3.254 \(\int (a+b \sec ^2(c+d x))^{5/2} \, dx\)
Optimal. Leaf size=166 \[ \frac {a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)+b}}\right )}{d}+\frac {\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)+b}}\right )}{8 d}+\frac {b \tan (c+d x) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}{4 d}+\frac {b (7 a+3 b) \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)+b}}{8 d} \]
[Out]
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
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Rubi [A] time = 0.17, antiderivative size = 166, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used =
{4128, 416, 528, 523, 217, 206, 377, 203} \[ \frac {\sqrt {b} \left (15 a^2+10 a b+3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)+b}}\right )}{8 d}+\frac {a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+b \tan ^2(c+d x)+b}}\right )}{d}+\frac {b \tan (c+d x) \left (a+b \tan ^2(c+d x)+b\right )^{3/2}}{4 d}+\frac {b (7 a+3 b) \tan (c+d x) \sqrt {a+b \tan ^2(c+d x)+b}}{8 d} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Sec[c + d*x]^2)^(5/2),x]
[Out]
(a^(5/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + b + b*Tan[c + d*x]^2]])/d + (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]])/(8*d) + (b*(7*a + 3*b)*Tan[c + d*x]*Sqrt[a +
b + b*Tan[c + d*x]^2])/(8*d) + (b*Tan[c + d*x]*(a + b + b*Tan[c + d*x]^2)^(3/2))/(4*d)
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 217
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 377
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Rule 416
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] && !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]
Rule 523
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]
Rule 528
Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
(f*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(n*(p + q + 1) + 1)), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]
Rule 4128
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[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]
Rubi steps
\begin {align*} \int \left (a+b \sec ^2(c+d x)\right )^{5/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b+b x^2\right )^{5/2}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {b \tan (c+d x) \left (a+b+b \tan ^2(c+d x)\right )^{3/2}}{4 d}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b+b x^2} \left ((a+b) (4 a+3 b)+b (7 a+3 b) x^2\right )}{1+x^2} \, dx,x,\tan (c+d x)\right )}{4 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}+\frac {\operatorname {Subst}\left (\int \frac {(a+b) \left (8 a^2+7 a b+3 b^2\right )+b \left (15 a^2+10 a b+3 b^2\right ) x^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (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}+\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (b \left (15 a^2+10 a b+3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (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}+\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (c+d x)}{\sqrt {a+b+b \tan ^2(c+d x)}}\right )}{d}+\frac {\left (b \left (15 a^2+10 a b+3 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (c+d x)}{\sqrt {a+b+b \tan ^2(c+d x)}}\right )}{8 d}\\ &=\frac {a^{5/2} \tan ^{-1}\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 ) \tanh ^{-1}\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}\\ \end {align*}
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Mathematica [C] time = 10.51, size = 706, normalized size = 4.25 \[ \frac {e^{i (c+d x)} \cos ^5(c+d x) \sqrt {4 b+a e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^2} \left (\frac {-4 i a^{5/2} \log \left (\sqrt {a} \sqrt {a \left (1+e^{2 i (c+d x)}\right )^2+4 b e^{2 i (c+d x)}}+a e^{2 i (c+d x)}+a+2 b\right )+4 i a^{5/2} \log \left (\sqrt {a} \sqrt {a \left (1+e^{2 i (c+d x)}\right )^2+4 b e^{2 i (c+d x)}}+a e^{2 i (c+d x)}+a+2 b e^{2 i (c+d x)}\right )+8 a^{5/2} d x-15 a^2 \sqrt {b} \log \left (\frac {4 i d \sqrt {a \left (1+e^{2 i (c+d x)}\right )^2+4 b e^{2 i (c+d x)}}-4 \sqrt {b} d \left (-1+e^{2 i (c+d x)}\right )}{b \left (15 a^2+10 a b+3 b^2\right ) \left (1+e^{2 i (c+d x)}\right )}\right )-10 a b^{3/2} \log \left (\frac {4 i d \sqrt {a \left (1+e^{2 i (c+d x)}\right )^2+4 b e^{2 i (c+d x)}}-4 \sqrt {b} d \left (-1+e^{2 i (c+d x)}\right )}{b \left (15 a^2+10 a b+3 b^2\right ) \left (1+e^{2 i (c+d x)}\right )}\right )-3 b^{5/2} \log \left (\frac {4 i d \sqrt {a \left (1+e^{2 i (c+d x)}\right )^2+4 b e^{2 i (c+d x)}}-4 \sqrt {b} d \left (-1+e^{2 i (c+d x)}\right )}{b \left (15 a^2+10 a b+3 b^2\right ) \left (1+e^{2 i (c+d x)}\right )}\right )}{\sqrt {a \left (1+e^{2 i (c+d x)}\right )^2+4 b e^{2 i (c+d x)}}}-\frac {i b \left (-1+e^{2 i (c+d x)}\right ) \left (9 a \left (1+e^{2 i (c+d x)}\right )^2+b \left (14 e^{2 i (c+d x)}+3 e^{4 i (c+d x)}+3\right )\right )}{\left (1+e^{2 i (c+d x)}\right )^4}\right ) \left (a+b \sec ^2(c+d x)\right )^{5/2}}{\sqrt {2} d (a \cos (2 c+2 d x)+a+2 b)^{5/2}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*Sec[c + d*x]^2)^(5/2),x]
[Out]
(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*Sq
rt[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*Sq
rt[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))))])/Sqrt[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))
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fricas [B] time = 2.25, size = 1611, normalized size = 9.70 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c)^2)^(5/2),x, algorithm="fricas")
[Out]
[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))*s
qrt(-b)*sqrt((a*cos(d*x + c)^2 + b)/cos(d*x + c)^2)/((a*b*cos(d*x + c)^2 + b^2)*sin(d*x + c)))*cos(d*x + c)^3
+ 2*(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), -1/32*(8*a^(5/2)*arctan(1/4*(8*a^2*cos(d*x + c)^5 - 8*(a^2 - a*b)*cos(d*x + c)^3 + (a^2 - 6*a*b +
