3.331 \(\int \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\)

Optimal. Leaf size=404 \[ -\frac {2 (a-b) \sqrt {a+b} \left (8 a^2-21 b^2\right ) \cot (c+d x) \sqrt {-\frac {b (\sec (c+d x)-1)}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b^4 d}+\frac {2 \sqrt {a+b} \left (-8 a^2+2 a b+21 b^2\right ) \cot (c+d x) \sqrt {-\frac {b (\sec (c+d x)-1)}{a+b}} \sqrt {\frac {b (\sec (c+d x)+1)}{b-a}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b^3 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{15 b^2 d}+\frac {2 \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a d} \]

[Out]

-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/d-2/15*(a-b)*(8*a^2-21*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)/b^4/d+2/15*(-8*a^2+2*a*b+21*b^2)*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-8/15*a*(a
+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/5*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d

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Rubi [A]  time = 0.76, antiderivative size = 610, normalized size of antiderivative = 1.51, number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3895, 3784, 3837, 3832, 4004, 3860, 4082, 4005} \[ -\frac {2 \sqrt {a+b} \left (8 a^2-2 a b+9 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b^3 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^2+9 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b^4 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{15 b^2 d}+\frac {4 (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}+\frac {2 \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}+\frac {4 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {2 \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{a};\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/Sqrt[a + b*Sec[c + d*x]],x]

[Out]

(4*(a - b)*Sqrt[a + b]*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S
qrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*(a - b)*Sqrt[a + b]*
(8*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*b^4*d) + (4*Sqrt[a + b]*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))])/(b*d) - (2*Sqrt[a + b]*(8*a^2 - 2*a*b + 9*b^2)*Cot[c + d*x]*Elliptic
F[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*b^3*d) - (2*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))])/(a*d) - (8*a*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*b^2*d) + (2*Sec[c + d*x]*Sqrt[a + b*S
ec[c + d*x]]*Tan[c + d*x])/(5*b*d)

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 3837

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

Rule 3860

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*d^2
*Cos[e + f*x]*(d*Csc[e + f*x])^(n - 2)*Sqrt[a + b*Csc[e + f*x]])/(b*f*(2*n - 3)), x] + Dist[d^3/(b*(2*n - 3)),
Int[((d*Csc[e + f*x])^(n - 3)*Simp[2*a*(n - 3) + b*(2*n - 5)*Csc[e + f*x] - 2*a*(n - 2)*Csc[e + f*x]^2, x])/S
qrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n
]

Rule 3895

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

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 \frac {\tan ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx &=\int \left (\frac {1}{\sqrt {a+b \sec (c+d x)}}-\frac {2 \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}}+\frac {\sec ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}}\right ) \, dx\\ &=-\left (2 \int \frac {\sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\right )+\int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx+\int \frac {\sec ^4(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\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 d}+\frac {2 \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b d}+2 \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx-2 \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {\int \frac {\sec (c+d x) \left (2 a+3 b \sec (c+d x)-4 a \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{5 b}\\ &=\frac {4 (a-b) \sqrt {a+b} \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}}}{b^2 d}+\frac {4 \sqrt {a+b} \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}}}{b 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 d}-\frac {8 a \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^2 d}+\frac {2 \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b d}+\frac {2 \int \frac {\sec (c+d x) \left (a b+\frac {1}{2} \left (8 a^2+9 b^2\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 b^2}\\ &=\frac {4 (a-b) \sqrt {a+b} \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}}}{b^2 d}+\frac {4 \sqrt {a+b} \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}}}{b 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 d}-\frac {8 a \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^2 d}+\frac {2 \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b d}+\frac {1}{15} \left (9+\frac {8 a^2}{b^2}\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx-\frac {\left (8 a^2-2 a b+9 b^2\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 b^2}\\ &=\frac {4 (a-b) \sqrt {a+b} \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}}}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 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 b^4 d}+\frac {4 \sqrt {a+b} \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}}}{b d}-\frac {2 \sqrt {a+b} \left (8 a^2-2 a b+9 b^2\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 b^3 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 d}-\frac {8 a \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^2 d}+\frac {2 \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b d}\\ \end {align*}

