3.67 \(\int \sqrt{a+b \sec ^2(e+f x)} \sin ^5(e+f x) \, dx\)

Optimal. Leaf size=139 \[ \frac{2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f} \]

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/f - (Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]
^2])/f + (2*(5*a + b)*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2))/(15*a^2*f) - (Cos[e + f*x]^5*(a + b*Sec[e +
f*x]^2)^(3/2))/(5*a*f)

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Rubi [A]  time = 0.146946, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4134, 462, 451, 277, 217, 206} \[ \frac{2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sec[e + f*x]^2]*Sin[e + f*x]^5,x]

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/f - (Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]
^2])/f + (2*(5*a + b)*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2))/(15*a^2*f) - (Cos[e + f*x]^5*(a + b*Sec[e +
f*x]^2)^(3/2))/(5*a*f)

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rule 462

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

Rule 451

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

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \sqrt{a+b \sec ^2(e+f x)} \sin ^5(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2 \sqrt{a+b x^2}}{x^6} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}+\frac{\operatorname{Subst}\left (\int \frac{\left (-2 (5 a+b)+5 a x^2\right ) \sqrt{a+b x^2}}{x^4} \, dx,x,\sec (e+f x)\right )}{5 a f}\\ &=\frac{2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{x^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f}\\ &=\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}\\ \end{align*}

Mathematica [A]  time = 0.889095, size = 152, normalized size = 1.09 \[ -\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)} \left (\frac{2 \left (a \cos ^2(e+f x)+b\right )^{5/2}}{5 a^2}-\frac{2 (2 a+b) \left (a \cos ^2(e+f x)+b\right )^{3/2}}{3 a^2}+2 \sqrt{a \cos ^2(e+f x)+b}-2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{a \cos ^2(e+f x)+b}}{\sqrt{b}}\right )\right )}{\sqrt{2} f \sqrt{a \cos (2 e+2 f x)+a+2 b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sec[e + f*x]^2]*Sin[e + f*x]^5,x]

[Out]

-((Cos[e + f*x]*(-2*Sqrt[b]*ArcTanh[Sqrt[b + a*Cos[e + f*x]^2]/Sqrt[b]] + 2*Sqrt[b + a*Cos[e + f*x]^2] - (2*(2
*a + b)*(b + a*Cos[e + f*x]^2)^(3/2))/(3*a^2) + (2*(b + a*Cos[e + f*x]^2)^(5/2))/(5*a^2))*Sqrt[a + b*Sec[e + f
*x]^2])/(Sqrt[2]*f*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]))

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Maple [B]  time = 0.745, size = 1840, normalized size = 13.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

-1/30/f/b^(1/2)/(a+b)^(3/2)/a^2*(-1+cos(f*x+e))^2*(30*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*4
^(1/2)*(a+b)^(3/2)*b^(1/2)*a^2-4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*(a+b)^(3/2)*b^(3/2)-16*cos(f*x+e)
^3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*(a+b)^(3/2)*b^(3/2)-24*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(
f*x+e))^2)^(3/2)*(a+b)^(3/2)*b^(3/2)-15*cos(f*x+e)*4^(1/2)*(a+b)^(3/2)*arctanh(1/8*b^(1/2)*4^(1/2)*(-1+cos(f*x
+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2))*a^2
*b-16*cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*(a+b)^(3/2)*b^(3/2)+15*4^(1/2)*b^(5/2)*ln(-4/(a+b
)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a
*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^2-15*4^(1/2)*b^(5/2)*ln(-2/(a+b)^(1/2)*(
-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+
e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^2+10*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*
(a+b)^(3/2)*b^(1/2)*a+15*4^(1/2)*b^(3/2)*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+
cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f
*x+e)^2)*a^3-15*4^(1/2)*b^(3/2)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e
))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*
a^3-4*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*(a+b)^(3/2)*b^(3/2)-15*((b+a*cos(f*x+e)^2)/(1+c
os(f*x+e))^2)^(1/2)*4^(1/2)*(a+b)^(3/2)*b^(3/2)*a+15*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*4^(1/2)*(a+b)
^(3/2)*b^(1/2)*a^2+6*cos(f*x+e)^6*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*(a+b)^(3/2)*b^(1/2)*a+24*cos(f*x
+e)^5*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*(a+b)^(3/2)*b^(1/2)*a+16*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1
+cos(f*x+e))^2)^(3/2)*(a+b)^(3/2)*b^(1/2)*a-56*cos(f*x+e)^3*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*(a+b)^
(3/2)*b^(1/2)*a-84*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*(a+b)^(3/2)*b^(1/2)*a+15*cos(f*x+e
)*4^(1/2)*b^(5/2)*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a
+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^2-15*cos(f*x
+e)*4^(1/2)*b^(5/2)*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*
(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^2-20*cos(f
*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*(a+b)^(3/2)*b^(1/2)*a+15*cos(f*x+e)*4^(1/2)*b^(3/2)*ln(-4/(a
+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b
+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^3-15*cos(f*x+e)*4^(1/2)*b^(3/2)*ln(-2/
(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+(
(b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a^3-15*4^(1/2)*(a+b)^(3/2)*arctanh(1/8
*b^(1/2)*4^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/
(1+cos(f*x+e))^2)^(1/2))*a^2*b)*cos(f*x+e)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)/((b+a*cos(f*x+e)^2)/(1+cos(
f*x+e))^2)^(1/2)/sin(f*x+e)^4

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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(sin(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.18855, size = 734, normalized size = 5.28 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{b} \log \left (\frac{a \cos \left (f x + e\right )^{2} + 2 \, \sqrt{b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \,{\left (3 \, a^{2} \cos \left (f x + e\right )^{5} -{\left (10 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{3} +{\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{30 \, a^{2} f}, -\frac{15 \, a^{2} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) +{\left (3 \, a^{2} \cos \left (f x + e\right )^{5} -{\left (10 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{3} +{\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, a^{2} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/30*(15*a^2*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x +
e) + 2*b)/cos(f*x + e)^2) - 2*(3*a^2*cos(f*x + e)^5 - (10*a^2 - a*b)*cos(f*x + e)^3 + (15*a^2 - 10*a*b - 2*b^2
)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^2*f), -1/15*(15*a^2*sqrt(-b)*arctan(sqrt(-b)*s
qrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + (3*a^2*cos(f*x + e)^5 - (10*a^2 - a*b)*cos(f*x +
e)^3 + (15*a^2 - 10*a*b - 2*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^2*f)]

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Sympy [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(sin(f*x+e)**5*(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(f*x + e)^2 + a)*sin(f*x + e)^5, x)