3.578 \(\int \frac {1}{(a+b \sec (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=535 \[ -\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^4 d}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{15 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{15 a^3 d \left (a^2-b^2\right )^3 \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 a^3 d (a-b)^2 (a+b)^{5/2}}-\frac {2 \left (45 a^4-13 a^3 b-36 a^2 b^2+5 a b^3+15 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 a^3 d (a-b)^2 (a+b)^{5/2}} \]

[Out]

2/15*(58*a^4-41*a^2*b^2+15*b^4)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(
b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/(a-b)^2/(a+b)^(5/2)/d-2/15*(45*a^4-13*a^3*b-
36*a^2*b^2+5*a*b^3+15*b^4)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-
sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/(a-b)^2/(a+b)^(5/2)/d-2*cot(d*x+c)*EllipticPi((a+
b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1
+sec(d*x+c))/(a-b))^(1/2)/a^4/d+2/5*b^2*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(5/2)+2/15*b^2*(13*a^2-5*b^2
)*tan(d*x+c)/a^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(3/2)+2/15*b^2*(58*a^4-41*a^2*b^2+15*b^4)*tan(d*x+c)/a^3/(a^2-
b^2)^3/d/(a+b*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.86, antiderivative size = 535, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3785, 4060, 4058, 3921, 3784, 3832, 4004} \[ \frac {2 b^2 \left (-41 a^2 b^2+58 a^4+15 b^4\right ) \tan (c+d x)}{15 a^3 d \left (a^2-b^2\right )^3 \sqrt {a+b \sec (c+d x)}}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{15 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \tan (c+d x)}{5 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{5/2}}-\frac {2 \left (-36 a^2 b^2-13 a^3 b+45 a^4+5 a b^3+15 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 a^3 d (a-b)^2 (a+b)^{5/2}}+\frac {2 \left (-41 a^2 b^2+58 a^4+15 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 a^3 d (a-b)^2 (a+b)^{5/2}}-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^(-7/2),x]

[Out]

(2*(58*a^4 - 41*a^2*b^2 + 15*b^4)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)
/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(15*a^3*(a - b)^2*(a +
b)^(5/2)*d) - (2*(45*a^4 - 13*a^3*b - 36*a^2*b^2 + 5*a*b^3 + 15*b^4)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b
*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]
))/(a - b))])/(15*a^3*(a - b)^2*(a + b)^(5/2)*d) - (2*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sq
rt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[
c + d*x]))/(a - b))])/(a^4*d) + (2*b^2*Tan[c + d*x])/(5*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(5/2)) + (2*b^2*(
13*a^2 - 5*b^2)*Tan[c + d*x])/(15*a^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x])^(3/2)) + (2*b^2*(58*a^4 - 41*a^2*b^
2 + 15*b^4)*Tan[c + d*x])/(15*a^3*(a^2 - b^2)^3*d*Sqrt[a + b*Sec[c + d*x]])

Rule 3784

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

Rule 3785

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

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3921

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[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]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4058

