3.537 \(\int \sec ^4(c+d x) (a+b \sec (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=405 \[ \frac {2 \left (8 a^2+49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{315 b^2 d}+\frac {2 a \left (8 a^2+39 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{315 b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^4+33 a^2 b^2+147 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3+6 a^2 b+39 a b^2-147 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^3 d}-\frac {8 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{63 b^2 d}+\frac {2 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d} \]

[Out]

-2/315*(a-b)*(8*a^4+33*a^2*b^2+147*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))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^4/d-2/315*(a-b)*(8*a^3+6*a
^2*b+39*a*b^2-147*b^3)*cot(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)/b^3/d+2/315*(8*a^2+49*b^2)*(a+b*sec(d*x+c))^(
3/2)*tan(d*x+c)/b^2/d-8/63*a*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b^2/d+2/9*sec(d*x+c)*(a+b*sec(d*x+c))^(5/2)*tan
(d*x+c)/b/d+2/315*a*(8*a^2+39*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d

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Rubi [A]  time = 0.84, antiderivative size = 405, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3865, 4082, 4002, 4005, 3832, 4004} \[ \frac {2 \left (8 a^2+49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{315 b^2 d}+\frac {2 a \left (8 a^2+39 b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{315 b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (6 a^2 b+8 a^3+39 a b^2-147 b^3\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^3 d}-\frac {2 (a-b) \sqrt {a+b} \left (33 a^2 b^2+8 a^4+147 b^4\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^4 d}-\frac {8 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{63 b^2 d}+\frac {2 \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(-2*(a - b)*Sqrt[a + b]*(8*a^4 + 33*a^2*b^2 + 147*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))])/(
315*b^4*d) - (2*(a - b)*Sqrt[a + b]*(8*a^3 + 6*a^2*b + 39*a*b^2 - 147*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))])/(315*b^3*d) + (2*a*(8*a^2 + 39*b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(315*b^2*d) + (2
*(8*a^2 + 49*b^2)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(315*b^2*d) - (8*a*(a + b*Sec[c + d*x])^(5/2)*Tan[c
+ d*x])/(63*b^2*d) + (2*Sec[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(9*b*d)

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 3865

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

Rule 4002

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 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 4005

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

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((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 + 1))/(b*f*(m
+ 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) (a+b \sec (c+d x))^{3/2} \, dx &=\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{9 b d}+\frac {2 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (a+\frac {7}{2} b \sec (c+d x)-2 a \sec ^2(c+d x)\right ) \, dx}{9 b}\\ &=-\frac {8 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{9 b d}+\frac {4 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (-\frac {3 a b}{2}+\frac {1}{4} \left (8 a^2+49 b^2\right ) \sec (c+d x)\right ) \, dx}{63 b^2}\\ &=\frac {2 \left (8 a^2+49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b^2 d}-\frac {8 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{9 b d}+\frac {8 \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (-\frac {3}{8} b \left (2 a^2-49 b^2\right )+\frac {3}{8} a \left (8 a^2+39 b^2\right ) \sec (c+d x)\right ) \, dx}{315 b^2}\\ &=\frac {2 a \left (8 a^2+39 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}+\frac {2 \left (8 a^2+49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b^2 d}-\frac {8 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{9 b d}+\frac {16 \int \frac {\sec (c+d x) \left (\frac {3}{8} a b \left (a^2+93 b^2\right )+\frac {3}{16} \left (8 a^4+33 a^2 b^2+147 b^4\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{945 b^2}\\ &=\frac {2 a \left (8 a^2+39 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}+\frac {2 \left (8 a^2+49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b^2 d}-\frac {8 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{9 b d}-\frac {\left ((a-b) \left (8 a^3+6 a^2 b+39 a b^2-147 b^3\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{315 b^2}+\frac {\left (8 a^4+33 a^2 b^2+147 b^4\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{315 b^2}\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (8 a^4+33 a^2 b^2+147 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}}}{315 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3+6 a^2 b+39 a b^2-147 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}}}{315 b^3 d}+\frac {2 a \left (8 a^2+39 b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}+\frac {2 \left (8 a^2+49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b^2 d}-\frac {8 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b^2 d}+\frac {2 \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{9 b d}\\ \end {align*}

