\(\int (a+b \tan (c+d x))^{7/2} \, dx\) [527]

Optimal result

Integrand size = 14, antiderivative size = 167 \[ \int (a+b \tan (c+d x))^{7/2} \, dx=-\frac {i (a-i b)^{7/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {i (a+i b)^{7/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {4 a b (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d} \] Output:

-I*(a-I*b)^(7/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+I*(a+I*b) 
^(7/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d+2*b*(3*a^2-b^2)*(a+ 
b*tan(d*x+c))^(1/2)/d+4/3*a*b*(a+b*tan(d*x+c))^(3/2)/d+2/5*b*(a+b*tan(d*x+ 
c))^(5/2)/d
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.86 \[ \int (a+b \tan (c+d x))^{7/2} \, dx=\frac {-15 i (a-i b)^{7/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+15 i (a+i b)^{7/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+2 b \sqrt {a+b \tan (c+d x)} \left (58 a^2-15 b^2+16 a b \tan (c+d x)+3 b^2 \tan ^2(c+d x)\right )}{15 d} \] Input:

Integrate[(a + b*Tan[c + d*x])^(7/2),x]
 

Output:

((-15*I)*(a - I*b)^(7/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + 
 (15*I)*(a + I*b)^(7/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + 
2*b*Sqrt[a + b*Tan[c + d*x]]*(58*a^2 - 15*b^2 + 16*a*b*Tan[c + d*x] + 3*b^ 
2*Tan[c + d*x]^2))/(15*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.92 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.87, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {3042, 3963, 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.

\(\displaystyle \int (a+b \tan (c+d x))^{7/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (c+d x))^{7/2}dx\)

\(\Big \downarrow \) 3963

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

\(\displaystyle \int \sqrt {a+b \tan (c+d x)} \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right )dx+\frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {4 a b (a+b \tan (c+d x))^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {a+b \tan (c+d x)} \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right )dx+\frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {4 a b (a+b \tan (c+d x))^{3/2}}{3 d}\)

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

\(\displaystyle \frac {1}{2} (a-i b)^4 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^4 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {4 a b (a+b \tan (c+d x))^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} (a-i b)^4 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^4 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {4 a b (a+b \tan (c+d x))^{3/2}}{3 d}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {i (a-i b)^4 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a+i b)^4 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {4 a b (a+b \tan (c+d x))^{3/2}}{3 d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {i (a-i b)^4 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i (a+i b)^4 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {4 a b (a+b \tan (c+d x))^{3/2}}{3 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a-i b)^4 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b)^4 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {4 a b (a+b \tan (c+d x))^{3/2}}{3 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 b \left (3 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {(a-i b)^{7/2} \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{7/2} \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d}+\frac {2 b (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {4 a b (a+b \tan (c+d x))^{3/2}}{3 d}\)

Input:

Int[(a + b*Tan[c + d*x])^(7/2),x]
 

Output:

((a - I*b)^(7/2)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d + ((a + I*b)^(7/2)* 
ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d + (2*b*(3*a^2 - b^2)*Sqrt[a + b*Tan[ 
c + d*x]])/d + (4*a*b*(a + b*Tan[c + d*x])^(3/2))/(3*d) + (2*b*(a + b*Tan[ 
c + d*x])^(5/2))/(5*d)
 

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 3963

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

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1551\) vs. \(2(139)=278\).

Time = 0.30 (sec) , antiderivative size = 1552, normalized size of antiderivative = 9.29

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1552\)
default	\(\text {Expression too large to display}\)	\(1552\)

Input:

int((a+b*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

2/5*b*(a+b*tan(d*x+c))^(5/2)/d+4/3*a*b*(a+b*tan(d*x+c))^(3/2)/d+6/d*b*a^2* 
(a+b*tan(d*x+c))^(1/2)-2/d*b^3*(a+b*tan(d*x+c))^(1/2)-4/d*b^3/(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+1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*ta 
n(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^4-3/2/d*b*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+1/d*b^3/(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)+4/d*b/(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^3-1/4/d/b 
*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^4+3/2/d*b*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^2-1/d*b^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arcta 
n(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(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)-4/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan 
(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/ 
2)-2*a)^(1/2))*a^3+4/d*b^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b 
^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1569 vs. \(2 (133) = 266\).

