3.488 \(\int \sec ^5(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=188 \[ -\frac{3 \left (4 a^2-2 a b-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 d \sqrt{a-b}}+\frac{3 \left (4 a^2+2 a b-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 d \sqrt{a+b}}+\frac{\sec ^4(c+d x) (a \sin (c+d x)+b) \sqrt{a+b \sin (c+d x)}}{4 d}-\frac{\sec ^2(c+d x) (b-6 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{16 d} \]

[Out]

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

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Rubi [A]  time = 0.318461, antiderivative size = 188, 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, 739, 823, 827, 1166, 206} \[ -\frac{3 \left (4 a^2-2 a b-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 d \sqrt{a-b}}+\frac{3 \left (4 a^2+2 a b-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 d \sqrt{a+b}}+\frac{\sec ^4(c+d x) (a \sin (c+d x)+b) \sqrt{a+b \sin (c+d x)}}{4 d}-\frac{\sec ^2(c+d x) (b-6 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(4*a^2 - 2*a*b - b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(32*Sqrt[a - b]*d) + (3*(4*a^2 + 2*a*
b - b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(32*Sqrt[a + b]*d) - (Sec[c + d*x]^2*(b - 6*a*Sin[c +
d*x])*Sqrt[a + b*Sin[c + d*x]])/(16*d) + (Sec[c + d*x]^4*(b + a*Sin[c + d*x])*Sqrt[a + b*Sin[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 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, 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, 1] && 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) (a+b \sin (c+d x))^{3/2} \, dx &=\frac{b^5 \operatorname{Subst}\left (\int \frac{(a+x)^{3/2}}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^4(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{4 d}-\frac{b^3 \operatorname{Subst}\left (\int \frac{\frac{1}{2} \left (-6 a^2+b^2\right )-\frac{5 a 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) (b-6 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{16 d}+\frac{\sec ^4(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{4 d}+\frac{b \operatorname{Subst}\left (\int \frac{\frac{3}{4} \left (4 a^4-5 a^2 b^2+b^4\right )+\frac{3}{2} a \left (a^2-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) (b-6 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{16 d}+\frac{\sec ^4(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{4 d}+\frac{b \operatorname{Subst}\left (\int \frac{-\frac{3}{2} a^2 \left (a^2-b^2\right )+\frac{3}{4} \left (4 a^4-5 a^2 b^2+b^4\right )+\frac{3}{2} a \left (a^2-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) (b-6 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{16 d}+\frac{\sec ^4(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{4 d}-\frac{\left (3 \left (4 a^2-2 a b-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a-b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{32 d}+\frac{\left (3 \left (4 a^2+2 a b-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{32 d}\\ &=-\frac{3 \left (4 a^2-2 a b-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{32 \sqrt{a-b} d}+\frac{3 \left (4 a^2+2 a b-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{32 \sqrt{a+b} d}-\frac{\sec ^2(c+d x) (b-6 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{16 d}+\frac{\sec ^4(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{4 d}\\ \end{align*}

Mathematica [A]  time = 2.48678, size = 297, normalized size = 1.58 \[ -\frac{-2 b \sqrt{a+b \sin (c+d x)} \left (\left (6 a^3 b-4 a b^3\right ) \sin (c+d x)-13 a^2 b^2+12 a^4+3 b^4\right )+3 \sqrt{a-b} (a+b)^2 \left (-6 a^2 b+4 a^3+a b^2+b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )-3 (a-b)^2 \sqrt{a+b} \left (6 a^2 b+4 a^3+a b^2-b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )+8 \left (b^2-a^2\right ) \sec ^4(c+d x) (a \sin (c+d x)-b) (a+b \sin (c+d x))^{5/2}+2 \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \left (\left (4 a b^2-6 a^3\right ) \sin (c+d x)+5 a^2 b-3 b^3\right )}{32 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(3*Sqrt[a - b]*(a + b)^2*(4*a^3 - 6*a^2*b + a*b^2 + b^3)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]] - 3*(a
- b)^2*Sqrt[a + b]*(4*a^3 + 6*a^2*b + a*b^2 - b^3)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + 8*(-a^2 +
b^2)*Sec[c + d*x]^4*(-b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(5/2) + 2*Sec[c + d*x]^2*(a + b*Sin[c + d*x])^(
5/2)*(5*a^2*b - 3*b^3 + (-6*a^3 + 4*a*b^2)*Sin[c + d*x]) - 2*b*Sqrt[a + b*Sin[c + d*x]]*(12*a^4 - 13*a^2*b^2 +
3*b^4 + (6*a^3*b - 4*a*b^3)*Sin[c + d*x]))/(32*(a^2 - b^2)^2*d)

