3.13.98 \(\int \frac {(a+b \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [1298]

Optimal. Leaf size=190 \[ -\frac {2 (5 a A+3 b B+3 a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (7 A b+7 a B+5 b C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (b B+a C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (7 A b+7 a B+5 b C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 a A+3 b B+3 a C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]

[Out]

-2/5*(5*A*a+3*B*b+3*C*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))
/d+2/21*(7*A*b+7*B*a+5*C*b)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/
2))/d+2/7*b*C*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/5*(B*b+C*a)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/21*(7*A*b+7*B*a+5*C*
b)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*(5*A*a+3*B*b+3*C*a)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.23, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4197, 3110, 3100, 2827, 2716, 2720, 2719} \begin {gather*} \frac {2 F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (7 a B+7 A b+5 b C)}{21 d}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) (5 a A+3 a C+3 b B)}{5 d}+\frac {2 \sin (c+d x) (7 a B+7 A b+5 b C)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \sin (c+d x) (5 a A+3 a C+3 b B)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 (a C+b B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Cos[c + d*x]^(3/2),x]

[Out]

(-2*(5*a*A + 3*b*B + 3*a*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(7*A*b + 7*a*B + 5*b*C)*EllipticF[(c + d*x)/
2, 2])/(21*d) + (2*b*C*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + (2*(b*B + a*C)*Sin[c + d*x])/(5*d*Cos[c + d*x]
^(5/2)) + (2*(7*A*b + 7*a*B + 5*b*C)*Sin[c + d*x])/(21*d*Cos[c + d*x]^(3/2)) + (2*(5*a*A + 3*b*B + 3*a*C)*Sin[
c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3100

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

Rule 3110

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

Rule 4197

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {(a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx &=\int \frac {(b+a \cos (c+d x)) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {2}{7} \int \frac {-\frac {7}{2} (b B+a C)-\frac {1}{2} (7 A b+7 a B+5 b C) \cos (c+d x)-\frac {7}{2} a A \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (b B+a C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {4}{35} \int \frac {-\frac {5}{4} (7 A b+7 a B+5 b C)-\frac {7}{4} (5 a A+3 b B+3 a C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (b B+a C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {1}{5} (-5 a A-3 b B-3 a C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx-\frac {1}{7} (-7 A b-7 a B-5 b C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (b B+a C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (7 A b+7 a B+5 b C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 a A+3 b B+3 a C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {1}{5} (5 a A+3 b B+3 a C) \int \sqrt {\cos (c+d x)} \, dx-\frac {1}{21} (-7 A b-7 a B-5 b C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 (5 a A+3 b B+3 a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (7 A b+7 a B+5 b C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (b B+a C) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (7 A b+7 a B+5 b C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (5 a A+3 b B+3 a C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 4.49, size = 173, normalized size = 0.91 \begin {gather*} \frac {-42 (5 a A+3 b B+3 a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 (7 A b+7 a B+5 b C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {(70 A b+70 a B+110 b C+21 (15 a A+13 b B+13 a C) \cos (c+d x)+10 (7 A b+7 a B+5 b C) \cos (2 (c+d x))+105 a A \cos (3 (c+d x))+63 b B \cos (3 (c+d x))+63 a C \cos (3 (c+d x))) \sin (c+d x)}{2 \cos ^{\frac {7}{2}}(c+d x)}}{105 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Cos[c + d*x]^(3/2),x]

[Out]

(-42*(5*a*A + 3*b*B + 3*a*C)*EllipticE[(c + d*x)/2, 2] + 10*(7*A*b + 7*a*B + 5*b*C)*EllipticF[(c + d*x)/2, 2]
+ ((70*A*b + 70*a*B + 110*b*C + 21*(15*a*A + 13*b*B + 13*a*C)*Cos[c + d*x] + 10*(7*A*b + 7*a*B + 5*b*C)*Cos[2*
(c + d*x)] + 105*a*A*Cos[3*(c + d*x)] + 63*b*B*Cos[3*(c + d*x)] + 63*a*C*Cos[3*(c + d*x)])*Sin[c + d*x])/(2*Co
s[c + d*x]^(7/2)))/(105*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(823\) vs. \(2(222)=444\).
time = 0.38, size = 824, normalized size = 4.34

method	result	size
default	\(\text {Expression too large to display}\)	\(824\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a*A/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2
-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(2*sin(1/2
*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*(A*b+B*a)*(-1/6*c
os(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*C*b*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/
2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2/5*(B*b+C*a)/(8*sin(
1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)
^6*cos(1/2*d*x+1/2*c)-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1
/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(
1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2
*c)^2*cos(1/2*d*x+1/2*c)-3*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x
+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*
c)^2-1)^(1/2)/d

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Maxima [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))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.51, size = 288, normalized size = 1.52 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (7 i \, B a + i \, {\left (7 \, A + 5 \, C\right )} b\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-7 i \, B a - i \, {\left (7 \, A + 5 \, C\right )} b\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (i \, {\left (5 \, A + 3 \, C\right )} a + 3 i \, B b\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (-i \, {\left (5 \, A + 3 \, C\right )} a - 3 i \, B b\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (21 \, {\left ({\left (5 \, A + 3 \, C\right )} a + 3 \, B b\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (7 \, B a + {\left (7 \, A + 5 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} + 15 \, C b + 21 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*(5*sqrt(2)*(7*I*B*a + I*(7*A + 5*C)*b)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d
*x + c)) + 5*sqrt(2)*(-7*I*B*a - I*(7*A + 5*C)*b)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*s
in(d*x + c)) + 21*sqrt(2)*(I*(5*A + 3*C)*a + 3*I*B*b)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInvers
e(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*sqrt(2)*(-I*(5*A + 3*C)*a - 3*I*B*b)*cos(d*x + c)^4*weierstrassZ
eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(21*((5*A + 3*C)*a + 3*B*b)*cos(d*x
+ c)^3 + 5*(7*B*a + (7*A + 5*C)*b)*cos(d*x + c)^2 + 15*C*b + 21*(C*a + B*b)*cos(d*x + c))*sqrt(cos(d*x + c))*s
in(d*x + c))/(d*cos(d*x + c)^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \sec {\left (c + d x \right )}\right ) \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(3/2),x)

[Out]

Integral((a + b*sec(c + d*x))*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)/cos(c + d*x)**(3/2), x)

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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))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)/cos(d*x + c)^(3/2), x)

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Mupad [B]
time = 7.80, size = 223, normalized size = 1.17 \begin {gather*} \frac {6\,C\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,a\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+10\,B\,a\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {30\,C\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+70\,A\,b\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+42\,B\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{105\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*C*a*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 30*A*a*cos(c + d*x)^2*sin(c + d*x)*hypergeo
m([-1/4, 1/2], 3/4, cos(c + d*x)^2) + 10*B*a*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x
)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (30*C*b*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4,
cos(c + d*x)^2) + 70*A*b*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 42*B*b*cos
(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(105*d*cos(c + d*x)^(7/2)*(1 - cos(c + d*
x)^2)^(1/2))

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