3.7.39 \(\int \frac {(a+b \sec (c+d x))^{3/2}}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\) [639]

Optimal. Leaf size=303 \[ \frac {2 \left (25 a^4-31 a^2 b^2+6 b^4\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{105 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 b \left (41 a^2-3 b^2\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{105 a^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {16 b \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (25 a^2+3 b^2\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 a d \sqrt {\sec (c+d x)}} \]

[Out]

2/105*(25*a^4-31*a^2*b^2+6*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2
^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)*sec(d*x+c)^(1/2)/a^2/d/(a+b*sec(d*x+c))^(1/2)+2/7*a*sin
(d*x+c)*(a+b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(5/2)+16/35*b*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(3/2)
+2/105*(25*a^2+3*b^2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a/d/sec(d*x+c)^(1/2)+4/105*b*(41*a^2-3*b^2)*(cos(1/2*d
*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*(a+b*sec(d*x+c))^(
1/2)/a^2/d/((b+a*cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)

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Rubi [A]
time = 0.59, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3949, 4189, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \begin {gather*} \frac {2 \left (25 a^2+3 b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{105 a d \sqrt {\sec (c+d x)}}+\frac {4 b \left (41 a^2-3 b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a^2 d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 \left (25 a^4-31 a^2 b^2+6 b^4\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {16 b \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{7 d \sec ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^(3/2)/Sec[c + d*x]^(7/2),x]

[Out]

(2*(25*a^4 - 31*a^2*b^2 + 6*b^4)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt
[Sec[c + d*x]])/(105*a^2*d*Sqrt[a + b*Sec[c + d*x]]) + (4*b*(41*a^2 - 3*b^2)*EllipticE[(c + d*x)/2, (2*a)/(a +
b)]*Sqrt[a + b*Sec[c + d*x]])/(105*a^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*a*Sqrt[a
+ b*Sec[c + d*x]]*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + (16*b*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(35*d
*Sec[c + d*x]^(3/2)) + (2*(25*a^2 + 3*b^2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(105*a*d*Sqrt[Sec[c + d*x]])

Rule 2732

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

Rule 2734

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

Rule 2740

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

Rule 2742

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

Rule 3941

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3943

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[Sqrt[d*C
sc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/Sqrt[a + b*Csc[e + f*x]]), Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3949

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2), x_Symbol] :> Simp[a*Cot
[e + f*x]*Sqrt[a + b*Csc[e + f*x]]*((d*Csc[e + f*x])^n/(f*n)), x] + Dist[1/(2*d*n), Int[((d*Csc[e + f*x])^(n +
1)/Sqrt[a + b*Csc[e + f*x]])*Simp[a*b*(2*n - 1) + 2*(b^2*n + a^2*(n + 1))*Csc[e + f*x] + a*b*(2*n + 3)*Csc[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegersQ[2*n]

Rule 4120

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 4189

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1
)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {(a+b \sec (c+d x))^{3/2}}{\sec ^{\frac {7}{2}}(c+d x)} \, dx &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}-\frac {1}{7} \int \frac {-8 a b-\left (5 a^2+7 b^2\right ) \sec (c+d x)-4 a b \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {16 b \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {\frac {1}{2} a \left (25 a^2+3 b^2\right )+22 a^2 b \sec (c+d x)+8 a b^2 \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{35 a}\\ &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {16 b \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (25 a^2+3 b^2\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 a d \sqrt {\sec (c+d x)}}-\frac {4 \int \frac {-\frac {1}{2} a b \left (41 a^2-3 b^2\right )-\frac {1}{4} a^2 \left (25 a^2+51 b^2\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{105 a^2}\\ &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {16 b \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (25 a^2+3 b^2\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 a d \sqrt {\sec (c+d x)}}+\frac {1}{105} \left (2 b \left (41-\frac {3 b^2}{a^2}\right )\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx+\frac {\left (25 a^4-31 a^2 b^2+6 b^4\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{105 a^2}\\ &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {16 b \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (25 a^2+3 b^2\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 a d \sqrt {\sec (c+d x)}}+\frac {\left (\left (25 a^4-31 a^2 b^2+6 b^4\right ) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{105 a^2 \sqrt {a+b \sec (c+d x)}}+\frac {\left (2 b \left (41-\frac {3 b^2}{a^2}\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{105 \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {16 b \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (25 a^2+3 b^2\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 a d \sqrt {\sec (c+d x)}}+\frac {\left (\left (25 a^4-31 a^2 b^2+6 b^4\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{105 a^2 \sqrt {a+b \sec (c+d x)}}+\frac {\left (2 b \left (41-\frac {3 b^2}{a^2}\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}} \, dx}{105 \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ &=\frac {2 \left (25 a^4-31 a^2 b^2+6 b^4\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{105 a^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 b \left (41-\frac {3 b^2}{a^2}\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{105 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 a \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {16 b \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (25 a^2+3 b^2\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 a d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.15, size = 237, normalized size = 0.78 \begin {gather*} \frac {(a+b \sec (c+d x))^{3/2} \left (16 b \left (41 a^3+41 a^2 b-3 a b^2-3 b^3\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )+8 \left (25 a^4-31 a^2 b^2+6 b^4\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )+2 a \left (178 a^2 b+12 b^3+a \left (145 a^2+108 b^2\right ) \cos (c+d x)+78 a^2 b \cos (2 (c+d x))+15 a^3 \cos (3 (c+d x))\right ) \sin (c+d x)\right )}{420 a^2 d (b+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^(3/2)/Sec[c + d*x]^(7/2),x]

