3.3.54 \(\int (a+b \sec ^2(c+d x))^{5/2} \, dx\) [254]

3.3.54.1 Optimal result

Integrand size = 16, antiderivative size = 166 \[ \int \left (a+b \sec ^2(c+d x)\right )^{5/2} \, dx=\frac {a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+b+b \tan ^2(c+d x)}}\right )}{d}+\frac {\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a+b+b \tan ^2(c+d x)}}\right )}{8 d}+\frac {b (7 a+3 b) \tan (c+d x) \sqrt {a+b+b \tan ^2(c+d x)}}{8 d}+\frac {b \tan (c+d x) \left (a+b+b \tan ^2(c+d x)\right )^{3/2}}{4 d} \]

output

a^(5/2)*arctan(a^(1/2)*tan(d*x+c)/(a+b+b*tan(d*x+c)^2)^(1/2))/d+1/8*(15*a^ 
2+10*a*b+3*b^2)*arctanh(b^(1/2)*tan(d*x+c)/(a+b+b*tan(d*x+c)^2)^(1/2))*b^( 
1/2)/d+1/8*b*(7*a+3*b)*(a+b+b*tan(d*x+c)^2)^(1/2)*tan(d*x+c)/d+1/4*b*tan(d 
*x+c)*(a+b+b*tan(d*x+c)^2)^(3/2)/d
 

3.3.54.2 Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 9.62 (sec) , antiderivative size = 706, normalized size of antiderivative = 4.25 \[ \int \left (a+b \sec ^2(c+d x)\right )^{5/2} \, dx=\frac {e^{i (c+d x)} \sqrt {4 b+a e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^2} \cos ^5(c+d x) \left (-\frac {i b \left (-1+e^{2 i (c+d x)}\right ) \left (9 a \left (1+e^{2 i (c+d x)}\right )^2+b \left (3+14 e^{2 i (c+d x)}+3 e^{4 i (c+d x)}\right )\right )}{\left (1+e^{2 i (c+d x)}\right )^4}+\frac {8 a^{5/2} d x-4 i a^{5/2} \log \left (a+2 b+a e^{2 i (c+d x)}+\sqrt {a} \sqrt {4 b e^{2 i (c+d x)}+a \left (1+e^{2 i (c+d x)}\right )^2}\right )+4 i a^{5/2} \log \left (a+a e^{2 i (c+d x)}+2 b e^{2 i (c+d x)}+\sqrt {a} \sqrt {4 b e^{2 i (c+d x)}+a \left (1+e^{2 i (c+d x)}\right )^2}\right )-15 a^2 \sqrt {b} \log \left (\frac {-4 \sqrt {b} d \left (-1+e^{2 i (c+d x)}\right )+4 i d \sqrt {4 b e^{2 i (c+d x)}+a \left (1+e^{2 i (c+d x)}\right )^2}}{b \left (15 a^2+10 a b+3 b^2\right ) \left (1+e^{2 i (c+d x)}\right )}\right )-10 a b^{3/2} \log \left (\frac {-4 \sqrt {b} d \left (-1+e^{2 i (c+d x)}\right )+4 i d \sqrt {4 b e^{2 i (c+d x)}+a \left (1+e^{2 i (c+d x)}\right )^2}}{b \left (15 a^2+10 a b+3 b^2\right ) \left (1+e^{2 i (c+d x)}\right )}\right )-3 b^{5/2} \log \left (\frac {-4 \sqrt {b} d \left (-1+e^{2 i (c+d x)}\right )+4 i d \sqrt {4 b e^{2 i (c+d x)}+a \left (1+e^{2 i (c+d x)}\right )^2}}{b \left (15 a^2+10 a b+3 b^2\right ) \left (1+e^{2 i (c+d x)}\right )}\right )}{\sqrt {4 b e^{2 i (c+d x)}+a \left (1+e^{2 i (c+d x)}\right )^2}}\right ) \left (a+b \sec ^2(c+d x)\right )^{5/2}}{\sqrt {2} d (a+2 b+a \cos (2 c+2 d x))^{5/2}} \]

