3.6.78 \(\int (d \sec (e+f x))^{7/2} (a+b \tan (e+f x)) \, dx\) [578]

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

[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*f*x+1/2*e)^2)^(1/2)/
cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),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 = 0.07, 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 = {3567, 3853, 3856, 2719} \begin {gather*} -\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} \end {gather*}

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 2719

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

Rule 3567

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 3853

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 3856

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 = 0.68, size = 69, normalized size = 0.57 \begin {gather*} \frac {(d \sec (e+f x))^{7/2} \left (40 b-168 a \cos ^{\frac {7}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+70 a \sin (2 (e+f x))+21 a \sin (4 (e+f x))\right )}{140 f} \end {gather*}

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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Maple [C] Result contains complex when optimal does not.
time = 8.41, size = 371, normalized size = 3.07

method	result	size
default	\(-\frac {2 \left (\cos \left (f x +e \right )+1\right )^{2} \left (\cos \left (f x +e \right )-1\right )^{2} \left (21 i \left (\cos ^{4}\left (f x +e \right )\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a -21 i \left (\cos ^{4}\left (f x +e \right )\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a +21 i \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a -21 i \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticE \left (\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) a +21 \left (\cos ^{4}\left (f x +e \right )\right ) a -14 \left (\cos ^{3}\left (f x +e \right )\right ) a -7 a \cos \left (f x +e \right )-5 b \sin \left (f x +e \right )\right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}}}{35 f \sin \left (f x +e \right )^{5}}\)	\(371\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 156, normalized size = 1.29 \begin {gather*} \frac {-21 i \, \sqrt {2} a d^{\frac {7}{2}} \cos \left (f x + e\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 21 i \, \sqrt {2} a d^{\frac {7}{2}} \cos \left (f x + e\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (5 \, b d^{3} + 7 \, {\left (3 \, a d^{3} \cos \left (f x + e\right )^{3} + a d^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{35 \, f \cos \left (f x + e\right )^{3}} \end {gather*}

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]

1/35*(-21*I*sqrt(2)*a*d^(7/2)*cos(f*x + e)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) +
I*sin(f*x + e))) + 21*I*sqrt(2)*a*d^(7/2)*cos(f*x + e)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos
(f*x + e) - I*sin(f*x + e))) + 2*(5*b*d^3 + 7*(3*a*d^3*cos(f*x + e)^3 + a*d^3*cos(f*x + e))*sin(f*x + e))*sqrt
(d/cos(f*x + e)))/(f*cos(f*x + e)^3)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 3878 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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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