3.6.53 \(\int (a+b \sec (c+d x))^{7/2} \, dx\) [553]

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

[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 = 0.35, 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 = {3867, 4141, 4143, 4006, 3869, 3917, 4089} \begin {gather*} -\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};\text {ArcSin}\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 (\text {ArcSin}\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 (\text {ArcSin}\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} \end {gather*}

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 3867

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 3869

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

Rule 3917

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

Rule 4006

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 4089

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 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4141

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), I
nt[(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 4143

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] Result contains complex when optimal does not.
time = 13.38, size = 1150, normalized size = 2.85 \begin {gather*} \frac {2 (a+b \sec (c+d x))^{7/2} \left (58 a^3 b \sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )+58 a^2 b^2 \sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )+9 a b^3 \sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )+9 b^4 \sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )-116 a^3 b \sqrt {\frac {-a+b}{a+b}} \tan ^3\left (\frac {1}{2} (c+d x)\right )-18 a b^3 \sqrt {\frac {-a+b}{a+b}} \tan ^3\left (\frac {1}{2} (c+d x)\right )+58 a^3 b \sqrt {\frac {-a+b}{a+b}} \tan ^5\left (\frac {1}{2} (c+d x)\right )-58 a^2 b^2 \sqrt {\frac {-a+b}{a+b}} \tan ^5\left (\frac {1}{2} (c+d x)\right )+9 a b^3 \sqrt {\frac {-a+b}{a+b}} \tan ^5\left (\frac {1}{2} (c+d x)\right )-9 b^4 \sqrt {\frac {-a+b}{a+b}} \tan ^5\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 {-a+b}{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+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+30 i a^4 \Pi \left (-\frac {a+b}{a-b};i \sinh ^{-1}\left (\sqrt {\frac {-a+b}{a+b}} \tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a+b}{a-b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+i b \left (-58 a^3+58 a^2 b-9 a b^2+9 b^3\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {-a+b}{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 (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-i \left (15 a^4-60 a^3 b+58 a^2 b^2-22 a b^3+9 b^4\right ) F\left (i \sinh ^{-1}\left (\sqrt {\frac {-a+b}{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 (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}\right )}{15 \sqrt {\frac {-a+b}{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 (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}}+\frac {\cos ^3(c+d x) (a+b \sec (c+d x))^{7/2} \left (\frac {2}{15} b \left (58 a^2+9 b^2\right ) \sin (c+d x)+\frac {32}{15} a b^2 \tan (c+d x)+\frac {2}{5} b^3 \sec (c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^3} \end {gather*}

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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2184\) vs. \(2(364)=728\).
time = 0.30, size = 2185, normalized size = 5.42

method	result	size
default	\(\text {Expression too large to display}\)	\(2185\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*(-60*sin(d*x+c)*cos(d*x+c)^2*(co
s(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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-22*sin(d*x+c)*co
s(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*s
in(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)*Ellipti
cE((-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+6*cos(d*x+c)^2*b^4+3*b^4-58*cos(d*x+c)^4*a^3*b-16*cos(d*x+c)^4*a^2*b^2-9*cos(d*x+c)^4*a*b^3+58*cos(d*
x+c)^3*a^3*b-10*cos(d*x+c)^3*a*b^3+74*cos(d*x+c)^2*a^2*b^2+19*cos(d*x+c)*a*b^3-58*cos(d*x+c)^3*a^2*b^2-9*cos(d
*x+c)^3*b^4-9*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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4+15*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)*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)^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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4-30*sin(d*x+c)*c
os(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+58*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))*a^3*b+5
8*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)*Elli
pticE((-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)^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))*a*b^3-60*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)*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)^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))*a^2*b^2-22*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+co
s(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)/(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 \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))^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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))**(7/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3061 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))^(7/2),x, algorithm="giac")

[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 \begin {gather*} \int {\left (a+\frac {b}{\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))^(7/2),x)

[Out]

int((a + b/cos(c + d*x))^(7/2), x)

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