input

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

output

(E^(I*(c + d*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(c + d*x)))^2)/E^((2*I)*(c + 
d*x))]*Cos[c + d*x]^5*(((-I)*b*(-1 + E^((2*I)*(c + d*x)))*(9*a*(1 + E^((2* 
I)*(c + d*x)))^2 + b*(3 + 14*E^((2*I)*(c + d*x)) + 3*E^((4*I)*(c + d*x)))) 
)/(1 + E^((2*I)*(c + d*x)))^4 + (8*a^(5/2)*d*x - (4*I)*a^(5/2)*Log[a + 2*b 
 + a*E^((2*I)*(c + d*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(c + d*x)) + a*(1 + E 
^((2*I)*(c + d*x)))^2]] + (4*I)*a^(5/2)*Log[a + a*E^((2*I)*(c + d*x)) + 2* 
b*E^((2*I)*(c + d*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(c + d*x)) + a*(1 + E^(( 
2*I)*(c + d*x)))^2]] - 15*a^2*Sqrt[b]*Log[(-4*Sqrt[b]*d*(-1 + E^((2*I)*(c 
+ d*x))) + (4*I)*d*Sqrt[4*b*E^((2*I)*(c + d*x)) + a*(1 + E^((2*I)*(c + d*x 
)))^2])/(b*(15*a^2 + 10*a*b + 3*b^2)*(1 + E^((2*I)*(c + d*x))))] - 10*a*b^ 
(3/2)*Log[(-4*Sqrt[b]*d*(-1 + E^((2*I)*(c + d*x))) + (4*I)*d*Sqrt[4*b*E^(( 
2*I)*(c + d*x)) + a*(1 + E^((2*I)*(c + d*x)))^2])/(b*(15*a^2 + 10*a*b + 3* 
b^2)*(1 + E^((2*I)*(c + d*x))))] - 3*b^(5/2)*Log[(-4*Sqrt[b]*d*(-1 + E^((2 
*I)*(c + d*x))) + (4*I)*d*Sqrt[4*b*E^((2*I)*(c + d*x)) + a*(1 + E^((2*I)*( 
c + d*x)))^2])/(b*(15*a^2 + 10*a*b + 3*b^2)*(1 + E^((2*I)*(c + d*x))))])/S 
qrt[4*b*E^((2*I)*(c + d*x)) + a*(1 + E^((2*I)*(c + d*x)))^2])*(a + b*Sec[c 
 + d*x]^2)^(5/2))/(Sqrt[2]*d*(a + 2*b + a*Cos[2*c + 2*d*x])^(5/2))
 

3.3.54.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {3042, 4616, 318, 403, 398, 224, 219, 291, 216}

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.

\(\displaystyle \int \left (a+b \sec ^2(c+d x)\right )^{5/2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4616

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

\(\Big \downarrow \) 318

\(\displaystyle \frac {\frac {1}{4} \int \frac {\sqrt {b \tan ^2(c+d x)+a+b} \left (b (7 a+3 b) \tan ^2(c+d x)+(a+b) (4 a+3 b)\right )}{\tan ^2(c+d x)+1}d\tan (c+d x)+\frac {1}{4} b \tan (c+d x) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 403

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {b \left (15 a^2+10 b a+3 b^2\right ) \tan ^2(c+d x)+(a+b) \left (8 a^2+7 b a+3 b^2\right )}{\left (\tan ^2(c+d x)+1\right ) \sqrt {b \tan ^2(c+d x)+a+b}}d\tan (c+d x)+\frac {1}{2} b (7 a+3 b) \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)+b}\right )+\frac {1}{4} b \tan (c+d x) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 398

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{\left (\tan ^2(c+d x)+1\right ) \sqrt {b \tan ^2(c+d x)+a+b}}d\tan (c+d x)+b \left (15 a^2+10 a b+3 b^2\right ) \int \frac {1}{\sqrt {b \tan ^2(c+d x)+a+b}}d\tan (c+d x)\right )+\frac {1}{2} b (7 a+3 b) \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)+b}\right )+\frac {1}{4} b \tan (c+d x) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{\left (\tan ^2(c+d x)+1\right ) \sqrt {b \tan ^2(c+d x)+a+b}}d\tan (c+d x)+b \left (15 a^2+10 a b+3 b^2\right ) \int \frac {1}{1-\frac {b \tan ^2(c+d x)}{b \tan ^2(c+d x)+a+b}}d\frac {\tan (c+d x)}{\sqrt {b \tan ^2(c+d x)+a+b}}\right )+\frac {1}{2} b (7 a+3 b) \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)+b}\right )+\frac {1}{4} b \tan (c+d x) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{\left (\tan ^2(c+d x)+1\right ) \sqrt {b \tan ^2(c+d x)+a+b}}d\tan (c+d x)+\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)+b}}\right )\right )+\frac {1}{2} b (7 a+3 b) \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)+b}\right )+\frac {1}{4} b \tan (c+d x) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^3 \int \frac {1}{\frac {a \tan ^2(c+d x)}{b \tan ^2(c+d x)+a+b}+1}d\frac {\tan (c+d x)}{\sqrt {b \tan ^2(c+d x)+a+b}}+\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)+b}}\right )\right )+\frac {1}{2} b (7 a+3 b) \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)+b}\right )+\frac {1}{4} b \tan (c+d x) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)+b}}\right )+\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)+b}}\right )\right )+\frac {1}{2} b (7 a+3 b) \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)+b}\right )+\frac {1}{4} b \tan (c+d x) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}{d}\)

