3.795 \(\int \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x)) \, dx\)

Optimal. Leaf size=111 \[ \frac{10 a \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 a \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d}+\frac{10 a \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{6 b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 b \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d} \]

[Out]

(6*b*EllipticE[(c + d*x)/2, 2])/(5*d) + (10*a*EllipticF[(c + d*x)/2, 2])/(21*d) + (10*a*Sqrt[Cos[c + d*x]]*Sin
[c + d*x])/(21*d) + (2*b*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*a*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*d)

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Rubi [A]  time = 0.088547, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4225, 2748, 2635, 2639, 2641} \[ \frac{10 a F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{2 a \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d}+\frac{10 a \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{6 b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 b \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*b*EllipticE[(c + d*x)/2, 2])/(5*d) + (10*a*EllipticF[(c + d*x)/2, 2])/(21*d) + (10*a*Sqrt[Cos[c + d*x]]*Sin
[c + d*x])/(21*d) + (2*b*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*a*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*d)

Rule 4225

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(B + A*Sin[a + b*x]))/Sin[a
+ b*x], x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

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 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]

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]

Rubi steps

\begin{align*} \int \cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x)) \, dx &=\int \cos ^{\frac{5}{2}}(c+d x) (b+a \cos (c+d x)) \, dx\\ &=a \int \cos ^{\frac{7}{2}}(c+d x) \, dx+b \int \cos ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{2 b \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{7} (5 a) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{5} (3 b) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{6 b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{10 a \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 b \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{21} (5 a) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{6 b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{10 a F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{10 a \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 b \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.489563, size = 77, normalized size = 0.69 \[ \frac{50 a \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\sin (c+d x) \sqrt{\cos (c+d x)} (15 a \cos (2 (c+d x))+65 a+42 b \cos (c+d x))+126 b E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(126*b*EllipticE[(c + d*x)/2, 2] + 50*a*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(65*a + 42*b*Cos[c + d*
x] + 15*a*Cos[2*(c + d*x)])*Sin[c + d*x])/(105*d)

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Maple [A]  time = 1.383, size = 290, normalized size = 2.6 \begin{align*} -{\frac{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 ( 240\,a\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+ \left ( -360\,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 ( 280\,a+168\,b \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -80\,a-42\,b \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +25\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}a-63\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}b \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+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)*(240*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+
(-360*a-168*b)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(280*a+168*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(
-80*a-42*b)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+25*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*
c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a-63*(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)^2-1)^(1/2)*b)/(-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 )} \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+b*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((b*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 \cos \left (d x + c\right )^{3} \sec \left (d x + c\right ) + 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+b*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((b*cos(d*x + c)^3*sec(d*x + c) + 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+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 )} \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+b*sec(d*x+c)),x, algorithm="giac")

[Out]

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