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

Optimal. Leaf size=132 \[ \frac{2 a (5 A+7 B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{6 a (A+B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a (A+B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 a (5 A+7 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{2 a A \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d} \]

[Out]

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

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Rubi [A]  time = 0.229494, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {2954, 2968, 3023, 2748, 2635, 2641, 2639} \[ \frac{2 a (5 A+7 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{6 a (A+B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a (A+B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 a (5 A+7 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{2 a A \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2954

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 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 2635

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

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

Mathematica [C]  time = 6.28382, size = 872, normalized size = 6.61 \[ a \left (\sqrt{\cos (c+d x)} (\cos (c+d x)+1) \left (-\frac{3 (A+B) \cot (c)}{5 d}+\frac{(23 A+28 B) \cos (d x) \sin (c)}{84 d}+\frac{(A+B) \cos (2 d x) \sin (2 c)}{10 d}+\frac{A \cos (3 d x) \sin (3 c)}{28 d}+\frac{(23 A+28 B) \cos (c) \sin (d x)}{84 d}+\frac{(A+B) \cos (2 c) \sin (2 d x)}{10 d}+\frac{A \cos (3 c) \sin (3 d x)}{28 d}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )-\frac{3 A (\cos (c+d x)+1) \csc (c) \left (\frac{\text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt{\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}} \sqrt{\tan ^2(c)+1}}-\frac{\frac{2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac{\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{\tan ^2(c)+1}}}{\sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}}}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{10 d}-\frac{3 B (\cos (c+d x)+1) \csc (c) \left (\frac{\text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt{\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}} \sqrt{\tan ^2(c)+1}}-\frac{\frac{2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac{\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt{\tan ^2(c)+1}}}{\sqrt{\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt{\tan ^2(c)+1}}}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{10 d}-\frac{5 A (\cos (c+d x)+1) \csc (c) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{-\sqrt{\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{21 d \sqrt{\cot ^2(c)+1}}-\frac{B (\cos (c+d x)+1) \csc (c) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{-\sqrt{\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d \sqrt{\cot ^2(c)+1}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*((-3*(A + B)*Cot[c])/(5*d) + ((23*A + 28*B)*Cos[
d*x]*Sin[c])/(84*d) + ((A + B)*Cos[2*d*x]*Sin[2*c])/(10*d) + (A*Cos[3*d*x]*Sin[3*c])/(28*d) + ((23*A + 28*B)*C
os[c]*Sin[d*x])/(84*d) + ((A + B)*Cos[2*c]*Sin[2*d*x])/(10*d) + (A*Cos[3*c]*Sin[3*d*x])/(28*d)) - (5*A*(1 + Co
s[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^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[Co
t[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(21*d*Sqrt[1 + Cot[c]^2]) - (B*(1 + Cos[c + d*x])*Csc[c]*Hyperge
ometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^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*Sqrt[1 + Cot[c]^2]) - (3*A*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^2*((Hypergeom
etricPFQ[{-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 + ArcTa
n[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]
]))/(10*d) - (3*B*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^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])/(Co
s[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(10*d))

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Maple [B]  time = 1.923, size = 383, normalized size = 2.9 \begin{align*} -{\frac{2\,a}{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 ( 240\,A\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+ \left ( -528\,A-168\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( 448\,A+308\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -122\,A-112\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +25\,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 ) +35\,B\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\,B\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))*(A+B*sec(d*x+c)),x)

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(240*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8+(-528*A-168*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(448*A+308*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)
+(-122*A-112*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+25*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+35*B*(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*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El
lipticE(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )} \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))*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )} \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))*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

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