[Out]

((a + b*Sec[c + d*x])^(3/2)*(16*b*(41*a^3 + 41*a^2*b - 3*a*b^2 - 3*b^3)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Ell
ipticE[(c + d*x)/2, (2*a)/(a + b)] + 8*(25*a^4 - 31*a^2*b^2 + 6*b^4)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Ellipt
icF[(c + d*x)/2, (2*a)/(a + b)] + 2*a*(178*a^2*b + 12*b^3 + a*(145*a^2 + 108*b^2)*Cos[c + d*x] + 78*a^2*b*Cos[
2*(c + d*x)] + 15*a^3*Cos[3*(c + d*x)])*Sin[c + d*x]))/(420*a^2*d*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2049\) vs. \(2(327)=654\).
time = 0.24, size = 2050, normalized size = 6.77

method	result	size
default	\(\text {Expression too large to display}\)	\(2050\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/105/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(39*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3*b+27*cos(d*x+c)^3*((a-b)
/(a+b))^(1/2)*a^2*b^2+68*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*b-3*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^3-82*co
s(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*b+55*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b^2+6*cos(d*x+c)*((a-b)/(a+b))^(1/2)*
a*b^3+25*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*Elliptic
F((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^4+82*((b+a*cos(d*x+c))/(1+cos(d*x+c))
/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))
^(1/2))*a^3*b*sin(d*x+c)-82*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((
-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2*sin(d*x+c)-6*((b+a*cos(d*x+c))/(1+
cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(
a+b)/(a-b))^(1/2))*a*b^3*sin(d*x+c)-82*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*
EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3*b*sin(d*x+c)+51*((b+a*cos(d
*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(
d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2*sin(d*x+c)+6*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+
c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^3*sin(d*x+c)+6*(
(b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^
(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^4*sin(d*x+c)+25*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+c
os(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^4*sin(d*x+c
)-25*((a-b)/(a+b))^(1/2)*a^3*b-82*((a-b)/(a+b))^(1/2)*a^2*b^2-3*((a-b)/(a+b))^(1/2)*a*b^3+6*((a-b)/(a+b))^(1/2
)*b^4+6*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE
((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^4+82*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d
*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(
d*x+c),(-(a+b)/(a-b))^(1/2))*a^3*b-82*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(
1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2-6*
cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE((-1+cos
(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^3-82*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))
/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c)
,(-(a+b)/(a-b))^(1/2))*a^3*b+51*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(
d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2+6*cos(d*
x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c
))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^3-6*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^4+15*cos(d*x+
c)^5*((a-b)/(a+b))^(1/2)*a^4+10*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^4-25*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^4)*co
s(d*x+c)^4*(1/cos(d*x+c))^(7/2)/sin(d*x+c)/(b+a*cos(d*x+c))/((a-b)/(a+b))^(1/2)/a^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^(3/2)/sec(d*x + c)^(7/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.91, size = 501, normalized size = 1.65 \begin {gather*} \frac {\sqrt {2} {\left (-75 i \, a^{4} + 11 i \, a^{2} b^{2} - 12 i \, b^{4}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) + \sqrt {2} {\left (75 i \, a^{4} - 11 i \, a^{2} b^{2} + 12 i \, b^{4}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) - 6 \, \sqrt {2} {\left (-41 i \, a^{3} b + 3 i \, a b^{3}\right )} \sqrt {a} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right ) - 6 \, \sqrt {2} {\left (41 i \, a^{3} b - 3 i \, a b^{3}\right )} \sqrt {a} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right ) + \frac {6 \, {\left (15 \, a^{4} \cos \left (d x + c\right )^{3} + 24 \, a^{3} b \cos \left (d x + c\right )^{2} + {\left (25 \, a^{4} + 3 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{315 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/315*(sqrt(2)*(-75*I*a^4 + 11*I*a^2*b^2 - 12*I*b^4)*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/2
7*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a) + sqrt(2)*(75*I*a^4 - 11*I*a^2*b
^2 + 12*I*b^4)*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(
d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a) - 6*sqrt(2)*(-41*I*a^3*b + 3*I*a*b^3)*sqrt(a)*weierstrassZeta(-4/3*(3*
a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*
b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a)) - 6*sqrt(2)*(41*I*a^3*b - 3*I*a*b^3)*sqrt(a)*w
eierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/
a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a)) + 6*(15*a^4*cos(d*x + c
)^3 + 24*a^3*b*cos(d*x + c)^2 + (25*a^4 + 3*a^2*b^2)*cos(d*x + c))*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sin
(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**(3/2)/sec(d*x+c)**(7/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4370 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(3/2)/sec(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^(3/2)/sec(d*x + c)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^(3/2)/(1/cos(c + d*x))^(7/2),x)

[Out]

int((a + b/cos(c + d*x))^(3/2)/(1/cos(c + d*x))^(7/2), x)

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