3.319 \(\int \sqrt{a+b \sec (c+d x)} \tan ^3(c+d x) \, dx\)
Optimal. Leaf size=100 \[ \frac{2 (a+b \sec (c+d x))^{5/2}}{5 b^2 d}-\frac{2 a (a+b \sec (c+d x))^{3/2}}{3 b^2 d}-\frac{2 \sqrt{a+b \sec (c+d x)}}{d}+\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{d} \]
[Out]
(2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/d - (2*Sqrt[a + b*Sec[c + d*x]])/d - (2*a*(a + b*Sec[c +
d*x])^(3/2))/(3*b^2*d) + (2*(a + b*Sec[c + d*x])^(5/2))/(5*b^2*d)
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Rubi [A] time = 0.112788, antiderivative size = 100, 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 =
{3885, 898, 1261, 207} \[ \frac{2 (a+b \sec (c+d x))^{5/2}}{5 b^2 d}-\frac{2 a (a+b \sec (c+d x))^{3/2}}{3 b^2 d}-\frac{2 \sqrt{a+b \sec (c+d x)}}{d}+\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x]^3,x]
[Out]
(2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/d - (2*Sqrt[a + b*Sec[c + d*x]])/d - (2*a*(a + b*Sec[c +
d*x])^(3/2))/(3*b^2*d) + (2*(a + b*Sec[c + d*x])^(5/2))/(5*b^2*d)
Rule 3885
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Rule 898
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 + a*e^2)/e^2 - (2*c
*d*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]
Rule 1261
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Rule 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \sqrt{a+b \sec (c+d x)} \tan ^3(c+d x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+x} \left (b^2-x^2\right )}{x} \, dx,x,b \sec (c+d x)\right )}{b^2 d}\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{x^2 \left (-a^2+b^2+2 a x^2-x^4\right )}{-a+x^2} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{b^2 d}\\ &=-\frac{2 \operatorname{Subst}\left (\int \left (b^2+a x^2-x^4+\frac{a b^2}{-a+x^2}\right ) \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{b^2 d}\\ &=-\frac{2 \sqrt{a+b \sec (c+d x)}}{d}-\frac{2 a (a+b \sec (c+d x))^{3/2}}{3 b^2 d}+\frac{2 (a+b \sec (c+d x))^{5/2}}{5 b^2 d}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{-a+x^2} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{d}\\ &=\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{d}-\frac{2 \sqrt{a+b \sec (c+d x)}}{d}-\frac{2 a (a+b \sec (c+d x))^{3/2}}{3 b^2 d}+\frac{2 (a+b \sec (c+d x))^{5/2}}{5 b^2 d}\\ \end{align*}
Mathematica [A] time = 6.2172, size = 194, normalized size = 1.94 \[ \frac{\sqrt{a+b \sec (c+d x)} \left (-\frac{2 \left (2 a^2+15 b^2\right )}{15 b^2}+\frac{2 a \sec (c+d x)}{15 b}+\frac{2}{5} \sec ^2(c+d x)\right )}{d}+\frac{\sin ^2(c+d x) \sqrt{a \cos (c+d x)} \sqrt{a+b \sec (c+d x)} \left (\log \left (\frac{\sqrt{a \cos (c+d x)+b}}{\sqrt{a \cos (c+d x)}}+1\right )-\log \left (1-\frac{\sqrt{a \cos (c+d x)+b}}{\sqrt{a \cos (c+d x)}}\right )\right )}{d \left (1-\cos ^2(c+d x)\right ) \sqrt{a \cos (c+d x)+b}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x]^3,x]
[Out]
(Sqrt[a + b*Sec[c + d*x]]*((-2*(2*a^2 + 15*b^2))/(15*b^2) + (2*a*Sec[c + d*x])/(15*b) + (2*Sec[c + d*x]^2)/5))
/d + (Sqrt[a*Cos[c + d*x]]*(-Log[1 - Sqrt[b + a*Cos[c + d*x]]/Sqrt[a*Cos[c + d*x]]] + Log[1 + Sqrt[b + a*Cos[c
+ d*x]]/Sqrt[a*Cos[c + d*x]]])*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x]^2)/(d*Sqrt[b + a*Cos[c + d*x]]*(1 - Cos[
c + d*x]^2))
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Maple [B] time = 0.497, size = 2342, normalized size = 23.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sec(d*x+c))^(1/2)*tan(d*x+c)^3,x)
[Out]
1/60/d/b^2/(a-b)^(1/2)*(cos(d*x+c)+1)*(-1+cos(d*x+c))^4*(30*cos(d*x+c)^4*ln(-1/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*
cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+
a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a*b^3-30*cos(d*x+c)^4*ln(-2/(a-b
)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos
(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a*b^3
+30*cos(d*x+c)^2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*b^2-8*cos(d*x+c)^4*((b+a*cos
(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*a^3-54*cos(d*x+c)^4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(
d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*b^3-8*cos(d*x+c)^5*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(
1/2)*a^3-54*cos(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*b^3-6*cos(d*x+c)^5*(