input

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

output

((b*Tan[c + d*x]*(a + b + b*Tan[c + d*x]^2)^(3/2))/4 + ((8*a^(5/2)*ArcTan[ 
(Sqrt[a]*Tan[c + d*x])/Sqrt[a + b + b*Tan[c + d*x]^2]] + Sqrt[b]*(15*a^2 + 
 10*a*b + 3*b^2)*ArcTanh[(Sqrt[b]*Tan[c + d*x])/Sqrt[a + b + b*Tan[c + d*x 
]^2]])/2 + (b*(7*a + 3*b)*Tan[c + d*x]*Sqrt[a + b + b*Tan[c + d*x]^2])/2)/ 
4)/d
 

3.3.54.3.1 Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 291

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

rule 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

rule 398

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

rule 403

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

rule 3042

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

rule 4616

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = 
FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Subst[Int[(a + b + b*ff^2*x^2)^p 
/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] 
&& NeQ[a + b, 0] && NeQ[p, -1]
 

3.3.54.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1136\) vs. \(2(144)=288\).

Time = 21.08 (sec) , antiderivative size = 1137, normalized size of antiderivative = 6.85

method	result	size
default	\(\text {Expression too large to display}\)	\(1137\)

input

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

output

1/16/d/(-a)^(1/2)/b^2*(a+b*sec(d*x+c)^2)^(5/2)/((a*cos(d*x+c)^2+b)/(cos(d* 
x+c)+1)^2)^(1/2)/(a*cos(d*x+c)^2+b)^2/(cos(d*x+c)+1)*(3*b^(9/2)*(-a)^(1/2) 
*cos(d*x+c)^5*ln(4*(-b^(1/2)*((a*cos(d*x+c)^2+b)/(cos(d*x+c)+1)^2)^(1/2)*c 
os(d*x+c)+sin(d*x+c)*a-b^(1/2)*((a*cos(d*x+c)^2+b)/(cos(d*x+c)+1)^2)^(1/2) 
-a-b)/(sin(d*x+c)-1))+3*b^(9/2)*(-a)^(1/2)*cos(d*x+c)^5*ln(-4*(-b^(1/2)*(( 
a*cos(d*x+c)^2+b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)+sin(d*x+c)*a-b^(1/2)* 
((a*cos(d*x+c)^2+b)/(cos(d*x+c)+1)^2)^(1/2)+a+b)/(sin(d*x+c)+1))+10*b^(7/2 
)*(-a)^(1/2)*cos(d*x+c)^5*ln(4*(-b^(1/2)*((a*cos(d*x+c)^2+b)/(cos(d*x+c)+1 
)^2)^(1/2)*cos(d*x+c)+sin(d*x+c)*a-b^(1/2)*((a*cos(d*x+c)^2+b)/(cos(d*x+c) 
+1)^2)^(1/2)-a-b)/(sin(d*x+c)-1))*a+10*b^(7/2)*(-a)^(1/2)*cos(d*x+c)^5*ln( 
-4*(-b^(1/2)*((a*cos(d*x+c)^2+b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)+sin(d* 
x+c)*a-b^(1/2)*((a*cos(d*x+c)^2+b)/(cos(d*x+c)+1)^2)^(1/2)+a+b)/(sin(d*x+c 
)+1))*a+15*b^(5/2)*(-a)^(1/2)*cos(d*x+c)^5*ln(4*(-b^(1/2)*((a*cos(d*x+c)^2 
+b)/(cos(d*x+c)+1)^2)^(1/2)*cos(d*x+c)+sin(d*x+c)*a-b^(1/2)*((a*cos(d*x+c) 
^2+b)/(cos(d*x+c)+1)^2)^(1/2)-a-b)/(sin(d*x+c)-1))*a^2+15*b^(5/2)*(-a)^(1/ 
2)*cos(d*x+c)^5*ln(-4*(-b^(1/2)*((a*cos(d*x+c)^2+b)/(cos(d*x+c)+1)^2)^(1/2 
)*cos(d*x+c)+sin(d*x+c)*a-b^(1/2)*((a*cos(d*x+c)^2+b)/(cos(d*x+c)+1)^2)^(1 
/2)+a+b)/(sin(d*x+c)+1))*a^2+18*(-a)^(1/2)*cos(d*x+c)^4*sin(d*x+c)*((a*cos 
(d*x+c)^2+b)/(cos(d*x+c)+1)^2)^(1/2)*a*b^3+6*(-a)^(1/2)*cos(d*x+c)^4*sin(d 
*x+c)*((a*cos(d*x+c)^2+b)/(cos(d*x+c)+1)^2)^(1/2)*b^4+16*cos(d*x+c)^5*l...
 