Time = 0.13 (sec) , antiderivative size = 1569, normalized size of antiderivative = 9.40 \[ \int (a+b \tan (c+d x))^{7/2} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

1/30*(15*d*sqrt(-(a^7 - 21*a^5*b^2 + 35*a^3*b^4 - 7*a*b^6 + d^2*sqrt(-(49* 
a^12*b^2 - 490*a^10*b^4 + 1519*a^8*b^6 - 1484*a^6*b^8 + 511*a^4*b^10 - 42* 
a^2*b^12 + b^14)/d^4))/d^2)*log(-(7*a^12*b - 14*a^10*b^3 - 63*a^8*b^5 - 36 
*a^6*b^7 + 25*a^4*b^9 + 18*a^2*b^11 - b^13)*sqrt(b*tan(d*x + c) + a) + ((a 
^3 - 3*a*b^2)*d^3*sqrt(-(49*a^12*b^2 - 490*a^10*b^4 + 1519*a^8*b^6 - 1484* 
a^6*b^8 + 511*a^4*b^10 - 42*a^2*b^12 + b^14)/d^4) + (21*a^8*b^2 - 112*a^6* 
b^4 + 98*a^4*b^6 - 24*a^2*b^8 + b^10)*d)*sqrt(-(a^7 - 21*a^5*b^2 + 35*a^3* 
b^4 - 7*a*b^6 + d^2*sqrt(-(49*a^12*b^2 - 490*a^10*b^4 + 1519*a^8*b^6 - 148 
4*a^6*b^8 + 511*a^4*b^10 - 42*a^2*b^12 + b^14)/d^4))/d^2)) - 15*d*sqrt(-(a 
^7 - 21*a^5*b^2 + 35*a^3*b^4 - 7*a*b^6 + d^2*sqrt(-(49*a^12*b^2 - 490*a^10 
*b^4 + 1519*a^8*b^6 - 1484*a^6*b^8 + 511*a^4*b^10 - 42*a^2*b^12 + b^14)/d^ 
4))/d^2)*log(-(7*a^12*b - 14*a^10*b^3 - 63*a^8*b^5 - 36*a^6*b^7 + 25*a^4*b 
^9 + 18*a^2*b^11 - b^13)*sqrt(b*tan(d*x + c) + a) - ((a^3 - 3*a*b^2)*d^3*s 
qrt(-(49*a^12*b^2 - 490*a^10*b^4 + 1519*a^8*b^6 - 1484*a^6*b^8 + 511*a^4*b 
^10 - 42*a^2*b^12 + b^14)/d^4) + (21*a^8*b^2 - 112*a^6*b^4 + 98*a^4*b^6 - 
24*a^2*b^8 + b^10)*d)*sqrt(-(a^7 - 21*a^5*b^2 + 35*a^3*b^4 - 7*a*b^6 + d^2 
*sqrt(-(49*a^12*b^2 - 490*a^10*b^4 + 1519*a^8*b^6 - 1484*a^6*b^8 + 511*a^4 
*b^10 - 42*a^2*b^12 + b^14)/d^4))/d^2)) - 15*d*sqrt(-(a^7 - 21*a^5*b^2 + 3 
5*a^3*b^4 - 7*a*b^6 - d^2*sqrt(-(49*a^12*b^2 - 490*a^10*b^4 + 1519*a^8*b^6 
 - 1484*a^6*b^8 + 511*a^4*b^10 - 42*a^2*b^12 + b^14)/d^4))/d^2)*log(-(7...
 

Sympy [F]

\[ \int (a+b \tan (c+d x))^{7/2} \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {7}{2}}\, dx \] Input:

integrate((a+b*tan(d*x+c))**(7/2),x)
 

Output:

Integral((a + b*tan(c + d*x))**(7/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int (a+b \tan (c+d x))^{7/2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*tan(d*x+c))^(7/2),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
 

Giac [F(-2)]

Exception generated. \[ \int (a+b \tan (c+d x))^{7/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*tan(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,19,7]%%%}+%%%{8,[0,17,7]%%%}+%%%{28,[0,15,7]%%%}+ 
%%%{56,[0
 

Mupad [B] (verification not implemented)

