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

Optimal. Leaf size=403 \[ -\frac {2 a^3 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} \left (58 a^2+9 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 d}+\frac {2 \sqrt {a+b} \left (60 a^3-58 a^2 b+22 a b^2-9 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 d}+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{15 d}+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d} \]

[Out]

-2/15*(a-b)*(58*a^2+9*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((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)/d+2/15*(60*a^3-58*a^2*b+22*a*b^2-9*b^3)*c
ot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((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)/d-2*a^3*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)/d+2/5*b^2*
(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d+26/15*a*b^2*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d

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Rubi [A]  time = 0.50, antiderivative size = 403, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3782, 4056, 4058, 3921, 3784, 3832, 4004} \[ \frac {2 \sqrt {a+b} \left (-58 a^2 b+60 a^3+22 a b^2-9 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 d}-\frac {2 (a-b) \sqrt {a+b} \left (58 a^2+9 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 d}-\frac {2 a^3 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}+\frac {26 a b^2 \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{15 d}+\frac {2 b^2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^(7/2),x]

[Out]

(-2*(a - b)*Sqrt[a + b]*(58*a^2 + 9*b^2)*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*d) + (2*Sqr
t[a + b]*(60*a^3 - 58*a^2*b + 22*a*b^2 - 9*b^3)*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*d) -
(2*a^3*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(a + b)/a, 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))])/d + (26*a*b^2*Sqrt[a + b
*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*b^2*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

Rule 3782

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) +
3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

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 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 4056

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

Rubi steps

\begin {align*} \int (a+b \sec (c+d x))^{7/2} \, dx &=\frac {2 b^2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {2}{5} \int \sqrt {a+b \sec (c+d x)} \left (\frac {5 a^3}{2}+\frac {3}{2} b \left (5 a^2+b^2\right ) \sec (c+d x)+\frac {13}{2} a b^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {26 a b^2 \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b^2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {15 a^4}{4}+\frac {1}{2} a b \left (30 a^2+11 b^2\right ) \sec (c+d x)+\frac {1}{4} b^2 \left (58 a^2+9 b^2\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {26 a b^2 \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b^2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {15 a^4}{4}+\left (-\frac {1}{4} b^2 \left (58 a^2+9 b^2\right )+\frac {1}{2} a b \left (30 a^2+11 b^2\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{15} \left (b^2 \left (58 a^2+9 b^2\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (58 a^2+9 b^2\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 d}+\frac {26 a b^2 \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b^2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+a^4 \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{15} \left (b \left (60 a^3-58 a^2 b+22 a b^2-9 b^3\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (58 a^2+9 b^2\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 d}+\frac {2 \sqrt {a+b} \left (60 a^3-58 a^2 b+22 a b^2-9 b^3\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 d}-\frac {2 a^3 \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}}}{d}+\frac {26 a b^2 \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 b^2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end {align*}