3.3.54.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (144) = 288\).

Time = 1.29 (sec) , antiderivative size = 1611, normalized size of antiderivative = 9.70 \[ \int \left (a+b \sec ^2(c+d x)\right )^{5/2} \, dx=\text {Too large to display} \]

input

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

output

[1/32*(4*sqrt(-a)*a^2*cos(d*x + c)^3*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 
 - a^3*b)*cos(d*x + c)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(d*x + c)^ 
4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a 
^2*b^2 - a*b^3)*cos(d*x + c)^2 - 8*(16*a^3*cos(d*x + c)^7 - 24*(a^3 - a^2* 
b)*cos(d*x + c)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(d*x + c)^3 - (a^3 - 
 7*a^2*b + 7*a*b^2 - b^3)*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c)^2 + 
b)/cos(d*x + c)^2)*sin(d*x + c)) + (15*a^2 + 10*a*b + 3*b^2)*sqrt(b)*cos(d 
*x + c)^3*log(((a^2 - 6*a*b + b^2)*cos(d*x + c)^4 + 8*(a*b - b^2)*cos(d*x 
+ c)^2 + 4*((a - b)*cos(d*x + c)^3 + 2*b*cos(d*x + c))*sqrt(b)*sqrt((a*cos 
(d*x + c)^2 + b)/cos(d*x + c)^2)*sin(d*x + c) + 8*b^2)/cos(d*x + c)^4) + 4 
*(3*(3*a*b + b^2)*cos(d*x + c)^2 + 2*b^2)*sqrt((a*cos(d*x + c)^2 + b)/cos( 
d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^3), 1/16*(2*sqrt(-a)*a^2*cos(d*x 
 + c)^3*log(128*a^4*cos(d*x + c)^8 - 256*(a^4 - a^3*b)*cos(d*x + c)^6 + 32 
*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(d*x + c)^4 + a^4 - 28*a^3*b + 70*a^2*b 
^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(d*x + c)^ 
2 - 8*(16*a^3*cos(d*x + c)^7 - 24*(a^3 - a^2*b)*cos(d*x + c)^5 + 2*(5*a^3 
- 14*a^2*b + 5*a*b^2)*cos(d*x + c)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos 
(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c)^2 + b)/cos(d*x + c)^2)*sin(d*x + 
c)) + (15*a^2 + 10*a*b + 3*b^2)*sqrt(-b)*arctan(-1/2*((a - b)*cos(d*x + c) 
^3 + 2*b*cos(d*x + c))*sqrt(-b)*sqrt((a*cos(d*x + c)^2 + b)/cos(d*x + c...
 

3.3.54.6 Sympy [F]

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

input

integrate((a+b*sec(d*x+c)**2)**(5/2),x)
 

output

Integral((a + b*sec(c + d*x)**2)**(5/2), x)
 

3.3.54.7 Maxima [F]

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

input

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

output

integrate((b*sec(d*x + c)^2 + a)^(5/2), x)
 

3.3.54.8 Giac [F]

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

input

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

output

integrate((b*sec(d*x + c)^2 + a)^(5/2), x)
 

3.3.54.9 Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sec ^2(c+d x)\right )^{5/2} \, dx=\int {\left (a+\frac {b}{{\cos \left (c+d\,x\right )}^2}\right )}^{5/2} \,d x \]

input

int((a + b/cos(c + d*x)^2)^(5/2),x)
 

output

int((a + b/cos(c + d*x)^2)^(5/2), x)