b^2)*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c)^2 + b)/cos(d*x + c)^2)/((2*a^3*cos(d*x + c)^4 - a^2*b + a*b^2
- (a^3 - 3*a^2*b)*cos(d*x + c)^2)*sin(d*x + c)))*cos(d*x + c)^3 - (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*(4*a^(5/2)*arctan(1/4*(8*a^2*cos(d*x + c)^5 - 8*(a^2 - a*b)*cos(d*x + c)^3 + (a^2 - 6*a*b + b^2)
*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c)^2 + b)/cos(d*x + c)^2)/((2*a^3*cos(d*x + c)^4 - a^2*b + a*b^2 - (a
^3 - 3*a^2*b)*cos(d*x + c)^2)*sin(d*x + c)))*cos(d*x + c)^3 - (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)^2)/((a*b*cos(d*x
+ c)^2 + b^2)*sin(d*x + c)))*cos(d*x + c)^3 - 2*(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)]
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right )^{2} + a\right )}^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c)^2)^(5/2),x, algorithm="giac")
[Out]
integrate((b*sec(d*x + c)^2 + a)^(5/2), x)
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maple [C] time = 1.80, size = 2231, normalized size = 13.44 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sec(d*x+c)^2)^(5/2),x)
[Out]
1/8/d*(16*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)+a*cos(d*x+c)+b)/(1+
cos(d*x+c))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)-a*cos(d*x+c)-b)/(1+cos(d*x+c))/(a
+b))^(1/2)*EllipticPi((-1+cos(d*x+c))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(d*x+c),-1/(2*I*a^(1/2)*b^(1/
2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^3+30*sin(d*x
+c)*cos(d*x+c)^4*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)+a*cos(d*x+c)+b)/(1+cos(d*x+c))/(a+b)
)^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)-a*cos(d*x+c)-b)/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipt
icPi((-1+cos(d*x+c))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(d*x+c),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2
*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a^2*b+20*sin(d*x+c)*cos(d*x+c)^4
*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)+a*cos(d*x+c)+b)/(1+cos(d*x+c))/(a+b))^(1/2)*(-2*(I*a
^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)-a*cos(d*x+c)-b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x
+c))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(d*x+c),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/
2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*a*b^2+6*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)*((I*a^(1
/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)+a*cos(d*x+c)+b)/(1+cos(d*x+c))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*co
s(d*x+c)-I*a^(1/2)*b^(1/2)-a*cos(d*x+c)-b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))*((2*I*a^(1/2
)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(d*x+c),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1
/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b^3-8*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(d*x
+c)-I*a^(1/2)*b^(1/2)+a*cos(d*x+c)+b)/(1+cos(d*x+c))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*
b^(1/2)-a*cos(d*x+c)-b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b)
)^(1/2)/sin(d*x+c),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^3-15*sin(d*x+c)
*cos(d*x+c)^4*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)+a*cos(d*x+c)+b)/(1+cos(d*x+c))/(a+b))^(
1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)-a*cos(d*x+c)-b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF
((-1+cos(d*x+c))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(d*x+c),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)
-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^2*b-10*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1
/2)*b^(1/2)+a*cos(d*x+c)+b)/(1+cos(d*x+c))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)-a*
cos(d*x+c)-b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/si
n(d*x+c),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b^2-3*sin(d*x+c)*cos(d*x+
c)^4*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)+a*cos(d*x+c)+b)/(1+cos(d*x+c))/(a+b))^(1/2)*(-2*
(I*a^(1/2)*b^(1/2)*cos(d*x+c)-I*a^(1/2)*b^(1/2)-a*cos(d*x+c)-b)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(
d*x+c))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(d*x+c),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*
b-b^2)/(a+b)^2)^(1/2))*b^3+9*cos(d*x+c)^5*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b+3*cos(d*x+c)^5*((2*I*a
^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2-9*cos(d*x+c)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-3*cos(d*x+
c)^4*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2+11*cos(d*x+c)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b
^2+3*cos(d*x+c)^3*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3-11*cos(d*x+c)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b)
)^(1/2)*a*b^2-3*cos(d*x+c)^2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3+2*cos(d*x+c)*((2*I*a^(1/2)*b^(1/2)+a-
b)/(a+b))^(1/2)*b^3-2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^3)*cos(d*x+c)*((a*cos(d*x+c)^2+b)/cos(d*x+c)^2
)^(5/2)*sin(d*x+c)/(-1+cos(d*x+c))/(a*cos(d*x+c)^2+b)^3/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right )^{2} + a\right )}^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c)^2)^(5/2),x, algorithm="maxima")
[Out]
integrate((b*sec(d*x + c)^2 + a)^(5/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {b}{{\cos \left (c+d\,x\right )}^2}\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b/cos(c + d*x)^2)^(5/2),x)
[Out]
int((a + b/cos(c + d*x)^2)^(5/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec ^{2}{\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c)**2)**(5/2),x)
[Out]
Integral((a + b*sec(c + d*x)**2)**(5/2), x)
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