a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*b^2-6*cos(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*x+c
)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*b^2+36*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(
a-b)^(1/2)*b^2+30*cos(d*x+c)^5*ln(-1/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(c
os(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)
^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a^2*b^2-15*cos(d*x+c)^5*ln(-1/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((
b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))
*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a*b^3-30*cos(d*x+c)^5*ln(-2/(a-b)^(1/2)*(-1+c
os(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos
(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a^2*b^2+15*cos(d*x
+c)^5*ln(-2/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-
b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin
(d*x+c)^2)*a*b^3-18*cos(d*x+c)^4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*b^2-54*cos(d
*x+c)^4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*a*b^2-8*cos(d*x+c)^4*((b+a*cos(d*x+c)
)*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*a^2*b-8*cos(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)
+1)^2)^(1/2)*(a-b)^(1/2)*a^2*b-54*cos(d*x+c)^5*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2
)*a*b^2+30*cos(d*x+c)^5*(a-b)^(1/2)*a^(3/2)*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/
2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*b^2+30*cos(d*x+c
)^4*(a-b)^(1/2)*a^(1/2)*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d
*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*b^3+12*cos(d*x+c)^3*((b+a*cos(d*x+c)
)*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a*b+4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^
2)^(3/2)*(a-b)^(1/2)*a*b+4*cos(d*x+c)^4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a*b+1
2*cos(d*x+c)^2*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*a*b-15*cos(d*x+c)^4*ln(-1/(a-b
)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos
(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*b^4+1
5*cos(d*x+c)^4*ln(-2/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^
(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/
2)-b)/sin(d*x+c)^2)*b^4+12*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*b^2)*((b+a*cos(d*x
+c))/cos(d*x+c))^(1/2)*4^(1/2)/sin(d*x+c)^8/cos(d*x+c)^2/((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 3.62269, size = 779, normalized size = 7.79 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{2} \cos \left (d x + c\right )^{2} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} - 4 \,{\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) + 4 \,{\left (a b \cos \left (d x + c\right ) -{\left (2 \, a^{2} + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{30 \, b^{2} d \cos \left (d x + c\right )^{2}}, -\frac{15 \, \sqrt{-a} b^{2} \arctan \left (\frac{2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) \cos \left (d x + c\right )^{2} - 2 \,{\left (a b \cos \left (d x + c\right ) -{\left (2 \, a^{2} + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{15 \, b^{2} d \cos \left (d x + c\right )^{2}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="fricas")
[Out]
[1/30*(15*sqrt(a)*b^2*cos(d*x + c)^2*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) - b^2 - 4*(2*a*cos(d*x + c
)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))) + 4*(a*b*cos(d*x + c) - (2*a^2 + 15*b^2
)*cos(d*x + c)^2 + 3*b^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(b^2*d*cos(d*x + c)^2), -1/15*(15*sqrt(-a)*
b^2*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b))*cos(d*x + c
)^2 - 2*(a*b*cos(d*x + c) - (2*a^2 + 15*b^2)*cos(d*x + c)^2 + 3*b^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/
(b^2*d*cos(d*x + c)^2)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sec{\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))**(1/2)*tan(d*x+c)**3,x)
[Out]
Integral(sqrt(a + b*sec(c + d*x))*tan(c + d*x)**3, x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (d x + c\right ) + a} \tan \left (d x + c\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="giac")
[Out]
integrate(sqrt(b*sec(d*x + c) + a)*tan(d*x + c)^3, x)