3.202 \(\int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=113 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac {4 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac {2 \tan (c+d x)}{a^2 d \sqrt {a \sec (c+d x)+a}} \]

[Out]

2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-4*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec
(d*x+c))^(1/2))*2^(1/2)/a^(5/2)/d+2*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 113, 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 = {3887, 479, 522, 203} \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac {4 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac {2 \tan (c+d x)}{a^2 d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d) - (4*Sqrt[2]*ArcTan[(Sqrt[a]*Tan[c + d
*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(a^(5/2)*d) + (2*Tan[c + d*x])/(a^2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 479

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(2*n
- 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q) + 1)), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 3887

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[(-2*a^(m/2 +
n + 1/2))/d, Subst[Int[(x^m*(2 + a*x^2)^(m/2 + n - 1/2))/(1 + a*x^2), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rubi steps

\begin {align*} \int \frac {\tan ^4(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac {2 \operatorname {Subst}\left (\int \frac {x^4}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {2 \tan (c+d x)}{a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {2 \operatorname {Subst}\left (\int \frac {2+3 a x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^2 d}\\ &=\frac {2 \tan (c+d x)}{a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^2 d}+\frac {8 \operatorname {Subst}\left (\int \frac {1}{2+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^2 d}\\ &=\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {4 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {2 \tan (c+d x)}{a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 23.74, size = 5491, normalized size = 48.59 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[c + d*x]^4/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

Result too large to show

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fricas [A]  time = 0.52, size = 414, normalized size = 3.66 \[ \left [\frac {2 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - \sqrt {-a} {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{a^{3} d \cos \left (d x + c\right ) + a^{3} d}, -\frac {2 \, {\left (\sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {2 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}} - \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{a^{3} d \cos \left (d x + c\right ) + a^{3} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[(2*sqrt(2)*(a*cos(d*x + c) + a)*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*
cos(d*x + c)*sin(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) - 1)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - sq
rt(-a)*(cos(d*x + c) + 1)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x
+ c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d
*x + c))/(a^3*d*cos(d*x + c) + a^3*d), -2*(sqrt(a)*(cos(d*x + c) + 1)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*sqrt(2)*(a*cos(d*x + c) + a)*arctan(sqrt(2)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a) - sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*s
in(d*x + c))/(a^3*d*cos(d*x + c) + a^3*d)]

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giac [A]  time = 3.05, size = 73, normalized size = 0.65 \[ \frac {2 \, \sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{2} d \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)*a^2*d*sgn(t
an(1/2*d*x + 1/2*c)^2 - 1))

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maple [B]  time = 1.21, size = 326, normalized size = 2.88 \[ -\frac {\left (\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \cos \left (d x +c \right )+\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )+4 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \cos \left (d x +c \right )+4 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )-2 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{d \left (1+\cos \left (d x +c \right )\right ) a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x)

[Out]

-1/d*(2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)
/cos(d*x+c)*2^(1/2))*cos(d*x+c)+2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-(-2*cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)+4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-
(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-2*sin(d*x+c))*(a*(1+cos(d*x+c))/co
s(d*x+c))^(1/2)/(1+cos(d*x+c))/a^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{4}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(tan(d*x + c)^4/(a*sec(d*x + c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^4}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^4/(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(tan(c + d*x)^4/(a + a/cos(c + d*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**4/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral(tan(c + d*x)**4/(a*(sec(c + d*x) + 1))**(5/2), x)

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