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Maple [B]  time = 0.474, size = 409, normalized size = 2.2 \begin{align*}{\frac{1}{32\,b \left ( \cos \left ( dx+c \right ) \right ) ^{4}d} \left ( 4\,\sqrt{-a+b}\sqrt{a+b}\sqrt{a+b\sin \left ( dx+c \right ) }b \left ( b \left ( \cos \left ( dx+c \right ) \right ) ^{2}+8\,a\sin \left ( dx+c \right ) -b \right ) +3\,b \left ( 4\,{\it Artanh} \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{a+b}}} \right ){a}^{2}\sqrt{-a+b}+2\,b{\it Artanh} \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{a+b}}} \right ) a\sqrt{-a+b}-{b}^{2}{\it Artanh} \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{a+b}}} \right ) \sqrt{-a+b}+4\,\arctan \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{-a+b}}} \right ){a}^{2}\sqrt{a+b}-2\,b\arctan \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{-a+b}}} \right ) a\sqrt{a+b}-{b}^{2}\arctan \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{-a+b}}} \right ) \sqrt{a+b} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+6\,\sqrt{-a+b}\sqrt{a+b}\sqrt{a+b\sin \left ( dx+c \right ) }b \left ( 2\,a\sin \left ( dx+c \right ) -b \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}-24\, \left ( a+b\sin \left ( dx+c \right ) \right ) ^{3/2}a\sqrt{-a+b}\sqrt{a+b}+24\,\sqrt{a+b\sin \left ( dx+c \right ) }{a}^{2}\sqrt{-a+b}\sqrt{a+b}+12\,\sqrt{a+b\sin \left ( dx+c \right ) }{b}^{2}\sqrt{-a+b}\sqrt{a+b} \right ){\frac{1}{\sqrt{-a+b}}}{\frac{1}{\sqrt{a+b}}}} \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))^(3/2),x)

[Out]

1/32*(4*(-a+b)^(1/2)*(a+b)^(1/2)*(a+b*sin(d*x+c))^(1/2)*b*(b*cos(d*x+c)^2+8*a*sin(d*x+c)-b)+3*b*(4*arctanh((a+
b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a^2*(-a+b)^(1/2)+2*b*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a*(-a+b)^(1/
2)-b^2*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*(-a+b)^(1/2)+4*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*
a^2*(a+b)^(1/2)-2*b*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a*(a+b)^(1/2)-b^2*arctan((a+b*sin(d*x+c))^(1/2
)/(-a+b)^(1/2))*(a+b)^(1/2))*cos(d*x+c)^4+6*(-a+b)^(1/2)*(a+b)^(1/2)*(a+b*sin(d*x+c))^(1/2)*b*(2*a*sin(d*x+c)-
b)*cos(d*x+c)^2-24*(a+b*sin(d*x+c))^(3/2)*a*(-a+b)^(1/2)*(a+b)^(1/2)+24*(a+b*sin(d*x+c))^(1/2)*a^2*(-a+b)^(1/2
)*(a+b)^(1/2)+12*(a+b*sin(d*x+c))^(1/2)*b^2*(-a+b)^(1/2)*(a+b)^(1/2))/(-a+b)^(1/2)/(a+b)^(1/2)/b/cos(d*x+c)^4/
d

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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))^(3/2),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 ({\left (b \sec \left (d x + c\right )^{5} \sin \left (d x + c\right ) + a \sec \left (d x + c\right )^{5}\right )} \sqrt{b \sin \left (d x + c\right ) + a}, 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))^(3/2),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c)^5*sin(d*x + c) + a*sec(d*x + c)^5)*sqrt(b*sin(d*x + c) + a), 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))**(3/2),x)

[Out]

Timed out

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Giac [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))^(3/2),x, algorithm="giac")

[Out]

Timed out