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Mathematica [C]  time = 15.70, size = 1150, normalized size = 2.85 \[ \frac {2 \left (-9 b^4 \sqrt {\frac {b-a}{a+b}} \tan ^5\left (\frac {1}{2} (c+d x)\right )+9 a b^3 \sqrt {\frac {b-a}{a+b}} \tan ^5\left (\frac {1}{2} (c+d x)\right )-58 a^2 b^2 \sqrt {\frac {b-a}{a+b}} \tan ^5\left (\frac {1}{2} (c+d x)\right )+58 a^3 b \sqrt {\frac {b-a}{a+b}} \tan ^5\left (\frac {1}{2} (c+d x)\right )-18 a b^3 \sqrt {\frac {b-a}{a+b}} \tan ^3\left (\frac {1}{2} (c+d x)\right )-116 a^3 b \sqrt {\frac {b-a}{a+b}} \tan ^3\left (\frac {1}{2} (c+d x)\right )+30 i a^4 \Pi \left (-\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {b-a}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a+b}{a-b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}} \tan ^2\left (\frac {1}{2} (c+d x)\right )+9 b^4 \sqrt {\frac {b-a}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )+9 a b^3 \sqrt {\frac {b-a}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )+58 a^2 b^2 \sqrt {\frac {b-a}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )+58 a^3 b \sqrt {\frac {b-a}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )+i b \left (-58 a^3+58 b a^2-9 b^2 a+9 b^3\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {b-a}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a+b}{a-b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right ) \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}}-i \left (15 a^4-60 b a^3+58 b^2 a^2-22 b^3 a+9 b^4\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {b-a}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a+b}{a-b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right ) \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}}+30 i a^4 \Pi \left (-\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {b-a}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a+b}{a-b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}}\right ) (a+b \sec (c+d x))^{7/2}}{15 \sqrt {\frac {b-a}{a+b}} d (b+a \cos (c+d x))^{7/2} \sec ^{\frac {7}{2}}(c+d x) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )-1\right ) \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^{3/2} \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}}}+\frac {\cos ^3(c+d x) \left (\frac {2}{5} \sec (c+d x) \tan (c+d x) b^3+\frac {32}{15} a \tan (c+d x) b^2+\frac {2}{15} \left (58 a^2+9 b^2\right ) \sin (c+d x) b\right ) (a+b \sec (c+d x))^{7/2}}{d (b+a \cos (c+d x))^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(a + b*Sec[c + d*x])^(7/2)*(58*a^3*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2] + 58*a^2*b^2*Sqrt[(-a + b)/(a
+ b)]*Tan[(c + d*x)/2] + 9*a*b^3*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2] + 9*b^4*Sqrt[(-a + b)/(a + b)]*Tan[(c
+ d*x)/2] - 116*a^3*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^3 - 18*a*b^3*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*
x)/2]^3 + 58*a^3*b*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 - 58*a^2*b^2*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)
/2]^5 + 9*a*b^3*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 - 9*b^4*Sqrt[(-a + b)/(a + b)]*Tan[(c + d*x)/2]^5 +
(30*I)*a^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)*a^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*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + I
*b*(-58*a^3 + 58*a^2*b - 9*a*b^2 + 9*b^3)*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^4 - 60*a^3*b + 58*a^2*b^2 - 22*a*b^3 + 9*b^4)*EllipticF[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)]))/(15*Sqrt[(-a + b)/(a + b)]*d*(b + a*Cos[c + d*x]
)^(7/2)*Sec[c + d*x]^(7/2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(-1 + Tan[(c + d*x)/2]^2)*(1 + Tan[(c + d*x)/2]
^2)^(3/2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]) + (Cos[c + d*x
]^3*(a + b*Sec[c + d*x])^(7/2)*((2*b*(58*a^2 + 9*b^2)*Sin[c + d*x])/15 + (32*a*b^2*Tan[c + d*x])/15 + (2*b^3*S
ec[c + d*x]*Tan[c + d*x])/5))/(d*(b + a*Cos[c + d*x])^3)

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fricas [F]  time = 1.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}\right )} \sqrt {b \sec \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 1.50, size = 2185, normalized size = 5.42 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/d*(1+cos(d*x+c))^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(74*cos(d*x+c)^2*a^2*b^2-16*cos(
d*x+c)^4*a^2*b^2-9*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2
)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*b^4+3*b^4+15*sin(d*x+c)*cos(d*x+c)^3*(c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*
x+c),((a-b)/(a+b))^(1/2))*a^4-58*cos(d*x+c)^4*a^3*b-9*cos(d*x+c)^4*a*b^3-10*cos(d*x+c)^3*a*b^3+19*cos(d*x+c)*a
*b^3+58*cos(d*x+c)^3*a^3*b-58*cos(d*x+c)^3*a^2*b^2-60*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*
b-58*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*E
llipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-22*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b
))^(1/2))*a*b^3+58*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/
(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b+58*sin(d*x+c)*cos(d*x+c)^2*(cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c)
,((a-b)/(a+b))^(1/2))*a^2*b^2+9*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1
+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3+9*sin(d*x+c)*cos(d*x
+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c)
)/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4-30*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d
*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*a^4+15*sin(d*
x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-
1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^4-9*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4+9*
sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipt
icE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4-30*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1
/2))*a^4-9*cos(d*x+c)^3*b^4-60*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))
/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3*b-58*cos(d*x+c)^3*(cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c
),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b^2-22*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/
(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b^3+58*cos(
d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x
+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3*b+58*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*
cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a
^2*b^2+9*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Elliptic
E((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a*b^3+6*cos(d*x+c)^2*b^4)/(b+a*cos(d*x+c))/cos(d*
x+c)^2/sin(d*x+c)^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 \[ \int {\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^(7/2),x)

[Out]

int((a + b/cos(c + d*x))^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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