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Mathematica [B]  time = 16.94, size = 835, normalized size = 2.07 \[ \frac {(b+a \cos (c+d x)) \sec (c+d x) \left (-\frac {2 \left (21 b^2-8 a^2\right ) \sin (c+d x)}{15 b^3}+\frac {2 \sec (c+d x) \tan (c+d x)}{5 b}-\frac {8 a \tan (c+d x)}{15 b^2}\right )}{d \sqrt {a+b \sec (c+d x)}}-\frac {2 \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)} \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (8 a^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )+21 b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )-21 a b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )-8 a^2 b \tan ^5\left (\frac {1}{2} (c+d x)\right )-16 a^3 \tan ^3\left (\frac {1}{2} (c+d x)\right )+42 a b^2 \tan ^3\left (\frac {1}{2} (c+d x)\right )-30 b^3 \Pi \left (-1;\sin ^{-1}\left (\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 )+8 a^3 \tan \left (\frac {1}{2} (c+d x)\right )-21 b^3 \tan \left (\frac {1}{2} (c+d x)\right )-21 a b^2 \tan \left (\frac {1}{2} (c+d x)\right )+8 a^2 b \tan \left (\frac {1}{2} (c+d x)\right )+\left (8 a^3+8 b a^2-21 b^2 a-21 b^3\right ) E\left (\sin ^{-1}\left (\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}}-2 b \left (4 a^2+b a-18 b^2\right ) F\left (\sin ^{-1}\left (\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 b^3 \Pi \left (-1;\sin ^{-1}\left (\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 )}{15 b^3 d \sqrt {a+b \sec (c+d x)} \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}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[c + d*x]^4/Sqrt[a + b*Sec[c + d*x]],x]

[Out]

(-2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(8*a^3*Tan[(c + d*x)/2] +
8*a^2*b*Tan[(c + d*x)/2] - 21*a*b^2*Tan[(c + d*x)/2] - 21*b^3*Tan[(c + d*x)/2] - 16*a^3*Tan[(c + d*x)/2]^3 + 4
2*a*b^2*Tan[(c + d*x)/2]^3 + 8*a^3*Tan[(c + d*x)/2]^5 - 8*a^2*b*Tan[(c + d*x)/2]^5 - 21*a*b^2*Tan[(c + d*x)/2]
^5 + 21*b^3*Tan[(c + d*x)/2]^5 - 30*b^3*EllipticPi[-1, ArcSin[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*b^3*EllipticPi[-1, A
rcSin[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)] + (8*a^3 + 8*a^2*b - 21*a*b^2 - 21*b^3)*EllipticE[ArcSin[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)] - 2*b*(4*a^2 + a*b - 18*b^2)*EllipticF[ArcSin[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*Ta
n[(c + d*x)/2]^2)/(a + b)]))/(15*b^3*d*Sqrt[a + b*Sec[c + d*x]]*(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)]) + ((b + a*Cos[c + d*x])*Sec[c + d*x]*((
-2*(-8*a^2 + 21*b^2)*Sin[c + d*x])/(15*b^3) - (8*a*Tan[c + d*x])/(15*b^2) + (2*Sec[c + d*x]*Tan[c + d*x])/(5*b
)))/(d*Sqrt[a + b*Sec[c + d*x]])

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\tan \left (d x + c\right )^{4}}{\sqrt {b \sec \left (d x + c\right ) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(tan(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 \[ \int \frac {\tan \left (d x + c\right )^{4}}{\sqrt {b \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)^4/sqrt(b*sec(d*x + c) + a), x)

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maple [B]  time = 1.80, size = 1780, normalized size = 4.41 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+b*sec(d*x+c))^(1/2),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*(-21*sin(d*x+c)*cos(d*x+c)^3*(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)*EllipticE((-1+cos(d*x+c))/sin(d*x
+c),((a-b)/(a+b))^(1/2))*b^3+4*cos(d*x+c)^2*a^2*b+3*b^3-30*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))*b^3-20*a*b^2*cos(d*x+c)^3+36*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^3+8*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^3-21*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^3-30*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+c
os(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*b^3+36*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^3+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)*Ellip
ticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3-cos(d*x+c)*a*b^2+4*cos(d*x+c)^4*a^2*b+21*cos(d*x+c)^4
*a*b^2-8*a^2*cos(d*x+c)^3*b+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+c
os(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-21*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^2-8*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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-2*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+c
os(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+8*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
1*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)*Elli
pticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-8*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*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^2+8*cos(d*x+c)^3*a^3-24*cos(d*x+c)^2*b^3+2
1*cos(d*x+c)^3*b^3-8*cos(d*x+c)^4*a^3)/(b+a*cos(d*x+c))/cos(d*x+c)^2/sin(d*x+c)^5/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(tan(d*x + c)^4/sqrt(b*sec(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4/(a + b/cos(c + d*x))^(1/2),x)

[Out]

int(tan(c + d*x)^4/(a + b/cos(c + d*x))^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4/(a+b*sec(d*x+c))**(1/2),x)

[Out]

Integral(tan(c + d*x)**4/sqrt(a + b*sec(c + d*x)), x)

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