3.5.94 \(\int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx\) [494]

Optimal. Leaf size=175 \[ \frac {16 a (6 A+7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {8 a (6 A+7 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (6 A+7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

8/105*a*(6*A+7*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/35*a*(6*A+7*B)*sec(d*x+c)^(5/2)*sin(d
*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/7*a*A*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+16/105*a*(6*A+7*B)
*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3040, 3059, 2851, 2850} \begin {gather*} \frac {2 a (6 A+7 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{35 d \sqrt {a \cos (c+d x)+a}}+\frac {8 a (6 A+7 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {16 a (6 A+7 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{105 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a A \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x]^(9/2),x]

[Out]

(16*a*(6*A + 7*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (8*a*(6*A + 7*B)*Sec[c +
d*x]^(3/2)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*(6*A + 7*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])
/(35*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a*A*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(7*d*Sqrt[a + a*Cos[c + d*x]])

Rule 2850

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b,
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2851

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x]
+ Dist[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 3040

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

Rule 3059

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)) \sec ^{\frac {9}{2}}(c+d x) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{7} \left ((6 A+7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a (6 A+7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{35} \left (4 (6 A+7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {8 a (6 A+7 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (6 A+7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {1}{105} \left (8 (6 A+7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {16 a (6 A+7 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {8 a (6 A+7 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a (6 A+7 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.50, size = 102, normalized size = 0.58 \begin {gather*} \frac {2 \sqrt {a (1+\cos (c+d x))} (27 A+14 B+9 (6 A+7 B) \cos (c+d x)+2 (6 A+7 B) \cos (2 (c+d x))+12 A \cos (3 (c+d x))+14 B \cos (3 (c+d x))) \sec ^{\frac {7}{2}}(c+d x) \tan \left (\frac {1}{2} (c+d x)\right )}{105 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x]^(9/2),x]

[Out]

(2*Sqrt[a*(1 + Cos[c + d*x])]*(27*A + 14*B + 9*(6*A + 7*B)*Cos[c + d*x] + 2*(6*A + 7*B)*Cos[2*(c + d*x)] + 12*
A*Cos[3*(c + d*x)] + 14*B*Cos[3*(c + d*x)])*Sec[c + d*x]^(7/2)*Tan[(c + d*x)/2])/(105*d)

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Maple [A]
time = 0.39, size = 116, normalized size = 0.66

method	result	size
default	\(-\frac {2 \left (-1+\cos \left (d x +c \right )\right ) \left (48 A \left (\cos ^{3}\left (d x +c \right )\right )+56 B \left (\cos ^{3}\left (d x +c \right )\right )+24 A \left (\cos ^{2}\left (d x +c \right )\right )+28 B \left (\cos ^{2}\left (d x +c \right )\right )+18 A \cos \left (d x +c \right )+21 B \cos \left (d x +c \right )+15 A \right ) \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{105 d \sin \left (d x +c \right )}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/105/d*(-1+cos(d*x+c))*(48*A*cos(d*x+c)^3+56*B*cos(d*x+c)^3+24*A*cos(d*x+c)^2+28*B*cos(d*x+c)^2+18*A*cos(d*x
+c)+21*B*cos(d*x+c)+15*A)*cos(d*x+c)*(1/cos(d*x+c))^(9/2)*(a*(1+cos(d*x+c)))^(1/2)/sin(d*x+c)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 568 vs. \(2 (151) = 302\).
time = 0.60, size = 568, normalized size = 3.25 \begin {gather*} \frac {2 \, {\left (\frac {3 \, A {\left (\frac {35 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {70 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {84 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {58 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {9 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{4}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (\frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {\sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 1\right )}} + \frac {7 \, B {\left (\frac {15 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {42 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {24 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {7 \, \sqrt {2} \sqrt {a} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{4}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (\frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {\sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + 1\right )}}\right )}}{105 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2/105*(3*A*(35*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 70*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 84*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 58*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*
x + c) + 1)^7 + 9*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(4*sin(d*x + c)
^2/(cos(d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin(
d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)) + 7*B*(15*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 40*sqrt(2)*s
qrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 42*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 24*sqrt(
2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 7*sqrt(2)*sqrt(a)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*(sin(d
*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c
) + 1) + 1)^(9/2)*(4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x +
c)^6/(cos(d*x + c) + 1)^6 + sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)))/d

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Fricas [A]
time = 0.37, size = 104, normalized size = 0.59 \begin {gather*} \frac {2 \, {\left (8 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (6 \, A + 7 \, B\right )} \cos \left (d x + c\right ) + 15 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )} \sqrt {\cos \left (d x + c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/105*(8*(6*A + 7*B)*cos(d*x + c)^3 + 4*(6*A + 7*B)*cos(d*x + c)^2 + 3*(6*A + 7*B)*cos(d*x + c) + 15*A)*sqrt(a
*cos(d*x + c) + a)*sin(d*x + c)/((d*cos(d*x + c)^4 + d*cos(d*x + c)^3)*sqrt(cos(d*x + c)))

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Sympy [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*cos(d*x+c))*sec(d*x+c)**(9/2)*(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [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*cos(d*x+c))*sec(d*x+c)^(9/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 4.55, size = 441, normalized size = 2.52 \begin {gather*} \frac {\sqrt {\frac {1}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}\right )}\,\left (96\,A+112\,B\right )\,1{}\mathrm {i}}{105\,d}-\frac {{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}\,\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}\right )}\,\left (96\,A+112\,B\right )\,1{}\mathrm {i}}{105\,d}+\frac {{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}\right )}\,\left (336\,A+392\,B\right )\,1{}\mathrm {i}}{105\,d}-\frac {{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}\right )}\,\left (336\,A+392\,B\right )\,1{}\mathrm {i}}{105\,d}-\frac {B\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}\right )}\,8{}\mathrm {i}}{3\,d}+\frac {B\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}\right )}\,8{}\mathrm {i}}{3\,d}\right )}{{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+3\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+3\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}+3\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}+3\,{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}+{\mathrm {e}}^{c\,6{}\mathrm {i}+d\,x\,6{}\mathrm {i}}+{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(9/2)*(a + a*cos(c + d*x))^(1/2),x)

[Out]

((1/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(((a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1
i)/2))^(1/2)*(96*A + 112*B)*1i)/(105*d) - (exp(c*7i + d*x*7i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*
1i)/2))^(1/2)*(96*A + 112*B)*1i)/(105*d) + (exp(c*2i + d*x*2i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x
*1i)/2))^(1/2)*(336*A + 392*B)*1i)/(105*d) - (exp(c*5i + d*x*5i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d
*x*1i)/2))^(1/2)*(336*A + 392*B)*1i)/(105*d) - (B*exp(c*3i + d*x*3i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i
+ d*x*1i)/2))^(1/2)*8i)/(3*d) + (B*exp(c*4i + d*x*4i)*(a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))
^(1/2)*8i)/(3*d)))/(exp(c*1i + d*x*1i) + 3*exp(c*2i + d*x*2i) + 3*exp(c*3i + d*x*3i) + 3*exp(c*4i + d*x*4i) +
3*exp(c*5i + d*x*5i) + exp(c*6i + d*x*6i) + exp(c*7i + d*x*7i) + 1)

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