3.6.73 \(\int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [573]

Optimal. Leaf size=214 \[ -\frac {2 \left (5 a^2 A+3 A b^2+6 a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2 A+3 A b^2+6 a b B\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]

[Out]

-2/5*(5*A*a^2+3*A*b^2+6*B*a*b)*(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*(14*A*a*b+7*B*a^2+5*B*b^2)*(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^2*B*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/5*b*(A*b+2*B*a)*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/2
1*(14*A*a*b+7*B*a^2+5*B*b^2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*(5*A*a^2+3*A*b^2+6*B*a*b)*sin(d*x+c)/d/cos(d*x+
c)^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3033, 3067, 3100, 2827, 2716, 2720, 2719} \begin {gather*} \frac {2 \left (7 a^2 B+14 a A b+5 b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 \left (5 a^2 A+6 a b B+3 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2 A+6 a b B+3 A b^2\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b (2 a B+A b) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 b^2 B \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])^2*(A + B*Sec[c + d*x]))/Cos[c + d*x]^(3/2),x]

[Out]

(-2*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(14*a*A*b + 7*a^2*B + 5*b^2*B)*Ellipti
cF[(c + d*x)/2, 2])/(21*d) + (2*b^2*B*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + (2*b*(A*b + 2*a*B)*Sin[c + d*x]
)/(5*d*Cos[c + d*x]^(5/2)) + (2*(14*a*A*b + 7*a^2*B + 5*b^2*B)*Sin[c + d*x])/(21*d*Cos[c + d*x]^(3/2)) + (2*(5
*a^2*A + 3*A*b^2 + 6*a*b*B)*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 3033

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]

Rule 3067

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

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]

Rubi steps

\begin {align*} \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx &=\int \frac {(b+a \cos (c+d x))^2 (B+A \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {2}{7} \int \frac {-\frac {7}{2} b (A b+2 a B)-\frac {1}{2} \left (14 a A b+7 a^2 B+5 b^2 B\right ) \cos (c+d x)-\frac {7}{2} a^2 A \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {4}{35} \int \frac {-\frac {5}{4} \left (14 a A b+7 a^2 B+5 b^2 B\right )-\frac {7}{4} \left (5 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {1}{5} \left (-5 a^2 A-3 A b^2-6 a b B\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx-\frac {1}{7} \left (-14 a A b-7 a^2 B-5 b^2 B\right ) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2 A+3 A b^2+6 a b B\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {1}{5} \left (5 a^2 A+3 A b^2+6 a b B\right ) \int \sqrt {\cos (c+d x)} \, dx-\frac {1}{21} \left (-14 a A b-7 a^2 B-5 b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 \left (5 a^2 A+3 A b^2+6 a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2 A+3 A b^2+6 a b B\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 4.80, size = 191, normalized size = 0.89 \begin {gather*} \frac {2 \left (-21 \left (5 a^2 A+3 A b^2+6 a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \left (14 a A b+7 a^2 B+5 b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\frac {15 b^2 B \sin (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)}+\frac {21 b (A b+2 a B) \sin (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}+\frac {5 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {21 \left (5 a^2 A+3 A b^2+6 a b B\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{105 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-21*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*EllipticE[(c + d*x)/2, 2] + 5*(14*a*A*b + 7*a^2*B + 5*b^2*B)*EllipticF[(
c + d*x)/2, 2] + (15*b^2*B*Sin[c + d*x])/Cos[c + d*x]^(7/2) + (21*b*(A*b + 2*a*B)*Sin[c + d*x])/Cos[c + d*x]^(
5/2) + (5*(14*a*A*b + 7*a^2*B + 5*b^2*B)*Sin[c + d*x])/Cos[c + d*x]^(3/2) + (21*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*
Sin[c + d*x])/Sqrt[Cos[c + d*x]]))/(105*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(831\) vs. \(2(246)=492\).
time = 8.92, size = 832, normalized size = 3.89

method	result	size
default	\(\text {Expression too large to display}\)	\(832\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^2*(A+B*sec(d*x+c))/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*(2*A*b+B*a)*(-1/6*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+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)))+2/5*b*(A*b+2*B*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*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1
/2*c)+12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*s
in(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)+2*
a^2*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)*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2)))+2*b^2*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))))/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))^2*(A+B*sec(d*x+c))/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 = 0.59, size = 314, normalized size = 1.47 \begin {gather*} -\frac {5 \, \sqrt {2} {\left (7 i \, B a^{2} + 14 i \, A a b + 5 i \, B b^{2}\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^{2} - 14 i \, A a b - 5 i \, B b^{2}\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 (5 i \, A a^{2} + 6 i \, B a b + 3 i \, A b^{2}\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 (-5 i \, A a^{2} - 6 i \, B a b - 3 i \, A b^{2}\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 (5 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, B b^{2} + 5 \, {\left (7 \, B a^{2} + 14 \, A a b + 5 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 21 \, {\left (2 \, B a b + A b^{2}\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))^2*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-1/105*(5*sqrt(2)*(7*I*B*a^2 + 14*I*A*a*b + 5*I*B*b^2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c)
+ I*sin(d*x + c)) + 5*sqrt(2)*(-7*I*B*a^2 - 14*I*A*a*b - 5*I*B*b^2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0,
cos(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(5*I*A*a^2 + 6*I*B*a*b + 3*I*A*b^2)*cos(d*x + c)^4*weierstrassZeta
(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*sqrt(2)*(-5*I*A*a^2 - 6*I*B*a*b - 3*I*
A*b^2)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(2
1*(5*A*a^2 + 6*B*a*b + 3*A*b^2)*cos(d*x + c)^3 + 15*B*b^2 + 5*(7*B*a^2 + 14*A*a*b + 5*B*b^2)*cos(d*x + c)^2 +
21*(2*B*a*b + A*b^2)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(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 )}\right )^{2}}{\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))**2*(A+B*sec(d*x+c))/cos(d*x+c)**(3/2),x)

[Out]

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

[Out]

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

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Mupad [B]
time = 4.98, size = 233, normalized size = 1.09 \begin {gather*} \frac {6\,A\,b^2\,\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^2\,{\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 )+20\,A\,a\,b\,\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\,B\,b^2\,\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\,B\,a^2\,{\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 )+84\,B\,a\,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))^2)/cos(c + d*x)^(3/2),x)

[Out]

(6*A*b^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 30*A*a^2*cos(c + d*x)^2*sin(c + d*x)*hype
rgeom([-1/4, 1/2], 3/4, cos(c + d*x)^2) + 20*A*a*b*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*B*b^2*sin(c + d*x)*hypergeom([-7/4, 1/2
], -3/4, cos(c + d*x)^2) + 70*B*a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2) +
84*B*a*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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