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

Optimal. Leaf size=127 \[ \frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{f}+\frac{(a-2 b) \sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{2 f}+\frac{(a+b) \tan (e+f x) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.145834, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3190, 413, 523, 217, 206, 377} \[ \frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{f}+\frac{(a-2 b) \sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{2 f}+\frac{(a+b) \tan (e+f x) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^(3/2)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/f + ((a - 2*b)*Sqrt[a + b]*ArcTanh[(Sqrt[
a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(2*f) + ((a + b)*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2]*Tan
[e + f*x])/(2*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 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, 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])

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]

Rubi steps

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

Mathematica [A]  time = 0.969861, size = 210, normalized size = 1.65 \[ \frac{2 \left (a^2-a b-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2 a+2 b} \sin (e+f x)}{\sqrt{2 a-b \cos (2 (e+f x))+b}}\right )-2 b^2 \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a+b} \sin (e+f x)}{\sqrt{2 a-b \cos (2 (e+f x))+b}}\right )+\sqrt{a+b} \left (4 b^{3/2} \log \left (\sqrt{2 a-b \cos (2 (e+f x))+b}+\sqrt{2} \sqrt{b} \sin (e+f x)\right )+\sqrt{2} (a+b) \tan (e+f x) \sec (e+f x) \sqrt{2 a-b \cos (2 (e+f x))+b}\right )}{4 f \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2*ArcTanh[(Sqrt[2]*Sqrt[a + b]*Sin[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]] + 2*(a^2 - a*b - b^2)*A
rcTanh[(Sqrt[2*a + 2*b]*Sin[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]] + Sqrt[a + b]*(4*b^(3/2)*Log[Sqrt[2*
a + b - b*Cos[2*(e + f*x)]] + Sqrt[2]*Sqrt[b]*Sin[e + f*x]] + Sqrt[2]*(a + b)*Sqrt[2*a + b - b*Cos[2*(e + f*x)
]]*Sec[e + f*x]*Tan[e + f*x]))/(4*Sqrt[a + b]*f)

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Maple [B]  time = 3.408, size = 402, normalized size = 3.2 \begin{align*}{\frac{1}{4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}f} \left ( 2\,\sin \left ( fx+e \right ) \sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( a+b \right ) ^{5/2}- \left ( -4\,{b}^{3/2}\ln \left ( \sin \left ( fx+e \right ) \sqrt{b}+\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \right ) \left ( a+b \right ) ^{3/2}-\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}^{3}+3\,\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{b}^{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 ){b}^{3}+\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}^{3}-3\,\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{b}^{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 ){b}^{3} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) \left ( a+b \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*sin(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)-(-4*b^(3/2)*ln(sin(f*x+e)*b^(1/2)+(a+b-b*cos(f*x+e)^2
)^(1/2))*(a+b)^(3/2)-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^3+3*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^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))*b^3+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^3-3*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^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))*b^3)*cos(f*x+e)^2)/(a+b)^(3/2)/c
os(f*x+e)^2/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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(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)^3, x)

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Fricas [B]  time = 7.19163, size = 3646, normalized size = 28.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(b^(3/2)*cos(f*x + e)^2*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2*b^4)*cos(f*x + e)^6 + 32*(5*a^2*b^2 +
24*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 + 32*a^3*b + 160*a^2*b^2 + 256*a*b^3 + 128*b^4 - 32*(a^3*b + 10*a^2*b
^2 + 24*a*b^3 + 16*b^4)*cos(f*x + e)^2 - 8*(16*b^3*cos(f*x + e)^6 - 24*(a*b^2 + 2*b^3)*cos(f*x + e)^4 - a^3 -
10*a^2*b - 24*a*b^2 - 16*b^3 + 2*(5*a^2*b + 24*a*b^2 + 24*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)
*sqrt(b)*sin(f*x + e)) - sqrt(a + b)*(a - 2*b)*cos(f*x + e)^2*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 8*(a
^2 + 3*a*b + 2*b^2)*cos(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*sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)
*sin(f*x + e))/(f*cos(f*x + e)^2), -1/8*(2*(a - 2*b)*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)^2 - b^(3/2)*cos(f*x + e)^2*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2*b^4)*cos(f*x + e)^6
+ 32*(5*a^2*b^2 + 24*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 + 32*a^3*b + 160*a^2*b^2 + 256*a*b^3 + 128*b^4 - 32*
(a^3*b + 10*a^2*b^2 + 24*a*b^3 + 16*b^4)*cos(f*x + e)^2 - 8*(16*b^3*cos(f*x + e)^6 - 24*(a*b^2 + 2*b^3)*cos(f*
x + e)^4 - a^3 - 10*a^2*b - 24*a*b^2 - 16*b^3 + 2*(5*a^2*b + 24*a*b^2 + 24*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*
x + e)^2 + a + b)*sqrt(b)*sin(f*x + e)) - 4*sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)*sin(f*x + e))/(f*cos(f*x +
e)^2), -1/8*(2*sqrt(-b)*b*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + 2*b^2)*cos(f*x + e)^2 + a^2 + 8*a*b + 8
*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)/((2*b^3*cos(f*x + e)^4 + a^2*b + 3*a*b^2 + 2*b^3 - (3*a*b^2 + 4
*b^3)*cos(f*x + e)^2)*sin(f*x + e)))*cos(f*x + e)^2 + sqrt(a + b)*(a - 2*b)*cos(f*x + e)^2*log(((a^2 + 8*a*b +
8*b^2)*cos(f*x + e)^4 - 8*(a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^2 + 4*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqr
t(-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*sqrt(-b*co
s(f*x + e)^2 + a + b)*(a + b)*sin(f*x + e))/(f*cos(f*x + e)^2), -1/4*((a - 2*b)*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)^2 + sqrt(-b)*b*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + 2*b^2
)*cos(f*x + e)^2 + a^2 + 8*a*b + 8*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)/((2*b^3*cos(f*x + e)^4 + a^2*
b + 3*a*b^2 + 2*b^3 - (3*a*b^2 + 4*b^3)*cos(f*x + e)^2)*sin(f*x + e)))*cos(f*x + e)^2 - 2*sqrt(-b*cos(f*x + e)
^2 + a + b)*(a + b)*sin(f*x + e))/(f*cos(f*x + e)^2)]

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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)**3*(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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3*(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)^3, x)