10*(A^2 - B^2)*a^3*b^2 - 5*(A^2 - B^2)*a*b^4 + d^2*sqrt(-(4*A^2*B^2* 
a^10 + 20*(A^3*B - A*B^3)*a^9*b + 5*(5*A^4 - 26*A^2*B^2 + 5*B^4)*a^8*b^2 - 
 240*(A^3*B - A*B^3)*a^7*b^3 - 20*(5*A^4 - 32*A^2*B^2 + 5*B^4)*a^6*b^4 + 5 
04*(A^3*B - A*B^3)*a^5*b^5 + 10*(11*A^4 - 62*A^2*B^2 + 11*B^4)*a^4*b^6 - 2 
40*(A^3*B - A*B^3)*a^3*b^7 - 20*(A^4 - 7*A^2*B^2 + B^4)*a^2*b^8 + 20*(A^3* 
B - A*B^3)*a*b^9 + (A^4 - 2*A^2*B^2 + B^4)*b^10)/d^4))/d^2)*log(-(2*(A^3*B 
 + A*B^3)*a^9 + 5*(A^4 - B^4)*a^8*b - 16*(A^3*B + A*B^3)*a^7*b^2 - 28*(A^3 
*B + A*B^3)*a^5*b^4 - 14*(A^4 - B^4)*a^4*b^5 - 8*(A^4 - B^4)*a^2*b^7 + 10* 
(A^3*B + A*B^3)*a*b^8 + (A^4 - B^4)*b^9)*sqrt(b*tan(d*x + c) + a) + ((A*a^ 
2 - 2*B*a*b - A*b^2)*d^3*sqrt(-(4*A^2*B^2*a^10 + 20*(A^3*B - A*B^3)*a^9*b 
+ 5*(5*A^4 - 26*A^2*B^2 + 5*B^4)*a^8*b^2 - 240*(A^3*B - A*B^3)*a^7*b^3 - 2 
0*(5*A^4 - 32*A^2*B^2 + 5*B^4)*a^6*b^4 + 504*(A^3*B - A*B^3)*a^5*b^5 + 10* 
(11*A^4 - 62*A^2*B^2 + 11*B^4)*a^4*b^6 - 240*(A^3*B - A*B^3)*a^3*b^7 - 20* 
(A^4 - 7*A^2*B^2 + B^4)*a^2*b^8 + 20*(A^3*B - A*B^3)*a*b^9 + (A^4 - 2*A^2* 
B^2 + B^4)*b^10)/d^4) - (2*A*B^2*a^7 + (9*A^2*B - 5*B^3)*a^6*b + 2*(5*A^3 
- 16*A*B^2)*a^5*b^2 - 5*(11*A^2*B - 3*B^3)*a^4*b^3 - 10*(2*A^3 - 5*A*B^2)* 
a^3*b^4 + (31*A^2*B - 11*B^3)*a^2*b^5 + 2*(A^3 - 6*A*B^2)*a*b^6 - (A^2*B - 
 B^3)*b^7)*d)*sqrt((10*A*B*a^4*b - 20*A*B*a^2*b^3 + 2*A*B*b^5 - (A^2 - B^2 
)*a^5 + 10*(A^2 - B^2)*a^3*b^2 - 5*(A^2 - B^2)*a*b^4 + d^2*sqrt(-(4*A^2...
 

3.4.34.6 Sympy [F]

\[ \int (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]

input

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

output

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**(5/2), x)
 

3.4.34.7 Maxima [F(-2)]

Exception generated. \[ \int (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Exception raised: ValueError} \]

input

integrate((a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")
 

output

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more 
details)Is
 

3.4.34.8 Giac [F(-1)]

Timed out. \[ \int (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.4.34.9 Mupad [B] (verification not implemented)

Time = 41.53 (sec) , antiderivative size = 3863, normalized size of antiderivative = 20.55 \[ \int (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

int((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2),x)
 

output

log(- ((((((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + B^2*a^5*d^2 
 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4)^(1/2)*(32*B*b^6 - 32*B*a^4*b 
^2 + 32*a*b^2*d*(((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + B^2* 
a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4)^(1/2)*(a + b*tan(c + 
d*x))^(1/2)))/(2*d) - (16*B^2*b^2*(a + b*tan(c + d*x))^(1/2)*(a^6 - b^6 + 
15*a^2*b^4 - 15*a^4*b^2))/d^2)*(((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^ 
2)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4)^(1/2)) 
/2 - (8*B^3*a*b^2*(a^2 - 3*b^2)*(a^2 + b^2)^3)/d^3)*((20*B^4*a^2*b^8*d^4 - 
 B^4*b^10*d^4 - 110*B^4*a^4*b^6*d^4 + 100*B^4*a^6*b^4*d^4 - 25*B^4*a^8*b^2 
*d^4)^(1/2)/(4*d^4) + (B^2*a^5)/(4*d^2) - (5*B^2*a^3*b^2)/(2*d^2) + (5*B^2 
*a*b^4)/(4*d^2))^(1/2) - log(((((((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2) 
^2)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4)^(1/2) 
*(32*B*a^4*b^2 - 32*B*b^6 + 32*a*b^2*d*(((-B^4*b^2*d^4*(5*a^4 + b^4 - 10*a 
^2*b^2)^2)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2*a*b^4*d^2)/d^4 
)^(1/2)*(a + b*tan(c + d*x))^(1/2)))/(2*d) - (16*B^2*b^2*(a + b*tan(c + d* 
x))^(1/2)*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2))/d^2)*(((-B^4*b^2*d^4*(5*a 
^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 + 5*B^2 
*a*b^4*d^2)/d^4)^(1/2))/2 - (8*B^3*a*b^2*(a^2 - 3*b^2)*(a^2 + b^2)^3)/d^3) 
*(((20*B^4*a^2*b^8*d^4 - B^4*b^10*d^4 - 110*B^4*a^4*b^6*d^4 + 100*B^4*a^6* 
b^4*d^4 - 25*B^4*a^8*b^2*d^4)^(1/2) + B^2*a^5*d^2 - 10*B^2*a^3*b^2*d^2 ...