3.478 \(\int \sec ^5(c+d x) \sqrt{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=207 \[ -\frac{\left (12 a^2-18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 d (a-b)^{3/2}}+\frac{\left (12 a^2+18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 d (a+b)^{3/2}}-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 d \left (a^2-b^2\right )}+\frac{\tan (c+d x) \sec ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{4 d} \]

[Out]

-((12*a^2 - 18*a*b + 5*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(32*(a - b)^(3/2)*d) + ((12*a^2 + 1
8*a*b + 5*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(32*(a + b)^(3/2)*d) - (Sec[c + d*x]^2*Sqrt[a +
b*Sin[c + d*x]]*(a*b - (6*a^2 - 5*b^2)*Sin[c + d*x]))/(16*(a^2 - b^2)*d) + (Sec[c + d*x]^3*Sqrt[a + b*Sin[c +
d*x]]*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.323333, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2668, 737, 823, 827, 1166, 206} \[ -\frac{\left (12 a^2-18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 d (a-b)^{3/2}}+\frac{\left (12 a^2+18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 d (a+b)^{3/2}}-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 d \left (a^2-b^2\right )}+\frac{\tan (c+d x) \sec ^3(c+d x) \sqrt{a+b \sin (c+d x)}}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

-((12*a^2 - 18*a*b + 5*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(32*(a - b)^(3/2)*d) + ((12*a^2 + 1
8*a*b + 5*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(32*(a + b)^(3/2)*d) - (Sec[c + d*x]^2*Sqrt[a +
b*Sin[c + d*x]]*(a*b - (6*a^2 - 5*b^2)*Sin[c + d*x]))/(16*(a^2 - b^2)*d) + (Sec[c + d*x]^3*Sqrt[a + b*Sin[c +
d*x]]*Tan[c + d*x])/(4*d)

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*a*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(d*(2*p + 3) + e*(m + 2*p + 3)*x)*(a + c*x^2
)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (LtQ[m, 1]
|| (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

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(c+d x) \sqrt{a+b \sin (c+d x)} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{\sqrt{a+x}}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \tan (c+d x)}{4 d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{-3 a-\frac{5 x}{2}}{\sqrt{a+x} \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \tan (c+d x)}{4 d}+\frac{b \operatorname{Subst}\left (\int \frac{\frac{1}{4} a \left (12 a^2-13 b^2\right )+\frac{1}{4} \left (6 a^2-5 b^2\right ) x}{\sqrt{a+x} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right ) d}\\ &=-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \tan (c+d x)}{4 d}+\frac{b \operatorname{Subst}\left (\int \frac{\frac{1}{4} a \left (12 a^2-13 b^2\right )-\frac{1}{4} a \left (6 a^2-5 b^2\right )+\frac{1}{4} \left (6 a^2-5 b^2\right ) x^2}{-a^2+b^2+2 a x^2-x^4} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \tan (c+d x)}{4 d}-\frac{\left (12 a^2-18 a b+5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{32 (a-b) d}+\frac{\left (12 a^2+18 a b+5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{32 (a+b) d}\\ &=-\frac{\left (12 a^2-18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 (a-b)^{3/2} d}+\frac{\left (12 a^2+18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 (a+b)^{3/2} d}-\frac{\sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \left (a b-\left (6 a^2-5 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right ) d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.48296, size = 224, normalized size = 1.08 \[ \frac{-\sqrt{a-b} (a+b)^2 \left (12 a^2-18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )+(a-b)^2 \sqrt{a+b} \left (12 a^2+18 a b+5 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )+\frac{1}{2} \left (a^2-b^2\right ) \sec ^4(c+d x) \sqrt{a+b \sin (c+d x)} \left (\left (22 a^2-21 b^2\right ) \sin (c+d x)+6 a^2 \sin (3 (c+d x))-2 a b \cos (2 (c+d x))-2 a b-5 b^2 \sin (3 (c+d x))\right )}{32 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(-(Sqrt[a - b]*(a + b)^2*(12*a^2 - 18*a*b + 5*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]]) + (a - b)^2*
Sqrt[a + b]*(12*a^2 + 18*a*b + 5*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + ((a^2 - b^2)*Sec[c + d*x
]^4*Sqrt[a + b*Sin[c + d*x]]*(-2*a*b - 2*a*b*Cos[2*(c + d*x)] + (22*a^2 - 21*b^2)*Sin[c + d*x] + 6*a^2*Sin[3*(
c + d*x)] - 5*b^2*Sin[3*(c + d*x)]))/2)/(32*(a^2 - b^2)^2*d)

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Maple [B]  time = 0.717, size = 509, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16/d/(b*sin(d*x+c)+b)^2*b/(a-b)*(a+b*sin(d*x+c))^(3/2)*a+5/32/d/(b*sin(d*x+c)+b)^2*b^2/(a-b)*(a+b*sin(d*x+c
))^(3/2)+3/16/d/(b*sin(d*x+c)+b)^2*b*(a+b*sin(d*x+c))^(1/2)*a-7/32/d/(b*sin(d*x+c)+b)^2*b^2*(a+b*sin(d*x+c))^(
1/2)+3/8/d/(a-b)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a^2-9/16/d/(a-b)/(-a+b)^(1/2)*arctan
((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a*b+5/32/d/(a-b)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2)
)*b^2-3/16/d/(b*sin(d*x+c)-b)^2*b/(a+b)*(a+b*sin(d*x+c))^(3/2)*a-5/32/d/(b*sin(d*x+c)-b)^2*b^2/(a+b)*(a+b*sin(
d*x+c))^(3/2)+3/16/d/(b*sin(d*x+c)-b)^2*b*(a+b*sin(d*x+c))^(1/2)*a+7/32/d/(b*sin(d*x+c)-b)^2*b^2*(a+b*sin(d*x+
c))^(1/2)+3/8/d/(a+b)^(3/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a^2+9/16/d/(a+b)^(3/2)*arctanh((a+b*si
n(d*x+c))^(1/2)/(a+b)^(1/2))*a*b+5/32/d/(a+b)^(3/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*b^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(sec(d*x+c)^5*(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sin(d*x + c) + a)*sec(d*x + c)^5, x)

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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(d*x+c)**5*(a+b*sin(d*x+c))**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sin(d*x + c) + a)*sec(d*x + c)^5, x)