3.1092 \(\int \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=164 \[ \frac{8 a^2 (3 A+7 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{4 a^2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 (33 A+35 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{105 d}+\frac{8 A \sin (c+d x) \sqrt{\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{35 d}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^2}{7 d} \]

[Out]

(4*a^2*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (8*a^2*(3*A + 7*C)*EllipticF[(c + d*x)/2, 2])/(21*d) + (
2*a^2*(33*A + 35*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*A*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*
Sin[c + d*x])/(7*d) + (8*A*Sqrt[Cos[c + d*x]]*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(35*d)

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Rubi [A]  time = 0.466033, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {4114, 3046, 2976, 2968, 3023, 2748, 2641, 2639} \[ \frac{8 a^2 (3 A+7 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a^2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 (33 A+35 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{105 d}+\frac{8 A \sin (c+d x) \sqrt{\cos (c+d x)} \left (a^2 \cos (c+d x)+a^2\right )}{35 d}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)} (a \cos (c+d x)+a)^2}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

(4*a^2*(3*A + 5*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (8*a^2*(3*A + 7*C)*EllipticF[(c + d*x)/2, 2])/(21*d) + (
2*a^2*(33*A + 35*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(105*d) + (2*A*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*
Sin[c + d*x])/(7*d) + (8*A*Sqrt[Cos[c + d*x]]*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(35*d)

Rule 4114

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

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*
sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Simp
[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b
, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-
1)] && NeQ[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 2968

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

Rule 3023

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

Rule 2748

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 2641

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

Rule 2639

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

Rubi steps

\begin{align*} \int \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\int \frac{(a+a \cos (c+d x))^2 \left (C+A \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 A \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac{2 \int \frac{(a+a \cos (c+d x))^2 \left (\frac{1}{2} a (A+7 C)+2 a A \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{7 a}\\ &=\frac{2 A \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac{8 A \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d}+\frac{4 \int \frac{(a+a \cos (c+d x)) \left (\frac{1}{4} a^2 (9 A+35 C)+\frac{1}{4} a^2 (33 A+35 C) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx}{35 a}\\ &=\frac{2 A \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac{8 A \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d}+\frac{4 \int \frac{\frac{1}{4} a^3 (9 A+35 C)+\left (\frac{1}{4} a^3 (9 A+35 C)+\frac{1}{4} a^3 (33 A+35 C)\right ) \cos (c+d x)+\frac{1}{4} a^3 (33 A+35 C) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{35 a}\\ &=\frac{2 a^2 (33 A+35 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 A \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac{8 A \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d}+\frac{8 \int \frac{\frac{5}{2} a^3 (3 A+7 C)+\frac{21}{4} a^3 (3 A+5 C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{105 a}\\ &=\frac{2 a^2 (33 A+35 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 A \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac{8 A \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d}+\frac{1}{5} \left (2 a^2 (3 A+5 C)\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{21} \left (4 a^2 (3 A+7 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{4 a^2 (3 A+5 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^2 (3 A+7 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a^2 (33 A+35 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{105 d}+\frac{2 A \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d}+\frac{8 A \sqrt{\cos (c+d x)} \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{35 d}\\ \end{align*}

Mathematica [C]  time = 6.37914, size = 1070, normalized size = 6.52 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

(Cos[c + d*x]^(9/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*((-2*(3*A + 5*C)*Cot[c]
)/(5*d) + ((51*A + 28*C)*Cos[d*x]*Sin[c])/(84*d) + (A*Cos[2*d*x]*Sin[2*c])/(5*d) + (A*Cos[3*d*x]*Sin[3*c])/(28
*d) + ((51*A + 28*C)*Cos[c]*Sin[d*x])/(84*d) + (A*Cos[2*c]*Sin[2*d*x])/(5*d) + (A*Cos[3*c]*Sin[3*d*x])/(28*d))
)/(A + 2*C + A*Cos[2*c + 2*d*x]) - (4*A*Cos[c + d*x]^4*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - A
rcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]
*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin
[d*x - ArcTan[Cot[c]]]])/(7*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]) - (4*C*Cos[c + d*x]^4*Csc[c]*
HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*
(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]
*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*S
qrt[1 + Cot[c]^2]) - (3*A*Cos[c + d*x]^4*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x
]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(
Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*
Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2
*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sq
rt[1 + Tan[c]^2]]))/(5*d*(A + 2*C + A*Cos[2*c + 2*d*x])) - (C*Cos[c + d*x]^4*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a +
a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]
*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sq
rt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[
c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt
[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(d*(A + 2*C + A*Cos[2*c + 2*d*x]))

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Maple [A]  time = 2.148, size = 380, normalized size = 2.3 \begin{align*} -{\frac{4\,{a}^{2}}{105\,d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 120\,A \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}\cos \left ( 1/2\,dx+c/2 \right ) -348\,A \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) + \left ( 378\,A+70\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -117\,A-35\,C \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +30\,A\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -63\,A\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +70\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -105\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x)

[Out]

-4/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(120*A*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2
*c)-348*A*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(378*A+70*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-117*A
-35*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+30*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1
/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-63*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticE(cos(1/2*d*x+1/2*c),2^(1/2))+70*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))-105*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(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)]  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(cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C a^{2} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{4} + 2 \, C a^{2} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{3} +{\left (A + C\right )} a^{2} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + 2 \, A a^{2} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right ) + A a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*a^2*cos(d*x + c)^3*sec(d*x + c)^4 + 2*C*a^2*cos(d*x + c)^3*sec(d*x + c)^3 + (A + C)*a^2*cos(d*x +
c)^3*sec(d*x + c)^2 + 2*A*a^2*cos(d*x + c)^3*sec(d*x + c) + A*a^2*cos(d*x + c)^3)*sqrt(cos(d*x + c)), x)

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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(cos(d*x+c)**(7/2)*(a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)^2*cos(d*x + c)^(7/2), x)