3.578 \(\int (d \sec (e+f x))^{7/2} (a+b \tan (e+f x)) \, dx\)

Optimal. Leaf size=121 \[ -\frac {6 a d^4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {6 a d^3 \sin (e+f x) \sqrt {d \sec (e+f x)}}{5 f}+\frac {2 a d \sin (e+f x) (d \sec (e+f x))^{5/2}}{5 f}+\frac {2 b (d \sec (e+f x))^{7/2}}{7 f} \]

[Out]

2/7*b*(d*sec(f*x+e))^(7/2)/f+2/5*a*d*(d*sec(f*x+e))^(5/2)*sin(f*x+e)/f-6/5*a*d^4*(cos(1/2*e+1/2*f*x)^2)^(1/2)/
cos(1/2*e+1/2*f*x)*EllipticE(sin(1/2*e+1/2*f*x),2^(1/2))/f/cos(f*x+e)^(1/2)/(d*sec(f*x+e))^(1/2)+6/5*a*d^3*sin
(f*x+e)*(d*sec(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.09, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3486, 3768, 3771, 2639} \[ \frac {6 a d^3 \sin (e+f x) \sqrt {d \sec (e+f x)}}{5 f}-\frac {6 a d^4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 a d \sin (e+f x) (d \sec (e+f x))^{5/2}}{5 f}+\frac {2 b (d \sec (e+f x))^{7/2}}{7 f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(7/2)*(a + b*Tan[e + f*x]),x]

[Out]

(-6*a*d^4*EllipticE[(e + f*x)/2, 2])/(5*f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]) + (2*b*(d*Sec[e + f*x])^(7/
2))/(7*f) + (6*a*d^3*Sqrt[d*Sec[e + f*x]]*Sin[e + f*x])/(5*f) + (2*a*d*(d*Sec[e + f*x])^(5/2)*Sin[e + f*x])/(5
*f)

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 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3768

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int (d \sec (e+f x))^{7/2} (a+b \tan (e+f x)) \, dx &=\frac {2 b (d \sec (e+f x))^{7/2}}{7 f}+a \int (d \sec (e+f x))^{7/2} \, dx\\ &=\frac {2 b (d \sec (e+f x))^{7/2}}{7 f}+\frac {2 a d (d \sec (e+f x))^{5/2} \sin (e+f x)}{5 f}+\frac {1}{5} \left (3 a d^2\right ) \int (d \sec (e+f x))^{3/2} \, dx\\ &=\frac {2 b (d \sec (e+f x))^{7/2}}{7 f}+\frac {6 a d^3 \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac {2 a d (d \sec (e+f x))^{5/2} \sin (e+f x)}{5 f}-\frac {1}{5} \left (3 a d^4\right ) \int \frac {1}{\sqrt {d \sec (e+f x)}} \, dx\\ &=\frac {2 b (d \sec (e+f x))^{7/2}}{7 f}+\frac {6 a d^3 \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac {2 a d (d \sec (e+f x))^{5/2} \sin (e+f x)}{5 f}-\frac {\left (3 a d^4\right ) \int \sqrt {\cos (e+f x)} \, dx}{5 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}\\ &=-\frac {6 a d^4 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 b (d \sec (e+f x))^{7/2}}{7 f}+\frac {6 a d^3 \sqrt {d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac {2 a d (d \sec (e+f x))^{5/2} \sin (e+f x)}{5 f}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 69, normalized size = 0.57 \[ \frac {(d \sec (e+f x))^{7/2} \left (70 a \sin (2 (e+f x))+21 a \sin (4 (e+f x))-168 a \cos ^{\frac {7}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+40 b\right )}{140 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(7/2)*(a + b*Tan[e + f*x]),x]

[Out]

((d*Sec[e + f*x])^(7/2)*(40*b - 168*a*Cos[e + f*x]^(7/2)*EllipticE[(e + f*x)/2, 2] + 70*a*Sin[2*(e + f*x)] + 2
1*a*Sin[4*(e + f*x)]))/(140*f)

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b d^{3} \sec \left (f x + e\right )^{3} \tan \left (f x + e\right ) + a d^{3} \sec \left (f x + e\right )^{3}\right )} \sqrt {d \sec \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(7/2)*(a+b*tan(f*x+e)),x, algorithm="fricas")

[Out]

integral((b*d^3*sec(f*x + e)^3*tan(f*x + e) + a*d^3*sec(f*x + e)^3)*sqrt(d*sec(f*x + e)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(7/2)*(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(7/2)*(b*tan(f*x + e) + a), x)

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maple [C]  time = 0.92, size = 371, normalized size = 3.07 \[ \frac {2 \left (1+\cos \left (f x +e \right )\right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2} \left (21 i \left (\cos ^{4}\left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a -21 i \left (\cos ^{4}\left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a +21 i \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a -21 i \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a -21 a \left (\cos ^{4}\left (f x +e \right )\right )+14 a \left (\cos ^{3}\left (f x +e \right )\right )+7 a \cos \left (f x +e \right )+5 b \sin \left (f x +e \right )\right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}}}{35 f \sin \left (f x +e \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(7/2)*(a+b*tan(f*x+e)),x)

[Out]

2/35/f*(1+cos(f*x+e))^2*(-1+cos(f*x+e))^2*(21*I*cos(f*x+e)^4*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e
)))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*a-21*I*cos(f*x+e)^4*(1/(1+cos(f*x+e)))^(1/2)*(c
os(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*a+21*I*cos(f*x+e)^3*(1/(1
+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*a-2
1*I*cos(f*x+e)^3*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(-1+cos(f*x+e))/sin(f*
x+e),I)*sin(f*x+e)*a-21*a*cos(f*x+e)^4+14*a*cos(f*x+e)^3+7*a*cos(f*x+e)+5*b*sin(f*x+e))*(d/cos(f*x+e))^(7/2)/s
in(f*x+e)^5

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(7/2)*(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(7/2)*(b*tan(f*x + e) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}\,\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(7/2)*(a + b*tan(e + f*x)),x)

[Out]

int((d/cos(e + f*x))^(7/2)*(a + b*tan(e + f*x)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(7/2)*(a+b*tan(f*x+e)),x)

[Out]

Timed out

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