Optimal. Leaf size=111 \[ -\frac {\sqrt {2} \cos (c+d x) \sqrt [3]{\frac {a+b \sin (c+d x)}{a+b}} F_1\left (\frac {1}{2};\frac {1}{2},\frac {4}{3};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right )}{d (a+b) \sqrt {\sin (c+d x)+1} \sqrt [3]{a+b \sin (c+d x)}} \]
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Rubi [A] time = 0.07, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2665, 139, 138} \[ -\frac {\sqrt {2} \cos (c+d x) \sqrt [3]{\frac {a+b \sin (c+d x)}{a+b}} F_1\left (\frac {1}{2};\frac {1}{2},\frac {4}{3};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right )}{d (a+b) \sqrt {\sin (c+d x)+1} \sqrt [3]{a+b \sin (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 138
Rule 139
Rule 2665
Rubi steps
\begin {align*} \int \frac {1}{(a+b \sin (c+d x))^{4/3}} \, dx &=\frac {\cos (c+d x) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} (a+b x)^{4/3}} \, dx,x,\sin (c+d x)\right )}{d \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}}\\ &=\frac {\left (\cos (c+d x) \sqrt [3]{-\frac {a+b \sin (c+d x)}{-a-b}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} \sqrt {1+x} \left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{4/3}} \, dx,x,\sin (c+d x)\right )}{(a+b) d \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)} \sqrt [3]{a+b \sin (c+d x)}}\\ &=-\frac {\sqrt {2} F_1\left (\frac {1}{2};\frac {1}{2},\frac {4}{3};\frac {3}{2};\frac {1}{2} (1-\sin (c+d x)),\frac {b (1-\sin (c+d x))}{a+b}\right ) \cos (c+d x) \sqrt [3]{\frac {a+b \sin (c+d x)}{a+b}}}{(a+b) d \sqrt {1+\sin (c+d x)} \sqrt [3]{a+b \sin (c+d x)}}\\ \end {align*}
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Mathematica [B] time = 1.93, size = 262, normalized size = 2.36 \[ -\frac {3 \sec (c+d x) \left (5 a \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}} (a+b \sin (c+d x)) F_1\left (\frac {2}{3};\frac {1}{2},\frac {1}{2};\frac {5}{3};\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )-2 \left (2 \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \sqrt {\frac {b (\sin (c+d x)+1)}{b-a}} (a+b \sin (c+d x))^2 F_1\left (\frac {5}{3};\frac {1}{2},\frac {1}{2};\frac {8}{3};\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )+5 b^2 \cos ^2(c+d x)\right )\right )}{10 b d \left (a^2-b^2\right ) \sqrt [3]{a+b \sin (c+d x)}} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +b \sin \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{4/3}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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