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

Rule 4060

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

Rubi steps

\begin {align*} \int \frac {1}{(a+b \sec (c+d x))^{7/2}} \, dx &=\frac {2 b^2 \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}-\frac {2 \int \frac {-\frac {5}{2} \left (a^2-b^2\right )+\frac {5}{2} a b \sec (c+d x)-\frac {3}{2} b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx}{5 a \left (a^2-b^2\right )}\\ &=\frac {2 b^2 \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}+\frac {4 \int \frac {\frac {15}{4} \left (a^2-b^2\right )^2-\frac {3}{2} a b \left (5 a^2-b^2\right ) \sec (c+d x)+\frac {1}{4} b^2 \left (13 a^2-5 b^2\right ) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{15 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {2 b^2 \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{15 a^3 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}}-\frac {8 \int \frac {-\frac {15}{8} \left (a^2-b^2\right )^3+\frac {1}{8} a b \left (45 a^4-23 a^2 b^2+10 b^4\right ) \sec (c+d x)+\frac {1}{8} b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3}\\ &=\frac {2 b^2 \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{15 a^3 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}}-\frac {8 \int \frac {-\frac {15}{8} \left (a^2-b^2\right )^3+\left (\frac {1}{8} a b \left (45 a^4-23 a^2 b^2+10 b^4\right )-\frac {1}{8} b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3}-\frac {\left (b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )^3}\\ &=\frac {2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 a^3 (a-b)^2 (a+b)^{5/2} d}+\frac {2 b^2 \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{15 a^3 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}}+\frac {\int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx}{a^3}-\frac {\left (b \left (45 a^4-13 a^3 b-36 a^2 b^2+5 a b^3+15 b^4\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a^3 (a-b)^2 (a+b)^3}\\ &=\frac {2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 a^3 (a-b)^2 (a+b)^{5/2} d}-\frac {2 \left (45 a^4-13 a^3 b-36 a^2 b^2+5 a b^3+15 b^4\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 a^3 (a-b)^2 (a+b)^{5/2} d}-\frac {2 \sqrt {a+b} \cot (c+d x) \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^4 d}+\frac {2 b^2 \tan (c+d x)}{5 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{5/2}}+\frac {2 b^2 \left (13 a^2-5 b^2\right ) \tan (c+d x)}{15 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))^{3/2}}+\frac {2 b^2 \left (58 a^4-41 a^2 b^2+15 b^4\right ) \tan (c+d x)}{15 a^3 \left (a^2-b^2\right )^3 d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 16.53, size = 2346, normalized size = 4.39 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^4*Sec[c + d*x]^4*((2*b*(58*a^4 - 41*a^2*b^2 + 15*b^4)*Sin[c + d*x])/(15*a^3*(-a^2 + b^2)
^3) + (2*b^4*Sin[c + d*x])/(5*a^3*(a^2 - b^2)*(b + a*Cos[c + d*x])^3) + (2*(-19*a^2*b^3*Sin[c + d*x] + 11*b^5*
Sin[c + d*x]))/(15*a^3*(a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + (2*(74*a^4*b^2*Sin[c + d*x] - 65*a^2*b^4*Sin[c
+ d*x] + 23*b^6*Sin[c + d*x]))/(15*a^3*(a^2 - b^2)^3*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(7/2)) +
(2*(b + a*Cos[c + d*x])^(7/2)*Sec[c + d*x]^(7/2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1
+ Tan[(c + d*x)/2]^2)]*(58*a^5*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2] + 58*a^4*b^2*Sqrt[(-a + b)/(a + b)]*
Tan[(c + d*x)/2] - 41*a^3*b^3*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2] - 41*a^2*b^4*Sqrt[(-a + b)/(a + b)]*Tan[
(c + d*x)/2] + 15*a*b^5*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2] + 15*b^6*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/
2] - 116*a^5*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^3 + 82*a^3*b^3*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^
3 - 30*a*b^5*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^3 + 58*a^5*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 -
58*a^4*b^2*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 - 41*a^3*b^3*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 +
41*a^2*b^4*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 + 15*a*b^5*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 - 15
*b^6*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 - (30*I)*a^6*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a
+ b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2
]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (90*I)*a^4*b^2*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(
a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 +
b*Tan[(c + d*x)/2]^2)/(a + b)] - (90*I)*a^2*b^4*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)
]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[
(c + d*x)/2]^2)/(a + b)] + (30*I)*b^6*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c +
d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/
2]^2)/(a + b)] - (30*I)*a^6*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]],
(a + b)/(a - b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(
c + d*x)/2]^2)/(a + b)] + (90*I)*a^4*b^2*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(
c + d*x)/2]], (a + b)/(a - b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2
]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - (90*I)*a^2*b^4*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(-a + b)/(
a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Ta
n[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (30*I)*b^6*EllipticPi[-((a + b)/(a - b)), I*ArcSinh[Sqrt[(
-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a +
b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + I*b*(-58*a^5 + 58*a^4*b + 41*a^3*b^2 - 41*a^2*b^3
- 15*a*b^4 + 15*b^5)*EllipticE[I*ArcSinh[Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 -
Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b
)] + I*(15*a^6 + 45*a^5*b - 103*a^4*b^2 - 23*a^3*b^3 + 86*a^2*b^4 + 10*a*b^5 - 30*b^6)*EllipticF[I*ArcSinh[Sqr
t[(-a + b)/(a + b)]*Tan[(c + d*x)/2]], (a + b)/(a - b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*
Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)]))/(15*a^3*Sqrt[(-a + b)/(a + b)]*(a^2 - b^
2)^3*d*(a + b*Sec[c + d*x])^(7/2)*(-1 + Tan[(c + d*x)/2]^2)*Sqrt[(1 + Tan[(c + d*x)/2]^2)/(1 - Tan[(c + d*x)/2
]^2)]*(a*(-1 + Tan[(c + d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2)))

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fricas [F]  time = 1.13, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (d x + c\right ) + a}}{b^{4} \sec \left (d x + c\right )^{4} + 4 \, a b^{3} \sec \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \sec \left (d x + c\right )^{2} + 4 \, a^{3} b \sec \left (d x + c\right ) + a^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(d*x + c) + a)/(b^4*sec(d*x + c)^4 + 4*a*b^3*sec(d*x + c)^3 + 6*a^2*b^2*sec(d*x + c)^2 + 4*
a^3*b*sec(d*x + c) + a^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 1.57, size = 7838, normalized size = 14.65 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/cos(c + d*x))^(7/2),x)

[Out]

int(1/(a + b/cos(c + d*x))^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sec(d*x+c))**(7/2),x)

[Out]

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

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