3.87 \(\int (a+a \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=136 \[ -\frac{a^2 (3 A-8 C) \sin (c+d x)}{3 d \sqrt{a \cos (c+d x)+a}}+\frac{3 a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d}-\frac{a (3 A-2 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{3 d}+\frac{A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d} \]

[Out]

(3*a^(3/2)*A*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (a^2*(3*A - 8*C)*Sin[c + d*x])/(3*d
*Sqrt[a + a*Cos[c + d*x]]) - (a*(3*A - 2*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + (A*(a + a*Cos[c + d
*x])^(3/2)*Tan[c + d*x])/d

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Rubi [A]  time = 0.440984, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3044, 2976, 2981, 2773, 206} \[ -\frac{a^2 (3 A-8 C) \sin (c+d x)}{3 d \sqrt{a \cos (c+d x)+a}}+\frac{3 a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d}-\frac{a (3 A-2 C) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{3 d}+\frac{A \tan (c+d x) (a \cos (c+d x)+a)^{3/2}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^(3/2)*A*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (a^2*(3*A - 8*C)*Sin[c + d*x])/(3*d
*Sqrt[a + a*Cos[c + d*x]]) - (a*(3*A - 2*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) + (A*(a + a*Cos[c + d
*x])^(3/2)*Tan[c + d*x])/d

Rule 3044

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

Rule 2976

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

Rule 2981

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

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{d}+\frac{\int (a+a \cos (c+d x))^{3/2} \left (\frac{3 a A}{2}-\frac{1}{2} a (3 A-2 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{a}\\ &=-\frac{a (3 A-2 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{3 d}+\frac{A (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{d}+\frac{2 \int \sqrt{a+a \cos (c+d x)} \left (\frac{9 a^2 A}{4}-\frac{1}{4} a^2 (3 A-8 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{3 a}\\ &=-\frac{a^2 (3 A-8 C) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}-\frac{a (3 A-2 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{3 d}+\frac{A (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{d}+\frac{1}{2} (3 a A) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=-\frac{a^2 (3 A-8 C) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}-\frac{a (3 A-2 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{3 d}+\frac{A (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{d}-\frac{\left (3 a^2 A\right ) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}\\ &=\frac{3 a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}-\frac{a^2 (3 A-8 C) \sin (c+d x)}{3 d \sqrt{a+a \cos (c+d x)}}-\frac{a (3 A-2 C) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{3 d}+\frac{A (a+a \cos (c+d x))^{3/2} \tan (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.341132, size = 106, normalized size = 0.78 \[ \frac{a \sec \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sqrt{a (\cos (c+d x)+1)} \left (2 \sin \left (\frac{1}{2} (c+d x)\right ) (3 A+10 C \cos (c+d x)+C \cos (2 (c+d x))+C)+9 \sqrt{2} A \cos (c+d x) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]*(9*Sqrt[2]*A*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos
[c + d*x] + 2*(3*A + C + 10*C*Cos[c + d*x] + C*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(6*d)

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Maple [B]  time = 0.089, size = 474, normalized size = 3.5 \begin{align*}{\frac{1}{3\,d}\sqrt{a}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 16\,C\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( -18\,A\ln \left ( 4\,{\frac{a\sqrt{2}\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{2\,\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{2}}} \right ) a-18\,A\ln \left ( -4\,{\frac{\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-a\sqrt{2}\cos \left ( 1/2\,dx+c/2 \right ) +2\,a}{-2\,\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{2}}} \right ) a-56\,C\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+6\,A\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a}+9\,A\ln \left ( 4\,{\frac{a\sqrt{2}\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{2\,\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{2}}} \right ) a+9\,A\ln \left ( -4\,{\frac{\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-a\sqrt{2}\cos \left ( 1/2\,dx+c/2 \right ) +2\,a}{-2\,\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{2}}} \right ) a+24\,C\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \right ) \left ( 2\,\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{2} \right ) ^{-1} \left ( 2\,\cos \left ( 1/2\,dx+c/2 \right ) -\sqrt{2} \right ) ^{-1} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{a \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^(1/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(
1/2)*sin(1/2*d*x+1/2*c)^4+(-18*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(
1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a-18*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin
(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a-56*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1
/2))*sin(1/2*d*x+1/2*c)^2+6*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+9*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^
(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a))*a+9*A*ln(-4/(-2*cos(
1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+2
4*C*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/(2*cos(1/2*d*x+1/2*c)+2^(1/2))/(2*cos(1/2*d*x+1/2*c)-2^(1/
2))/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [B]  time = 1.95261, size = 1828, normalized size = 13.44 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*C*sqrt(a) - 3*(2*sqrt(2)*a*cos(7/2
*d*x + 7/2*c)*sin(2*d*x + 2*c) + 6*sqrt(2)*a*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) + (2*sqrt(2)*a*sin(3/2*d*x
+ 3/2*c) + 6*sqrt(2)*a*sin(1/2*d*x + 1/2*c) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*
sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(
1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*
d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) +
2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*
sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 + (2*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 6*sqrt(2)*a*sin(1/2*d*x +
1/2*c) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt
(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/
2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*
c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2
+ 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(2*d*x
+ 2*c)^2 - 4*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 4*sqrt(2)*a*sin(1/2*d*x + 1/2*c) - 2*(sqrt(2)*a*sin(3/2*d*x + 3/
2*c) - 5*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt
(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*
d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x
+ 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) -
3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(
1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt
(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*
d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x
+ 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) +
3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(
1/2*d*x + 1/2*c) + 2) - 2*(sqrt(2)*a*cos(2*d*x + 2*c) + sqrt(2)*a)*sin(7/2*d*x + 7/2*c) - 6*(sqrt(2)*a*cos(2*d
*x + 2*c) + sqrt(2)*a)*sin(5/2*d*x + 5/2*c) + 2*(3*sqrt(2)*a*cos(3/2*d*x + 3/2*c) + sqrt(2)*a*cos(1/2*d*x + 1/
2*c))*sin(2*d*x + 2*c))*A*sqrt(a)/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1))/d

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Fricas [A]  time = 1.73008, size = 463, normalized size = 3.4 \begin{align*} \frac{9 \,{\left (A a \cos \left (d x + c\right )^{2} + A a \cos \left (d x + c\right )\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \,{\left (2 \, C a \cos \left (d x + c\right )^{2} + 10 \, C a \cos \left (d x + c\right ) + 3 \, A a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{12 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(9*(A*a*cos(d*x + c)^2 + A*a*cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*
cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(2*C*a
*cos(d*x + c)^2 + 10*C*a*cos(d*x + c) + 3*A*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^2 + d*co
s(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 3.96866, size = 423, normalized size = 3.11 \begin{align*} \frac{3 \, \sqrt{2} A a^{\frac{7}{2}}{\left (\frac{3 \, \sqrt{2} \log \left (\frac{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}\right )}{a{\left | a \right |}} + \frac{8 \,{\left (3 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )} a}\right )} + \frac{16 \,{\left (2 \, \sqrt{2} C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, \sqrt{2} C a^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*sqrt(2)*A*a^(7/2)*(3*sqrt(2)*log(abs(2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 +
a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 +
4*sqrt(2)*abs(a) - 6*a))/(a*abs(a)) + 8*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)
)^2 - a)/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(1/2*d*x + 1/
2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)*a)) + 16*(2*sqrt(2)*C*a^3*tan(1/2*d*x + 1/2*c)^2 + 3*sqr
t(2)*C*a^3)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2))/d