Time = 9.44 (sec) , antiderivative size = 2862, normalized size of antiderivative = 17.14 \[ \int (a+b \tan (c+d x))^{7/2} \, dx=\text {Too large to display} \] Input:

int((a + b*tan(c + d*x))^(7/2),x)
 

Output:

((8*a^2*b)/d - (2*b*(a^2 + b^2))/d)*(a + b*tan(c + d*x))^(1/2) - atan((((( 
32*(2*a^2*b^5*d^2 - b^7*d^2 + 3*a^4*b^3*d^2))/d^3 - 64*a*b^2*(a + b*tan(c 
+ d*x))^(1/2)*((7*a*b^6 - a^6*b*7i - a^7 + b^7*1i - a^2*b^5*21i - 35*a^3*b 
^4 + a^4*b^3*35i + 21*a^5*b^2)/(4*d^2))^(1/2))*((7*a*b^6 - a^6*b*7i - a^7 
+ b^7*1i - a^2*b^5*21i - 35*a^3*b^4 + a^4*b^3*35i + 21*a^5*b^2)/(4*d^2))^( 
1/2) - (16*(a + b*tan(c + d*x))^(1/2)*(b^10 - 28*a^2*b^8 + 70*a^4*b^6 - 28 
*a^6*b^4 + a^8*b^2))/d^2)*((7*a*b^6 - a^6*b*7i - a^7 + b^7*1i - a^2*b^5*21 
i - 35*a^3*b^4 + a^4*b^3*35i + 21*a^5*b^2)/(4*d^2))^(1/2)*1i - (((32*(2*a^ 
2*b^5*d^2 - b^7*d^2 + 3*a^4*b^3*d^2))/d^3 + 64*a*b^2*(a + b*tan(c + d*x))^ 
(1/2)*((7*a*b^6 - a^6*b*7i - a^7 + b^7*1i - a^2*b^5*21i - 35*a^3*b^4 + a^4 
*b^3*35i + 21*a^5*b^2)/(4*d^2))^(1/2))*((7*a*b^6 - a^6*b*7i - a^7 + b^7*1i 
 - a^2*b^5*21i - 35*a^3*b^4 + a^4*b^3*35i + 21*a^5*b^2)/(4*d^2))^(1/2) + ( 
16*(a + b*tan(c + d*x))^(1/2)*(b^10 - 28*a^2*b^8 + 70*a^4*b^6 - 28*a^6*b^4 
 + a^8*b^2))/d^2)*((7*a*b^6 - a^6*b*7i - a^7 + b^7*1i - a^2*b^5*21i - 35*a 
^3*b^4 + a^4*b^3*35i + 21*a^5*b^2)/(4*d^2))^(1/2)*1i)/((((32*(2*a^2*b^5*d^ 
2 - b^7*d^2 + 3*a^4*b^3*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*(( 
7*a*b^6 - a^6*b*7i - a^7 + b^7*1i - a^2*b^5*21i - 35*a^3*b^4 + a^4*b^3*35i 
 + 21*a^5*b^2)/(4*d^2))^(1/2))*((7*a*b^6 - a^6*b*7i - a^7 + b^7*1i - a^2*b 
^5*21i - 35*a^3*b^4 + a^4*b^3*35i + 21*a^5*b^2)/(4*d^2))^(1/2) - (16*(a + 
b*tan(c + d*x))^(1/2)*(b^10 - 28*a^2*b^8 + 70*a^4*b^6 - 28*a^6*b^4 + a^...
 

Reduce [F]

\[ \int (a+b \tan (c+d x))^{7/2} \, dx=\left (\int \sqrt {a +\tan \left (d x +c \right ) b}d x \right ) a^{3}+\left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{3}d x \right ) b^{3}+3 \left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}d x \right ) a \,b^{2}+3 \left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )d x \right ) a^{2} b \] Input:

int((a+b*tan(d*x+c))^(7/2),x)
 

Output:

int(sqrt(tan(c + d*x)*b + a),x)*a**3 + int(sqrt(tan(c + d*x)*b + a)*tan(c 
+ d*x)**3,x)*b**3 + 3*int(sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2,x)*a*b* 
*2 + 3*int(sqrt(tan(c + d*x)*b + a)*tan(c + d*x),x)*a**2*b