3.338 \(\int \sec ^5(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=122 \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{8 f \sqrt{a+b}}+\frac{\tan (e+f x) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f}+\frac{3 a \tan (e+f x) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{8 f} \]

[Out]

(3*a^2*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(8*Sqrt[a + b]*f) + (3*a*Sec[e + f*x]*S
qrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x])/(8*f) + (Sec[e + f*x]^3*(a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x])/(4*
f)

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Rubi [A]  time = 0.120778, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3190, 378, 377, 206} \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{8 f \sqrt{a+b}}+\frac{\tan (e+f x) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f}+\frac{3 a \tan (e+f x) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{8 f} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^5*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(3*a^2*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(8*Sqrt[a + b]*f) + (3*a*Sec[e + f*x]*S
qrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x])/(8*f) + (Sec[e + f*x]^3*(a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x])/(4*
f)

Rule 3190

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

Rule 378

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

Rule 377

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

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

Mathematica [C]  time = 0.130156, size = 63, normalized size = 0.52 \[ \frac{a^2 \sin (e+f x) \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{(a+b) \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}\right )}{f \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^5*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(a^2*Hypergeometric2F1[1/2, 3, 3/2, ((a + b)*Sin[e + f*x]^2)/(a + b*Sin[e + f*x]^2)]*Sin[e + f*x])/(f*Sqrt[a +
b*Sin[e + f*x]^2])

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Maple [B]  time = 2.654, size = 406, normalized size = 3.3 \begin{align*}{\frac{1}{16\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}f} \left ( 2\,\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( a+b \right ) ^{5/2} \left ( 3\,a-2\,b \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+4\,\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( a+b \right ) ^{7/2}\sin \left ( fx+e \right ) -3\,{a}^{2} \left ( \ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ){a}^{2}+2\,\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ) ab+\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ){b}^{2}-\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ){a}^{2}-2\,\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ) ab-\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ){b}^{2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4} \right ) \left ( a+b \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^5*(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

1/16*(2*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*(3*a-2*b)*sin(f*x+e)*cos(f*x+e)^2+4*(a+b-b*cos(f*x+e)^2)^(1/2)*
(a+b)^(7/2)*sin(f*x+e)-3*a^2*(ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2
+2*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b+ln(2/(1+sin(f*x+e))*((a+b)
^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^2-ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^
(1/2)+b*sin(f*x+e)+a))*a^2-2*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b
-ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^2)*cos(f*x+e)^4)/(a+b)^(5/2)/
cos(f*x+e)^4/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sec \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^5*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*sec(f*x + e)^5, x)

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Fricas [A]  time = 6.28442, size = 1014, normalized size = 8.31 \begin{align*} \left [\frac{3 \, \sqrt{a + b} a^{2} \cos \left (f x + e\right )^{4} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \,{\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \,{\left ({\left (3 \, a^{2} + a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{32 \,{\left (a + b\right )} f \cos \left (f x + e\right )^{4}}, -\frac{3 \, a^{2} \sqrt{-a - b} \arctan \left (\frac{{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a - b}}{2 \,{\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{4} - 2 \,{\left ({\left (3 \, a^{2} + a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{16 \,{\left (a + b\right )} f \cos \left (f x + e\right )^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^5*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/32*(3*sqrt(a + b)*a^2*cos(f*x + e)^4*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 8*(a^2 + 3*a*b + 2*b^2)*co
s(f*x + e)^2 - 4*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b)*sin(f*x +
e) + 8*a^2 + 16*a*b + 8*b^2)/cos(f*x + e)^4) + 4*((3*a^2 + a*b - 2*b^2)*cos(f*x + e)^2 + 2*a^2 + 4*a*b + 2*b^2
)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a + b)*f*cos(f*x + e)^4), -1/16*(3*a^2*sqrt(-a - b)*arctan(1
/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(((a*b + b^2)*cos(f*x +
e)^2 - a^2 - 2*a*b - b^2)*sin(f*x + e)))*cos(f*x + e)^4 - 2*((3*a^2 + a*b - 2*b^2)*cos(f*x + e)^2 + 2*a^2 + 4
*a*b + 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a + b)*f*cos(f*x + e)^4)]

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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(sec(f*x+e)**5*(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sec \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^5*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*sec(f*x + e)^5, x)