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Mathematica [A]  time = 17.83, size = 550, normalized size = 1.36 \[ \frac {\cos (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac {8 \sec (c+d x) \left (22 a b^2 \sin (c+d x)-a^3 \sin (c+d x)\right )}{315 b^2}+\frac {2 \sec ^2(c+d x) \left (3 a^2 \sin (c+d x)+49 b^2 \sin (c+d x)\right )}{315 b}+\frac {2 \left (8 a^4+33 a^2 b^2+147 b^4\right ) \sin (c+d x)}{315 b^3}+\frac {20}{63} a \tan (c+d x) \sec ^2(c+d x)+\frac {2}{9} b \tan (c+d x) \sec ^3(c+d x)\right )}{d (a \cos (c+d x)+b)}-\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (\left (8 a^4+33 a^2 b^2+147 b^4\right ) \cos (c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)-2 b \left (8 a^4+2 a^3 b+33 a^2 b^2+186 a b^3+147 b^4\right ) \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 \left (8 a^5+8 a^4 b+33 a^3 b^2+33 a^2 b^3+147 a b^4+147 b^5\right ) \sqrt {\frac {\cos (c+d x)}{\cos (c+d x)+1}} \sqrt {\frac {a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )\right )}{315 b^3 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(-2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(2*(8*a^5 + 8*a^4*b + 33*a^3*b^2 + 33*a^2
*b^3 + 147*a*b^4 + 147*b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[
c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(8*a^4 + 2*a^3*b + 33*a^2*b^2 + 186*a*b
^3 + 147*b^4)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*El
lipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (8*a^4 + 33*a^2*b^2 + 147*b^4)*Cos[c + d*x]*(b + a*Cos[c
+ d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(315*b^3*d*(b + a*Cos[c + d*x])^2*Sqrt[Sec[(c + d*x)/2]^2]*Sec[c
+ d*x]^(3/2)) + (Cos[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*((2*(8*a^4 + 33*a^2*b^2 + 147*b^4)*Sin[c + d*x])/(31
5*b^3) + (2*Sec[c + d*x]^2*(3*a^2*Sin[c + d*x] + 49*b^2*Sin[c + d*x]))/(315*b) + (8*Sec[c + d*x]*(-(a^3*Sin[c
+ d*x]) + 22*a*b^2*Sin[c + d*x]))/(315*b^2) + (20*a*Sec[c + d*x]^2*Tan[c + d*x])/63 + (2*b*Sec[c + d*x]^3*Tan[
c + d*x])/9))/(d*(b + a*Cos[c + d*x]))

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fricas [F]  time = 1.26, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (d x + c\right )^{5} + a \sec \left (d x + c\right )^{4}\right )} \sqrt {b \sec \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c)^5 + a*sec(d*x + c)^4)*sqrt(b*sec(d*x + c) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 2.12, size = 2522, normalized size = 6.23 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315/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*(-8*cos(d*x+c)^5*(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^5-4*cos(d*x+c)^6*a^4*b-147*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*co
s(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)*b^5
-53*cos(d*x+c)^2*a^2*b^3+8*cos(d*x+c)^6*a^5-8*cos(d*x+c)^5*a^5+147*cos(d*x+c)^5*b^5-98*cos(d*x+c)^4*b^5-14*cos
(d*x+c)^2*b^5+cos(d*x+c)^3*a^3*b^2-52*cos(d*x+c)^3*a*b^4-35*b^5+33*cos(d*x+c)^6*a^3*b^2+88*cos(d*x+c)^6*a^2*b^
3+147*cos(d*x+c)^6*a*b^4+8*cos(d*x+c)^5*a^4*b-34*cos(d*x+c)^5*a^3*b^2+33*cos(d*x+c)^5*a^2*b^3-10*cos(d*x+c)^5*
a*b^4-4*cos(d*x+c)^4*a^4*b-68*cos(d*x+c)^4*a^2*b^3-85*cos(d*x+c)*a*b^4+8*cos(d*x+c)^5*(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^4*b+2*cos(d*x+c)^5*(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^2+33*cos(d*x+c)^5*(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^3+186*cos(d*x+c)^5*(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))*sin(d*x+c)*a*b^4-8*cos(d*x+c)
^5*(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))/s
in(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^4*b-33*cos(d*x+c)^5*(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^2
-33*cos(d*x+c)^5*(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^3-147*cos(d*x+c)^5*(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))*s
in(d*x+c)*a*b^4+8*cos(d*x+c)^4*(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^4*b+2*cos(d*x+c)^4*(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^2+33*cos(d*x+c)^4*(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^3+186*cos(d*x+c)^4*(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))*sin(d*x+c)*a*b^4-8*cos(d*x+c)^4*(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^4*b-33*cos(
d*x+c)^4*(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^2-33*cos(d*x+c)^4*(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^3-147*cos(d*x+c)^4*(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))*sin(d*x+c)*a*b^4+147*cos(d*x+c)^4*(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^5-8*cos(d*x+c)^4*(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^5-147*cos(d*x+c)^4*(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)*b^5+147*cos(d*x+c)^5*(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^5)/(b+a*cos(d*x+c))/cos
(d*x+c)^4/sin(d*x+c)^5/b^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(c + d*x))^(3/2)/cos(c + d*x)^4,x)

[Out]

int((a + b/cos(c + d*x))^(3/2)/cos(c + d*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral((a + b*sec(c + d*x))**(3/2)*sec(c + d*x)**4, x)

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