3.178 \(\int \frac{\tan ^4(c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=125 \[ \frac{2 a^2 \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}+\frac{2 a \tan ^3(c+d x)}{3 d (a \sec (c+d x)+a)^{3/2}}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 \tan (c+d x)}{d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d) - (2*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c
+ d*x]]) + (2*a*Tan[c + d*x]^3)/(3*d*(a + a*Sec[c + d*x])^(3/2)) + (2*a^2*Tan[c + d*x]^5)/(5*d*(a + a*Sec[c +
d*x])^(5/2))

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Rubi [A]  time = 0.0838666, antiderivative size = 125, normalized size of antiderivative = 1., 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, 459, 302, 203} \[ \frac{2 a^2 \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}+\frac{2 a \tan ^3(c+d x)}{3 d (a \sec (c+d x)+a)^{3/2}}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 \tan (c+d x)}{d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d) - (2*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c
+ d*x]]) + (2*a*Tan[c + d*x]^3)/(3*d*(a + a*Sec[c + d*x])^(3/2)) + (2*a^2*Tan[c + d*x]^5)/(5*d*(a + a*Sec[c +
d*x])^(5/2))

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]

Rule 459

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

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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])

Rubi steps

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

Mathematica [C]  time = 2.62243, size = 238, normalized size = 1.9 \[ \frac{16 \sqrt{2} \tan ^5(c+d x) \left (\frac{1}{\sec (c+d x)+1}\right )^{9/2} \left (-\frac{4}{9} \tan ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \text{Hypergeometric2F1}\left (2,\frac{9}{2},\frac{11}{2},-2 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )-\frac{\cos (c+d x) (5 \cos (c+d x)+9) \csc ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left ((22 \cos (c+d x)-23 \cos (2 (c+d x))-29) \sqrt{1-\sec (c+d x)}+30 \cos ^2(c+d x) \tanh ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )\right )}{480 \sqrt{1-\sec (c+d x)}}\right )}{5 d \left (1-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )^{7/2} \sqrt{a (\sec (c+d x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[c + d*x]^4/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(16*Sqrt[2]*((1 + Sec[c + d*x])^(-1))^(9/2)*(-(Cos[c + d*x]*(9 + 5*Cos[c + d*x])*Csc[(c + d*x)/2]^6*Sec[(c + d
*x)/2]^2*(30*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Cos[c + d*x]^2 + (-29 + 22*Cos[c + d*x] - 23*Cos[2*(c + d*x)])*Sq
rt[1 - Sec[c + d*x]]))/(480*Sqrt[1 - Sec[c + d*x]]) - (4*Hypergeometric2F1[2, 9/2, 11/2, -2*Sec[c + d*x]*Sin[(
c + d*x)/2]^2]*Sec[c + d*x]*Tan[(c + d*x)/2]^2)/9)*Tan[c + d*x]^5)/(5*d*Sqrt[a*(1 + Sec[c + d*x])]*(1 - Tan[(c
+ d*x)/2]^2)^(7/2))

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Maple [B]  time = 0.237, size = 231, normalized size = 1.9 \begin{align*}{\frac{1}{30\,ad \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 15\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +15\,\sqrt{2} \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +68\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}-64\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-16\,\cos \left ( dx+c \right ) +12 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^4/(a+a*sec(d*x+c))^(1/2),x)

[Out]

1/30/d/a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(15*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)^2*sin(d*x+c)+15*2^(1/2)*(-2*cos(d*x
+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(
d*x+c)*cos(d*x+c)+68*cos(d*x+c)^3-64*cos(d*x+c)^2-16*cos(d*x+c)+12)/cos(d*x+c)^2/sin(d*x+c)

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Maxima [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(tan(d*x+c)^4/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.76342, size = 832, normalized size = 6.66 \begin{align*} \left [-\frac{15 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt{-a} \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 \,{\left (17 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) - 3\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{15 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}, -\frac{2 \,{\left (15 \,{\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \sqrt{a} \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 ) +{\left (17 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right ) - 3\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{15 \,{\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[-1/15*(15*(cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(-a)*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*(17*cos(d*x + c)^
2 + cos(d*x + c) - 3)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x
+ c)^2), -2/15*(15*(cos(d*x + c)^3 + cos(d*x + c)^2)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*co
s(d*x + c)/(sqrt(a)*sin(d*x + c))) + (17*cos(d*x + c)^2 + cos(d*x + c) - 3)*sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*sin(d*x + c))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\left (c + d x \right )}}{\sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^4/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError