3.116 \(\int \frac{\sin ^5(e+f x)}{\sqrt{a+b \tan ^2(e+f x)}} \, dx\)

Optimal. Leaf size=144 \[ -\frac{\left (15 a^2-10 a b+3 b^2\right ) \cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{15 f (a-b)^3}-\frac{\cos ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{5 f (a-b)}+\frac{2 (5 a-3 b) \cos ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{15 f (a-b)^2} \]

[Out]

-((15*a^2 - 10*a*b + 3*b^2)*Cos[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(15*(a - b)^3*f) + (2*(5*a - 3*b)*Cos
[e + f*x]^3*Sqrt[a - b + b*Sec[e + f*x]^2])/(15*(a - b)^2*f) - (Cos[e + f*x]^5*Sqrt[a - b + b*Sec[e + f*x]^2])
/(5*(a - b)*f)

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Rubi [A]  time = 0.144111, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3664, 462, 453, 264} \[ -\frac{\left (15 a^2-10 a b+3 b^2\right ) \cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{15 f (a-b)^3}-\frac{\cos ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{5 f (a-b)}+\frac{2 (5 a-3 b) \cos ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{15 f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*a^2 - 10*a*b + 3*b^2)*Cos[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2])/(15*(a - b)^3*f) + (2*(5*a - 3*b)*Cos
[e + f*x]^3*Sqrt[a - b + b*Sec[e + f*x]^2])/(15*(a - b)^2*f) - (Cos[e + f*x]^5*Sqrt[a - b + b*Sec[e + f*x]^2])
/(5*(a - b)*f)

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

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 453

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[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 264

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

Rubi steps

\begin{align*} \int \frac{\sin ^5(e+f x)}{\sqrt{a+b \tan ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2}{x^6 \sqrt{a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{5 (a-b) f}+\frac{\operatorname{Subst}\left (\int \frac{-2 (5 a-3 b)+5 (a-b) x^2}{x^4 \sqrt{a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=\frac{2 (5 a-3 b) \cos ^3(e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{5 (a-b) f}+\frac{\left (15 a^2-10 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^2 f}\\ &=-\frac{\left (15 a^2-10 a b+3 b^2\right ) \cos (e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{15 (a-b)^3 f}+\frac{2 (5 a-3 b) \cos ^3(e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{5 (a-b) f}\\ \end{align*}

Mathematica [A]  time = 2.02296, size = 112, normalized size = 0.78 \[ \frac{\cos (e+f x) \left (4 \left (7 a^2-10 a b+3 b^2\right ) \cos (2 (e+f x))-89 a^2-3 (a-b)^2 \cos (4 (e+f x))+34 a b-9 b^2\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{120 \sqrt{2} f (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(-89*a^2 + 34*a*b - 9*b^2 + 4*(7*a^2 - 10*a*b + 3*b^2)*Cos[2*(e + f*x)] - 3*(a - b)^2*Cos[4*(e +
f*x)])*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(120*Sqrt[2]*(a - b)^3*f)

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Maple [A]  time = 0.286, size = 169, normalized size = 1.2 \begin{align*} -{\frac{ \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}-6\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}ab+3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{2}-10\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}+16\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}ab-6\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{2}+15\,{a}^{2}-10\,ab+3\,{b}^{2} \right ) }{15\,f \left ( a-b \right ) ^{3}\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15/f/(a-b)^3*(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)*(3*cos(f*x+e)^4*a^2-6*cos(f*x+e)^4*a*b+3*cos(f*x+e)^4*b^2-10
*cos(f*x+e)^2*a^2+16*cos(f*x+e)^2*a*b-6*cos(f*x+e)^2*b^2+15*a^2-10*a*b+3*b^2)/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+
b)/cos(f*x+e)^2)^(1/2)/cos(f*x+e)

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Maxima [A]  time = 1.03985, size = 289, normalized size = 2.01 \begin{align*} -\frac{\frac{15 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a - b} + \frac{3 \,{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{5}{2}} \cos \left (f x + e\right )^{5} - 10 \,{\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} b \cos \left (f x + e\right )^{3} + 15 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac{10 \,{\left ({\left (a - b + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3} - 3 \, \sqrt{a - b + \frac{b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{2} - 2 \, a b + b^{2}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)/(a - b) + (3*(a - b + b/cos(f*x + e)^2)^(5/2)*cos(f*x +
e)^5 - 10*(a - b + b/cos(f*x + e)^2)^(3/2)*b*cos(f*x + e)^3 + 15*sqrt(a - b + b/cos(f*x + e)^2)*b^2*cos(f*x +
e))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 10*((a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 3*sqrt(a - b + b/c
os(f*x + e)^2)*b*cos(f*x + e))/(a^2 - 2*a*b + b^2))/f

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Fricas [A]  time = 2.04331, size = 292, normalized size = 2.03 \begin{align*} -\frac{{\left (3 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 2 \,{\left (5 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{3} +{\left (15 \, a^{2} - 10 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 2*(5*a^2 - 8*a*b + 3*b^2)*cos(f*x + e)^3 + (15*a^2 - 10*a*b + 3*
b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*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*tan(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 \frac{\sin \left (f x + e\right )^{5}}{\sqrt{b \tan \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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